matsefi/Octogon/volumes/octogon_2018_2_proposed... · 812 Octogon Mathematical Magazine, Vol. 26,...

812 Octogon Mathematical Magazine, Vol. 26, No.2, October, 2018

Proposed problems

PP28069. 27 If λ > 0 then Γ2 (λ) =

�∞R0

tλ−1e−xtdt

��∞R0

tλ−1e−tx dt

�for all

x > 0.

Mihaly Bencze

PP28070. If x, y > 0 then(n−1)!

∞�

0

tn−1e−(x+y)tdt= xn + yn + n!

nPk=1

1

k

�∞�

0

tn−ke−xtdt

��∞�

0

tk−1e−ytdt

� .

Mihaly Bencze

PP28071. If xk > 0 (k = 1, 2, ..., n) and λ > 0 thenP

cyclic

∞R0

tλ−1e−x1x2tdx ≤ Γ2(n)Γ(2n)

nPk=1

∞R0

t2λ−1e−xktdt.

Mihaly Bencze

PP28072. Solve in Z the equation x+y−zy+z−x + y+z−x

z+x−y + z+x−yx+y−z = 3.

Mihaly Bencze

PP28073. Let be G = (a, a+ 1) ∪ (a+ 1,+∞) where a > 0 andlogb ((x ◦ y)− a) = logb (x− a) logb (y − a) when b ∈ (0, 1) ∪ (1,+∞)1). Prove that (G, ◦) is abelian group2). Let be xk > a+ 1 (k = 1, 2, ..., n) and denote {y1, y2, ..., yn} arearrangement of the set {x1, x2, ..., xn} . Prove thatnP

k=1

logb ((xk ◦ yk)− a) ≤nP

k=1

log2b (xk − a) .

Mihaly Bencze

PP28074. Prove thatnP

k=1

1(k−1)!

∞R0

tk−1e−xtdt =�1 + 1

x

�n − 1 for all x > 0.

Mihaly Bencze

27Solution should be mailed to editor until 30.12.2021. No problem is ever permanently

closed. The editor is always pleased to consider for publication new solutions or new in sights

on past problems.

Proposed Problems 813

PP28075. If a, b, c > 0 thenP 1

Γ(a)

∞R0

ta−1e−xtdt+ 3Γ(a+b+c

3 )

∞R0

ta+b+c

3−1e−xtdt ≥ 2

P 1Γ(a+b

2 )

∞R0

ta+b2

−1e−xtdt.

Mihaly Bencze

PP28076. If xij > 0 (i = 1, 2, ..., n) (j = 1, 2, ..., n) and λ > 0 then�nP

i=1

∞R0

tλ−1e−xi1xi2...ximtdt

�2

≤mQj=1

�nP

i=1

∞R0

tλ−1e−xmij tdt

�.

Mihaly Bencze

PP28077. If xk, yk > 0 (k = 1, 2, ..., n) then�nP

k=1

∞R0

tλ−1e−xkyktdt

�2

≤�

nPk=1

∞R0

tλ−1e−x2ktdt

��nP

k=1

∞R0

tλ−1e−y2ktdt

�for all

λ > 0.

Mihaly Bencze


k=1

�1

Γ(λ)

∞R0

tλ−1e−k(k+1)tdt

� 1λ

= nn+1 for all λ > 0.

Mihaly Bencze

PP28079. In all triangle ABC holdsQ�

1Γ(A)

∞R0

tA−1e−xtdt

�= 1

xπ for all

x > 0.

Mihaly Bencze

PP28080. If xk > 0 (k = 1, 2, ..., n) and λ > 0 thennP

k=1

1Γ(λ)

∞R0

tλ−1e−xktdt ≥ n1−λ

�nP

k=1

xk

�λ

.

Mihaly Bencze


k=1

1(k−1)!

∞R0

tk−1e−xtdt = xn−1(x−1)xn for all x > 0, x 6= 1.

Mihaly Bencze



k=1

k2(k2+k+1)2

(k6+k3+3k2+3k+1)(k2−k+1)≤ 9n

n+1 .

Mihaly Bencze

PP28083. In all triangle ABC holds

1).P A3+B3

A2+AB+B2 ≥ 2π3 2).

Ptg

�π(A2+AB+B2)9(A2−AB+B2)

�≤ 3

√3

Mihaly Bencze


k=1

k2+k+1k2−k+1

≤ 3nn+1 .

Mihaly Bencze

PP28085. Prove that maxx,y∈[0,1]

�√sinπx−√

sinπy�2

≥ 2π − 1

2 .

Mihaly Bencze

PP28086. Prove that (2n+ 1)he2+12e (2n)!

i+he2−12e (2n+ 1)!

i= [e (2n+ 1)!]

for all n ∈ N∗, where [·] denote the integer part.

Mihaly Bencze


k=0

(2k)!

4k(k!)2(2k+1)=

�

4n−1(n!)2(2n+1)π(2n)!

�

4n(n!)2[2n+1][2n]!

where [·] denote

the integer part.

Mihaly Bencze

PP28088. Prove thatnQ

k=1

[ek!]k ≤�2[e(n+1)!]−n−2

n(n+1)

�n(n+1)2

where [·] denote the

integer part.

Mihaly Bencze

PP28089. Prove that

1). 1 + 13! +

15! + ...+ 1

(2n+1)! =

�

e2−12e

(2n+1)!�

(2n+1)!

2). 1 + 12! +

14! + ...+ 1

(2n)! =

�

e2+12e

(2n)!�

(2n)! where [·] denote the integer part.

Mihaly Bencze


PP28090. Prove that 1!2!3!...n! ≥�(n+1)![en!]

�n+1where [·] denote the integer

part.

Mihaly Bencze


k=0

k! ≥ (n+1)(n+1)![en!] where [·] denote the integer part.

Mihaly Bencze


k=0

�1k!

�λ ≥ n1−λ�[e·n!]n!

�λfor all

λ ∈ (−∞, 0] ∪ [1,+∞) where [·] denote the integer part.

Mihaly Bencze


k=1

(−1)k kn�1 + 1

k

�k �nk

�= (−1)n n!

nPk=1

1k! .

Mihaly Bencze


i=0

nPj=0

�4i2i

��4n−4i2n−2i

��4j+22j+1

��4n−2j−22n−2j−1

�= 44n−1 − 42n−1

�2nn

�2.

Mihaly Bencze

PP28095. Determine all λ > 0 for whichnP

k=1

k [λ · k!] = [λ (n+ 1)!]− n− 2

where [·] denote the integer part.

Mihaly Bencze

PP28096. Determine all x, y ∈ N such thatnP

k=0

�nk

�x�yn+kxn

�=

�xnn

�y.

Mihaly Bencze


k=0

�2nk

��nk

��3n+k3n

�=

�3nn

�2.

Mihaly Bencze


PP28098. Prove that∞Pk=1

1k2(k+1)(k+2)(k+3)

= 136

�π2 − 49

6

�.

Mihaly Bencze


k=0

(2n− 2k + 1)�2n+1k

�≥q(n+ 1) (2n+ 1)

�4n2n

�.

Mihaly Bencze


k=1

�nk

�2 ≤�

1n

�2nn+1

��n.

Mihaly Bencze


k=0

�4n+22k+1

�≤r(n+ 1)

�14

�8n+44n+2

�+ 1

4

�4n+22n+1

�+ 1

2

�4n+22n+1

�2�.

Mihaly Bencze

PP28102. Determine all ak ∈ Q+

�k = 0, 1, ...,

�n2

��such that

2n[n2 ]Pk=0

�n2k

��2kk

�ak =

�2nn

�where [·] denote the integer part.

Mihaly Bencze

PP28103. Prove thatP

0≤i<j≤n

�ni

��nj

�= 1

2

�4n −

�2nn

��.

Mihaly Bencze

PP28104. Determine all ak ∈ N (k = 0, 1, 2, ..., 2n) for which2nPk=0

(−1)k�2nk

��4n−k2n

�ak = (−1)n

�2nn

�.

Mihaly Bencze and Marius Dragan

PP28105. Determine all x, y ∈ N for whichnP

k=1

kx (n− k + 1)y =�n+x+yx+y+1

�.

Mihaly Bencze


PP28106. Determine all ak > 0 (k = 0, 1, ..., n) for whichnP

k=0

ak�nk

�= 3n.

Mihaly Bencze

PP28107. Denote Fk and Lk the kth Fibonacci respective Lucas numbers.

Prove that 2nP

k=0

3kpLkFk+1 ≤ 3n+1Fn+1.

Mihaly Bencze

PP28108. Determine all x ∈ N for whichnP

k=0

�nk

�x=

�nxn

�.

Mihaly Bencze

PP28109. If ak > 0 (k = 1, 2, ..., n) , thenP�

a21a21−a1a2+a22

�λ≤ n for all

λ ∈ [0, 1] .

Mihaly Bencze

PP28110. If a > 1 then

maxx,y∈[0,1]

�√1− 2a cosπx+ a2 −

p1− 2a cosπy + a2

�2≥ 1.

Mihaly Bencze

PP28111. If a, b, c > 0 thenP 4

q(a+b)2

a2+b2+c2+ab+bc−ca≤ 3.

Mihaly Bencze

PP28112. If xk > 0 (k = 1, 2, ..., n) then determine all λ ∈ R for whichP x41+x2

1x22+x4

2

x21+λx1x2+x2

2≥Px1x2.

Mihaly Bencze

PP28113. In all triangle ABC holds�s2 + r2 + 2Rr

� �s2 + r2 + 4Rr

�≥ 4Rr

�5s2 + r2 + 4Rr

�.

Mihaly Bencze


PP28114. If x, y ∈ R then

��3

qcosx+ i sinx+

p(cos 2x− cos 3y) + i (sin 2x− sin 3y) +

+ cos y+i sin y3�

cosx+i sinx+√

(cos 2x−cos 3y)+i(sin 2x−sin 3y)

�� ≤ 5 + 2√3

9

Mihaly Bencze

PP28115. In all triangle ABC holds 4R+rs ≥

q2�2− r

R

�.

Mihaly Bencze

PP28116. In all acute triangle ABC holdsP cos(B+A

2 ) sinC cos(B−A2 )

cos(C−A2 ) sinB cos(C+A

2 )= s2+r2−4R2

2(s2−(4R+r)2).

Mihaly Bencze

PP28117. If x ∈ (0, 1) then π3 cosπxsin3 πx

≥ 1x3 + 2

∞Pk=1

1

(x2−k2)32.

Mihaly Bencze

PP28118. If a, b, c > 0 then1R0

eax2+bx+cdx ≤ e

a3+ b

2+c + max

x,y∈[0,1]

�√eax2+bx+c −

√eay2+by+c

�2.

Mihaly Bencze

PP28119. If a, b, c > 0, and b2 > 4ac then

exp

�−

1R0

ln�ax2 + bx+ c

�dx

�+ max

x,y∈[0,1]

�1√

ax2+bx+c− 1√

ay2+by+c

�2

≥

≥ 1√b2−4ac

ln(b+

√b2−4ac)(2a+b−

√b2−4ac)

(b−√b2−4ac)(2a+b+

√b2−4ac)

.

Mihaly Bencze

PP28120. If x, y ∈ [0, 1] then

max�√

x2 + 3x+ 2−py2 + 3y + 2

�2≥ 23

6 − 8e2.

Mihaly Bencze



�1√x+1

− 1√y+1

�2≥ ln 2− e

4 .

Mihaly Bencze

PP28122. If 0 < x ≤ y ≤ z < 1 then√1− z lnxy ≤

�√1− x+

√1− y

�ln z.

Mihaly Bencze

PP28123. If x, y ∈ [0, 1] then max�√

ex −√ey�2 ≥ e−√

e.

Mihaly Bencze

PP28124. If x ∈�0, π4

�then (sinx)sinx ≤ (cosx)cosx .

Mihaly Bencze

PP28125. Let be ak > 0 (k = 1, 2, ..., n) and A = 1n

nPk=1

ak, G = n

snQ

k=1

ak,

H = nn�

k=1ak

. Prove that�GA

�√A−H ≥�HA

�√A−G.

Mihaly Bencze

PP28126. If xk ∈�0, π4

�(k = 1, 2, ..., n) , then

nPk=1

ctg2xk ≥ (logb a)2 when

a =nQ

k=1

sinxk and b =nQ

k=1

cosxk.

Mihaly Bencze


1). s4 − r4 ≥P�r2a − r2

�rbrc

2).(4s4−R2r2)r2

R2 ≥P�h2a − r2

�hbhc

Mihaly Bencze

PP28128. If 2 < a ≤ b thenbRa

(3x2−5x+4) lnxdx

x(x−2)2≤ 2

�b− a+ ln b−2

a−2

�+ 3

2 lna(b−2)b(a−2) − 3 ln bb−2

aa−2 + 12 ln ab ln

ba .

Mihaly Bencze


PP28129. If xi > 0 (i = 1, 2, ..., n) andnP

i=1x2i = 1 then

Pcyclic

x2k−11

x2+x3+...+xn≥ 1

nk−2(n−1)for all k ∈ N, k ≥ 1.

Mihaly Bencze

PP28130. In all triangle ABC holds 5P r2a

r2−rra+r2a+ 2

P h2a

r2−rha+h2a≤ 27.

Mihaly Bencze

PP28131. In all triangle ABC holdsP b+c−a

5a2+4b2+4c2−6bc+ab+ca≤ 1

a+b+c .

Mihaly Bencze


1).P 1

r3a(r2br2c+r2rbrc+4r2(r2b+r2c))≥ 1

2s4r3

2).P 1

h3a(h2

bh2c+r2hbhc+4r2(h2

b+h2

c))≥ R2

8s4r5

Mihaly Bencze


k=1

1�

sin�

k(k+1)n(n+1)

�2 ≤ n2 + n�1− 4

π2

�.

Mihaly Bencze

PP28134. If x ∈�0, π

2n

�then

nPk=1

1k2

≥ 2n2x2 sinxn sinx−sinnx cos(n+1)x − n

�1− 4

π2

�.

Mihaly Bencze

PP28135. If x ∈�0, π

2n

�then

nPk=1

q1− 4

π2 + 1k2x2 ≥ n2 sin x

2

sin nx2

sin(n+1)x

2

.

Mihaly Bencze

PP28136. If�0, π2

�then 1

sin2 x cos2 x≤ 2

�1− 4

π2

�+ 8x2−4πx+π2

x2(π−2x)2.

Mihaly Bencze


PP28137. If 0 < a ≤ b ≤ 1 thenbRa

x+1�√

5−3x+2√

(x+1)(4x+5)�2dx ≤ 1

6√5arctg 6(b−a)

√5

45+4ab .

Mihaly Bencze

PP28138. Solve in Z the equation x3+y2z

(y+z)3+ y3+z2x

(z+x)3+ z3+x2y

(x+y)3= 3

4 .

Mihaly Bencze


1).P�r

2+3 cos2 A2

1+sin2 A2

+ 2q5 + 4 sin2 A

2

�2

≥ 139 + r2−s2

2R2

2).P�r

2+3 sin2 A2

1+cos2 A2

+ 2q5 + 4 cos2 A

2

�2

≥ 143 + 8Rr+r2−s2

2R2

Mihaly Bencze

PP28140. In all triangle ABC holdsP�q

5−3 sinA1+sinA + 2

√5 + 4 sinA

�2

≥ 135 +2(s2−r(4R+r))

R2 .

Mihaly Bencze

PP28141. In all acute triangle ABC holdsP�q

5−3 cosA1+cosA + 2

√5 + 4 cosA

�2

≥ 147− 2(s2−4Rr−r2)R2 .

Mihaly Bencze

PP28142. If ak > 0 (k = 1, 2, ..., n) , thenP

a21�a42 − a22 + 1

�≥ 1

n3

�nP

k=1

ak

�4

.

Mihaly Bencze

PP28143. If x ∈�0, π2

�then�q

5−3 sinx1+sinx + 2

√4 sinx+ 5

�2

+�q

5−3 cosx1+cosx + 2

√4 cosx+ 5

�2

≥ 94.

Mihaly Bencze


PP28144. Let (Fn)n≥0 and (Ln)n≥0 be the Fibonacci and the Lucassequence, respectively. Compute the following limits:

a). limn→∞

�2n+2p(n+ 1)!!Fn+1 − 2n

√n!Fn

�√n

b). limn→∞

�2n+2p(n+ 1)!!Ln+1 − 2n

√n!Ln

�√n

D.M. Batinetu-Giurgiu and Neculai Stanciu

PP28145. Let a, b, c ≥ 0. Prove that3√4 + 17a2b+ 3

√4 + 17b2c+ 3

√4 + 17c2a+ 10

�127 − abc

�≥ 5

Paolo Perfetti

PP28146. Let (an)n be a sequence, given by therecurence:man+1 − (m− 2) an + an−1 = 0, where m ∈ R is a parameter andthe first two terms of (an)n are known real numbers. Find m ∈ R, so thatlimn→∞

an = 0.

Laurentiu Modan

PP28147. If a, b, c are real numbers for which ab+ bc+ ca = 1 prove that4a2 + 5b2 + 6c2 > 2

√6

Dorin Marghidanu

PP28148. Let u, v, w complex numbers such that:u+ v + w = 1, u2 + v2 + w2 = 3, uvw = 1Prove that:a). u, v, w are distinct numbers two by two;b). if S(k) := uk + vk + wk, then S(k) is an odd natural number;

c). the expression u2n+1−v2n+1

u−v + v2n+1−w2n+1

v−w + w2n+1−u2n+1

w−u is an integernumber.

Dorin Marghidanu

PP28149. If a, b, c are strictly positive real numbers, prove that:a

a2bc+b4+c4+ b

b2ca+c4+a4+ c

c2ab+a4+b4≤ 1

abc

Dorin Marghidanu


PP28150. If a, b, c > 0 and m,n, p ∈ (2,+∞) then prove the inequality

mp

ab +

n

qbc +

pp

ca ≥ m+n+p

m+n+p√mmnnpp

Dorin Marghidanu

PP28151. If 1 < a < b < c, prove that√a2 − 1 +

√b2 − 1 +

√c2 − 1 ≤ ab+bc+ca

2 and specify when equality holds.Generalization.

Dorin Marghidanu

PP28152. Prove thata2018

b2018+ b2018

c2018+ c2018

d2018+ d2018

a2018≥ a2017

b2017+ b2017

c2017+ c2017

d2017+ d2017

a2017.

Ovidiu Bagdasar

PP28153. Find the cardinal of the setB = {ij : (i, j) ∈ N× N, 0 ≤ i, j ≤ 2018},

Ovidiu Bagdasar

PP28154. Find the cardinal of the set containing all integers1 ≤ x ≤ 20182018, such that their highest prime factor is 2017.

Ovidiu Bagdasar

PP28155. Let a, b, c be non-negative real numbers. Find all positive real

k such that the inequality kabc+ a2b+ b2c+ c2a ≤ (k + 3) (a+ b+ c)3

27holds

for any nonnegative real a, b.c.

Arkady Alt

PP28156. Prove that inequality

2�xa+

y

b+

z

c

�· x+ y + z

a+ b+ c−�xyab

+yz

bc+

zx

ca

�≥ 3

�x+ y + z

a+ b+ c

�2

holds for

any triangle with sidelengths a, b, c and any positive real x, y, z anddetermine when equality occurs.

Arkady Alt


PP28157. Prove that:R π

20

Qnk=1

�sin2k+1 x+ cos2k+1 x

�dx ≥ π

2(√2)n

2 ;n ∈ N∗

Daniel Sitaru

PP28158. If 2 ≤ a ≤ b then 2 ln ba ≥

bRa

3x2+x+4x4+2

dx.

Mihaly Bencze

PP28159. If 2 ≤ a ≤ b then 2 ln b(a+1)a(b+1) ≥

bRa

3x2+x+4(x+1)(x4+2)

dx

Mihaly Bencze


1).�s2+2Rr+5r2

2

�s≥Q (ma)

a

2).

�s(s2+2Rr+5r2)3(s2−r2−4Rr)

� 32(s

2−r2−4Rr)≥Q

am2a

Mihaly Bencze

PP28161. In all triangle ABC holdsQ

whaa ≥

�r(s2+r2+4Rr)

R+2r

� s2+r2+4Rr4R

.

Mihaly Bencze

PP28162. If x ≥ 2 and λ ≥ 0 then 2x4

λ+5 + 4λ+1 ≥ 3x3

λ+4 + x2

λ+3 + 4xλ+2 .

Mihaly Bencze

PP28163. In all triangle ABC holdsP

a2m4a ≥ s2(s2+2Rr+5r2)

12 .

Mihaly Bencze

PP28164. Solve in R the following system

2�x41 + 2

�= 3x32 + x23 + 4x4

2�x42 + 2

�= 3x33 + x24 + 4x5

−−−−−−−−−−−−2�x4n + 2

�= 3x31 + x22 + 4x3

Mihaly Bencze


PP28165. If x ≥ 2 then48

�x4 + 2

� �x4 + 10

�≥ 5x2

�3x2 + x+ 4

� �9x2 + 4x+ 24

�.

Mihaly Bencze


1). 7Pq

73 + 2 sin2 A2 ≥ 201− s2+8Rr−r2

R2

2). 7Pq

73 + 2 cos2 A2 ≥ 183 + s2−r2

R2

Mihaly Bencze

PP28167. If x ∈�0, π2

�then

7�√

73 + 2 sinx+√73 + 2 cosx

�≥ 110 + 16 (sinx+ cosx) .

Mihaly Bencze

PP28168. In all triangle ABC holds:

1). 7P√

73 + 2 sinA ≥ 177 +4(4sR−s2+r(4R+r))

R2

2). 7P√

73 + 2 cosA ≥ 177 +4(s2−6R2−3Rr−r2)

R2

Mihaly Bencze

PP28169. If ak > 0 (k = 1, 2, .., n) thennP

k=1

(2a3k+1)2

ak≥ 1

n

�n+

nPk=1

ak +nP

k=1

a2k

�2

.

Mihaly Bencze

PP28170. Prove thateR1

q1 + 1

x2dx >√e2 − 2e+ 2.

Mihaly Bencze

PP28171. In all triangle ABC holds 3 (2R+ r)2 + 2r2 ≥ s2 + 8R2.

Mihaly Bencze


(ma)λ ≥ 3

�2r(s2−r(4R+r))

3R2

�λ

for all

λ ∈ (−∞, 0] ∪ [1,+∞) .

Mihaly Bencze


PP28173. If ak > 0 (k = 1, 2, ..., n) andnP

k=1

ak = 1 then

nPk=1

�ak − 1

2

�3 ≥ 3(2−n)3

8n3 .

Mihaly Bencze

PP28174. In all triangle ABC holdsP �

a cos B−C2

�λ ≥ 3�s(R+2r)

3R

�λfor all

λ ∈ (−∞, 0] ∪ [1,+∞) .

Mihaly Bencze

PP28175. If x ∈�0, π2

�then

4π

�arctg

�12 sin

2 x�+ arctg

�12 cos

2 x��

+ 59100 > 8

14−sin2 x cos2 x.

Mihaly Bencze

PP28176. In all triangle ABC holdsP a2 cos2(B−C)

sin 2B+sin 2C = 4sr.

Mihaly Bencze

PP28177. In all triangle ABC holds�

a3 cosA�

a cos3 A=

4R2(s2−r2−4Rr−3R2)5R2−s2+r2+4Rr

.

Mihaly Bencze

PP28178. In all acute triangle ABC holds

9 ≤ (R+r)(s2+r2−4R2)R(s2−(2R+r)2)

≤ 9 +P �� 1

cosA − 1cosB

�� .

Mihaly Bencze


1). 9 ≤ (2R−r)(s2+r2−8Rr)2Rr2

≤ 9 +P ��

1sin2 A

2

− 1sin2 B

2

��

2). 9 ≤ (4R+r)(s2+(4R+r)2)2Rs2

≤ 9 +P ��

1cos2 A

2

− 1cos2 B

2

��

Mihaly Bencze

PP28180. Let a, b ∈ N , determine all n ∈ N for which a2n+ b2 andb2n+ a2 are both perfect squares.

Mihaly Bencze


PP28181. In all triangle ABC holds 9 ≤ s2+r2+4Rr2Rr ≤ 9 +

P �� 1sinA − 1

sinB

�� .

Mihaly Bencze


AH3 ≥ 12(4+3√3)(sr−R2)R11 .

Mihaly Bencze

PP28183. In all triangle ABC holds 1r

P 1rarb

≤ 1rarbrc

+ 827r3

.

Mihaly Bencze

PP28184. In all triangle ABC holdsP a(a2+bc)

(b+c)2≤ 4s(s2−r2−Rr)

9(s2+r2+2Rr)

P a2+bca(b+c) .

Mihaly Bencze

PP28185. In all triangle ABC holdsP�

ara

�λ≥ 3

�2(4R+r)

3s

�λfor all

λ ∈ (−∞, 0] ∪ [1,+∞) .

Mihaly Bencze

PP28186. In all triangle ABC holds�P a

ra

�2+�

�

a�

ra

�2≥ 8.

Mihaly Bencze

PP28187. Determine all xk ≥ 3 (k = 1, 2, ..., n) for which

nQk=1

(3xk − 1)2xk ≤

3n�

k=1xk−n

n�

k=1xk−n

n

.

Mihaly Bencze

PP28188. Solve in R the equation (3x + 1)y + 2 · 3x = 3z + 1.

Mihaly Bencze

PP28189. If a, b, c, d ∈ (0,∞), prove that:ln a2 + ln b

4 + ln c6 + ln d

12 ≤ ln�a2 + b

4 + c6 + d

12

�.

Dorin Marghidanu


PP28190. Solve the equation tg πx tg

π2x =

√x− 2

Ionel Tudor

PP28191. Prove that there are only finitely many n, k, p ∈ N∗ such thatnQ

r=1

�nk

rp + rp�∈ N.

Mihaly Bencze

PP28192. If ak ∈ R (k = 1, 2, ..., n) such thatP a1

a2+a3+...+an= 1 then

determine all r ∈ N for which�P (a2+a3+...+an)

r

a2r1

� 1r ∈ Q.

Mihaly Bencze

PP28193. Find all n, k ∈ N for which an − 1 is divisible by knk for alla ∈ N, (a, n) = 1.

Mihaly Bencze

PP28194. Let M be the set of all rational numbers expressible in the formnQ

k=1

a2k−ak+1

b2k−bk+1

where ak, bk ∈ N (k = 1, 2, ..., n) . Prove that M contain

infinitely many prime numbers.

Mihaly Bencze

PP28195. If x, y, z > 0 and x+ y + z = 12λ then

2λ2P 1

3λ(x+y)+1 ≤ 1�

(x+y)(y+z) ≤λ3

P 1x+y .

Mihaly Bencze

PP28196. If a, b, c > 0 andP

a = 3 thenP

a2 ≥ 3 + 19 (P |a− b|)2 .

Mihaly Bencze

PP28197. In all triangle ABC holdsP hλ

b+hλ

c

rλa≥ 6 for all

λ ∈ (−∞, 0] ∪ [1,+∞) .

Mihaly Bencze


PP28198. Denote Fk the kth Fibonacci number. Prove that�nP

k=1

n−k+1FkFk+2

��nP

k=1

1Fk+1Fk+2

�≤

�n2

�2.

Mihaly Bencze

PP28199. In all triangle ABC holdsn√r2 +

Pnp

r2a +P n

√a2 ≤ n

q7n−1 (4R)2, for all n ∈ N∗.

Mihaly Bencze


(ha + hb) ≤ 8s2r.

Mihaly Bencze

PP28201. Let A1A2...An be a convex polyhedron and B1B2...Bn a regularpolyhedron, both inscribed in the same sphere. Prove thatP1≤i<j≤n

(BiBj −AiAj) ≥ 0.

Mihaly Bencze

PP28202. In all triangle ABC holdsP n

√a ·AI2 ≤ n

√3n−1abc for all

n ∈ N∗.

Mihaly Bencze

PP28203. Let A1A2...An be a convex polygon and B1B2...Bn a regularpolygon, both inscribed in the same circle. Prove thatP1≤i<j≤n

(BiBj −AiAj) ≥ 0.

Mihaly Bencze


k=1

5

r(k14+k7+1)2

3(k18+k9+1)≥ n(n+1)

2 .

Mihaly Bencze

PP28205. Let ABC be a triangle, and M ∈ Int (ABC) . The linesAM,BM,CM cut the circumcircle in points D,E and F . Prove thatAB ·BC · CA ·DE · EF · FD ≥ 27BD2 · CE2 ·AF 2.

Mihaly Bencze


PP28206. If x > 0 then�x14 + x7 + 1

�≥ 3x5

�x18 + x9 + 1

�.

Mihaly Bencze

PP28207. In all triangle ABC holds r2λ +P

r2λa +P

a2λ ≥ 71−λ (4R)2λ forall λ ∈ (−∞, 0] ∪ [1,+∞) .

Mihaly Bencze


ha+hb

rc

�λ≥ 3 · 2λ for all

λ ∈ (−∞, 0] ∪ [1,+∞) .

Mihaly Bencze

PP28209. In all triangle ABC holds (2Rr)s ≥ (AI)a (BI)b (CI)c .

Mihaly Bencze


(IA)λ +P

(IaA)λ ≥ 6 (abc)λ3 for all

λ ≥ 0.

Mihaly Bencze

PP28211. Let ABC be a triangle. Determine all point M∈ Int (ABC) forwhich aMA2 + bMB2 + cMC2 = abc.

Mihaly Bencze


IA4 ≥ (abc)2�

a2.

Mihaly Bencze

PP28213. If x, y, z > 0 then 3xyz +P

x2 (y + z) ≤ (P

x)�P

x2�.

Mihaly Bencze

PP28214. If x, y, z > 0 then (P

x)2 (P

xy)2 ≥ 4xyz (x+ y) (y + z) (z + x) .

Mihaly Bencze

PP28215. If x, y, z > 0 then (P

x) (P

xy) ≤ 98

Q(x+ y) .

Mihaly Bencze


PP28216. If x, y, z > 0 then (P

x) (P

xy) ≥ 3xyz + 13 (P

x√y + z)2 .

Mihaly Bencze

PP28217. If x, y, z > 0 then (P

x) (P

xy) ≤ 3xyz + 127

P(2x+ y + z)3

Mihaly Bencze

PP28218. Solve in N the equationnP

k=1

xtk + n− 1 = LCM�xt1, x

t2, ..., x

tn

�.

Mihaly Bencze

PP28219. Let A1A2...An be a polygon circumscribed to the circle withradius r and I the incenter, and denote G his centroid,λ = min |AiAj −AkAp| , i 6= j, k 6= p, i 6= k, j 6= p, i, j, k, p ∈ {1, 2, ..., n} .Prove that

nPk=1

Area [AkIG] ≥ (n−1)λrn .

Mihaly Bencze

PP28220. Solve in C the following system:

6x− 2y + 9z2 − 12z = −8x2 − 4y2 + 2x+ 4y − 6z = −46x2 − 12y2 − 18z2 + 12x+ 10y + 24z = 4

Mihaly Bencze

PP28221. Let be f : N → N such that f (3k) = k, f (3k + 1) = 3k + 2,f (3k + 2) = 3k + 3. Prove that for each n ∈ N, there exist k ∈ N such thatf ◦ f ◦ ... ◦ f| {z }

k−time

(n) = 1.

Mihaly Bencze

PP28222. Determine all pentagons A1A2A3A4A5 in whichA2A2

5A1A2

= A1A2 +A2A5.

Mihaly Bencze

PP28223. In all triangle ABC holdsP tgA

1+tg2A = srR2 .

Mihaly Bencze


PP28224. Determine all r, s ∈ N for which (nr)!n−1Qk=1

k!(ks+n)! ∈ N.

Mihaly Bencze

PP28225. If a, b, c > 0 thenP a+3b+2c

3a+5b+4c ≤ 32 .

Mihaly Bencze

PP28226. If a, b, c > 0 thenP a2

3a+ 3√

9(a+b+c)bc≥ a+b+c

6 .

Mihaly Bencze

PP28227. If a, b > 0 then�a√b+b

√a√

a+b+ a√

2

��√a+

√b�√

a+ b+ a√2�+

+�a√b+b

√a√

a+b+ b√

2

��√a+

√b�√

a+ b+ b√2�≤ 5a2 + 8ab+ 5b2.

Mihaly Bencze

PP28228. In all triangle ABC holds�4R+rsr

�2 ≥ 2r

�2r − 1

R

�.

Mihaly Bencze

PP28229. Prove that there is a constant d > 0 with the following property:If a, b, c, n ∈ N∗ such that gcd (a+ i, b+ j, c+ k) > 1 for alli, j, k ∈ {1, 2, ..., n} then min {a, b, c} > (dn)n .

Mihaly Bencze

PP28230. If xk > 0 (k = 1, 2, ..., n) , then P

cyclic

1x2+x3+...+xn

! P

1≤i<j≤nxixj

!≤ n

2

nPk=1

xk.

Mihaly Bencze

PP28231. Find all functions f : R → R such that1 + f (x) f (y) f (z) = f (x+ y + z) + f (xy + yz + zx) + xyz for allx, y, z ∈ R.

Mihaly Bencze


PP28232. If ak > 0 (k = 1, 2, ..., n) , then

Pcyclic

a41+4a31a2+6a21a22+4a1a32+a42

5a21+2a1a2+5a22≥ 2

Pcyclic

a1a2

Mihaly Bencze

PP28233. Solve in Z the equation x5

x+4y + y5

y+4z + z5

z+4x = 35 .

Mihaly Bencze


1).P 5a2+2ab+5b2

a4+4a3b+6a2b2+4ab3+b4≤ 1

4Rr

2).P 5h2

a+2hahb+5h2b

h4a+4h3

ahb+6h2ah

2b+4hah3

b+h4

b

≤ s2+r2+4Rr8s2r2

3).P 5r2a+2rarb+5r2

b

r4a+4r3arb+6r2ar2b+4rar3b+r4

b

≤ (4R+r)2−2s2

s4r2

Mihaly Bencze


mλa ≥ 3

�2r(s2−4Rr−r2)

3R2

�λ

for all

λ ∈ (−∞, 0] ∪ [1,+∞) .

Mihaly Bencze

PP28236. If a, b > 0 then(2a+b)(a2+2b2)

2a2+b2+

(2b+a)(b2+2a2)2b2+a2

≥ 3 (a+ b) .

Mihaly Bencze


k=1

k

|2k√k2+1−2k2−1| ≥ n2 (n+ 1)2 .

Mihaly Bencze


Pq(s−a)(s−b)

ac ≥ 14

�s2

s2−2r2−8Rr

�2− s2+r2−8Rr

8Rr .

Mihaly Bencze


PP28239. Solve in R the following system:

x31x2 = (x3 + x4)3

x32x3 = (x4 + x5)3

−−−−−−−−x3nx1 = (x2 + x3)

3

.

Mihaly Bencze

PP28240. Let be a ∈ N∗. Prove thatnP

k=1

��a− k (k + 1)√2�� ≥ (

√3−

√2)n

n+1 .

Mihaly Bencze


1).P 1

cos2 A2+sin4 A

2

≤ 108R2

28R2+2Rr+r2

2).P 1

sin2 A2+cos4 A

2

≤ 108R2

28R2+2Rr+r2

Mihaly Bencze

PP28242. Denote Fk, Lk, Pk the kth Fibonacci, Lucas, Pell numbers, and

xn =

Fn if n = 3kLn if n = 3k + 1Pn if n = 3k + 2

. Compute∞Pk=1

1x2k

.

Mihaly Bencze


1).P 1

1−sin2 A2+sin4 A

2

+ 108R2

28R2+2Rr+r2≤ 8

P 1

2(cos2 A2+cos2 B

2 )+(sin2 A

2+sin2 B

2 )2

2).P 1

1−cos2 A2+cos4 A

2

+ 108R2

28R2+2Rr+r2≤ 8

P 1

2(sin2 A2+sin2 B

2 )+(cos2A2+cos2 B

2 )2

Mihaly Bencze

PP28244. In all acute triangle ABC holdsP 11−cosA+cos2 A

+ 27R2

7R2−Rr+r2≤ 8

P 14−2(cosA+cosB)+(cosA+cosB)2

.

Mihaly Bencze


x+2y + y3

y+2z + z3

z+2x = 1.

Mihaly Bencze


PP28246. Let A1A2A3A4A5 be a pentagon with sides a1, a2, a3, a4, a5. Thediagonals of pentagon A1A2A3A4A5 determine the pentagon B1B2B3B4B5

with sides b1, b2, b3, b4, b5. Prove thatP

cyclic

�a1

a2+a3+ b1

b2+b3

�≥ 5.

Mihaly Bencze

PP28247. In all triangle ABC holdsP 1

1−sinA+sin2 A≤ 27R2

s2−3sR+9R2 .

Mihaly Bencze

PP28248. In all acute triangle ABC holdsP 1

1−cosA+cos2 A≤ 27R2

7R2−Rr+r2.

Mihaly Bencze

PP28249. In all triangle ABC holds1).

P 11−sin4 A

2+sin8 A

2

+ 1728R4

(8R2+r2−s2)2−24R2(r2−s2)+320R4≤

8P 1

4−2(sin4 A2+sin4 B

2 )+(sin4 A

2+sin4 B

2 )2

2).P 1

1−cos4 A2+cos8 A

2

+ 1728R4

((4R+r)2−s2)(8Rr−s2−8R2)+576R4≤

8P 1

4−2(cos4 A2+cos4 B

2 )+(cos4A2+cos4 B

2 )2

Mihaly Bencze

PP28250. If 0 < a ≤ b < π2 then ctga−ctgb

b−a ≤ 1ab + 1− 4

π2 .

Mihaly Bencze

PP28251. If x ∈�0, π2

�then 8

π2 + 1sin2 x cos2 x

≤ 1x2+ 1

(π2 −x)2+ 2.

Mihaly Bencze

PP28252. In all acute triangle ABC holds12π2 +

�s2+r2+Rr

2sr

�2≤ 3 +

P 1A2 + 4R

r .

Mihaly Bencze

PP28253. In all acute triangle ABC holds�s2+r2−4R2

s2−(2R+r)2

�2+ 12

π2 ≤ 3 +P 1

(π2−A)

2 + 8R(R+r)

s2−(2R+r)2.

Mihaly Bencze



1).�sr

�2+ 12

π2 ≤ 2 + 4P 1

A2 + 8Rr

2).�4R+r

s

�2+ 12

π2 ≤ 2 + 4P 1

(π2−A)

2

Mihaly Bencze


1).Q �

4A2 + 1− 4

π2

�≥

�4Rr

�2

2).Q�

4

(π2−A)

2 + 1− 4π2

�≥

�4Rs

�2

Mihaly Bencze

PP28256. If x, y, z ≥ 1 thenP 1x2−x+1

+ 27(�

x)2−3�

x−19≥ 8

P 1(x+y)2−2(x+y)+4

. If x, y, z ∈ (0, 1) then

holds the reverse inequality.

Mihaly Bencze

PP28257. In all triangle ABC holdsP 11−sinA+sin2 A

+ 27R2

9R2−3sR+s2≤ 8

P 14−2(sinA+sinB)+(sinA+sinB)2

.

Mihaly Bencze

PP28258. Prove that

1).nP

k=1

1k2−k+1

≥ 4nn2+3

2).nP

k=1

1k4−k2+1

≥ 36n4n4+12n3+n2−12n+31

Mihaly Bencze


k=1

k2(k+1)2

k4+2k3−k+1≤ n(n+1)2

n2+n+1.

Mihaly Bencze

PP28260. If xk ≥ 1 (k = 1, 2, ..., n) thennP

k=1

1x2k−xk+1

≥ n3�

n�

k=1xk

�2

−nn�

k=1xk+n2

.

Mihaly Bencze


PP28261. Denote Fk and Lk the kth Fibonacci respective Lucas numbers.Prove that:

1).nP

k=1

1F 2k−Fk+1

≥ n3

F 2n+2−(n+2)Fn+2+n2+n+1

2).nP

k=1

1L2k−Lk+1

≥ n3

L2n+2−(n+6)Ln+2+n2+3n+9

Mihaly Bencze

PP28262. If a, b, c > 0 and a+ b+ c ≥ 12 then

P(a+ b)4 + 7

P �ab+ bc+ ca+ b2

�2 ≥p8Q

(a2 + 2b2 + c2 + 2 (a+ c) b).

Mihaly Bencze


k=1

1

(nk)2−(nk)+1

≥ (n+1)3

4n−(n+1)2n+(n+1)2.

Mihaly Bencze

PP28264. In all acute triangle ABC holds 12P tgA

2A ≥ 18 +

PAtgA

2 .

Mihaly Bencze

PP28265. If x, y, z > 0 and x+ y + z = 1λ+1 where λ > 0 thenP x+λy+1

(x+λy)3+1≤ 27

7 .

Mihaly Bencze

PP28266. If ak ∈ R (k = 1, 2, ..., n) such that |ai − aj | ≥ πi−j for all i > j.Find the minimal value of

P1≤i<j≤n

(ai − aj)2 .

Mihaly Bencze

PP28267. If x, y, z > 0 then

Pp(x+ y) (y + z) (x+ 2y + z) ≥

pQ(x+ 2y + z)+2 (

Px)q

2�

(x+y)3�

x2+9�

xy.

Mihaly Bencze


PP28268. If xk > 0 (k = 1, 2, ..., n) and

�nP

k=1

x2k

��nP

k=1

xk

�2

= 1 then

Pcyclic

x51

x2+x3+...+xn≥ 1

n(n−1)

�

n�

k=1xk

�4 .

Mihaly Bencze

PP28269. If Fk denote the nth Fibonacci number, thennP

k=1

�Fk

Fk+Fk+2+

Fk+1

Fk+1+Fk+3− Fk+Fk+2

Fk+2Fk+2

�≤ 0.

Mihaly Bencze

PP28270. If a, b, c > 0 and a+ b+ c = 12 then

P a+b2+8(a2+b2)+10ab+13c

≥ 14 .

Mihaly Bencze

PP28271. Let be ak ∈ N∗ (k = 1, 2, ..., n) . Prove thatP

cyclic

lcm(a1,a2)+gcd(a1,a2)a1+a2

≤ 12

nPk=1

ak.

Mihaly Bencze

PP28272. Prove that if there exist sets M1,M2, ...,Mm satisfying thefollowing conditions:

1). For all 1 ≤ i ≤ m, Mi ⊆ {F1, F2, ..., Fn}2). For all 1 ≤ i < j ≤ m, min {|Fi − Fj | , |Fj − Fi|} = 1then m ≤ n, where Fk denote the kth Fibonacci number.

Mihaly Bencze


ctgA2 ctg

B2 − tgA

2 tgB2

�2 ≥ 4.

Mihaly Bencze

PP28274. Let ABC be a triangle, and D ∈ Int (ABC) such that I, B,C,Dare concyclic. The line through C parallel to BD meets AD at point E.Determine all triangles ABC in which CD2 = BD · CE.

Mihaly Bencze


PP28275. If a > 0 then solve in R the equationax + 2a2x + ...+ nanx + n(n−1)

2 = n2ax.

Mihaly Bencze

PP28276. Let A,B ∈ Mn (C) . Determine all x, y ∈ R for which(A+ xB)A∗ (A− yB) = (A− xB)A∗ (A+ yB) .

Mihaly Bencze

PP28277. In all triangle ABC holdsP sin 7A

sinA − 8Q

cos 2A = 5 + 2rR − 2(s2−(2R+r)2)

R2 .

Mihaly Bencze


knm�

a2

k+1 + b2

n+1 + c2

m+1

�≥

≥ 4√3Area [ABC]

kPu=1

nPv=1

mPw=1

q1

u(u+1)v(v+1) +1

v(v+1)w(w+1) +1

u(u+1)w(w+1) .

Mihaly Bencze

PP28279. Let be A,B ∈ Mn (C). Determine all x, y ∈ Z such thatdet (In +AxByA∗) = det (In +A∗BxAy) .

Mihaly Bencze

PP28280. Prove that the equation x3 + 3x2 + (3− 2n)x− n+ 1 = 0 havethree real roots x1 (n) < x2 (n) < x3 (n) and compute lim

n→∞x2 (n) .

Mihaly Bencze

PP28281. In all triangle ABC holdsPcos (sinA) +

Psin (cos (tgA)) > 3

p2−

√2.

Mihaly Bencze

PP28282. Determine all injective functions f : R → R for whichnP

k=1

f (axk) =nQ

k=1

ak when ak > 0 (k = 1, 2, ..., n) .

Mihaly Bencze


PP28283. Let ABC be a triangle in which b2 + c2 = 5a2. Prove that

a�

12c+

√10a

+ 12b+

√10a

�+ b

5a+√10a

+ c5a+

√10b

= 2ab+c .

Mihaly Bencze

PP28284. Determine all n ∈ N for which�112n − 26n

� �n4 − 1

�is divisible

by 285.

Mihaly Bencze

PP28285. If x ∈�0, π2

�then

tg (cos (sinx)) + tg (sin (cos (tgx))) > 2tg�π2

p2−

√2�.

Mihaly Bencze

PP28286. Let ABC be a triangle in which A∡ = 90◦. Prove that

1wa(wa+BD)CD + 1

wa(wa+CD)BD + 1(wa+BD)CD2 + 1

(wa+CD)BD2 = 4aw2

a.

Mihaly Bencze

PP28287. If x, y, z ∈ [0, 1] then

x+ y + z ≤r�

1−√1− x2

��1 +

p1− y2

��1 +

√1− z2

�+

+

r�1 +

√1− x2

��1−

p1− y2

��1 +

√1− z2

�+

+

r�1−

√1− x2

��1−

p1− y2

��1 +

√1− z2

�.

Mihaly Bencze

PP28288. Prove that the equation {xn}+ {yn} = {zn} (n ∈ N,n ≥ 2) havean infinitely many solutions in Q\Z, when {·} denote the fractional part.

Mihaly Bencze

PP28289. Determine all differentiable functions f : R → R for which

f ′ �ax3 + bx�= c and f (0) = 0 and

2R0

f (x) dx = 197210 .

Mihaly Bencze


PP28290. Determine all n ∈ N∗ for whichcard

�nx ∈ N |x3−x2−x−1

n ∈ No∩ {1, 2, 3, ..., n}

�= 1.

Mihaly Bencze

PP28291. Let be f : R → [0,+∞) a continuous function with period T > 0and gk : [0, T ] → R continuous function (k = 1, 2, ..., n) . Prove that

limn→∞

TR0

f (nx)nQ

k=1

gk (x) dx = 1TN−1

TR0

f (x) dx

! nQ

k=1

TR0

gk (x) dx

!.

Mihaly Bencze

PP28292. Let be 0 < x1 < x2 < x3, α,β, γ ∈ (0, 1) and x3n+1 = xα3n−1x1−α3n ;

x3n+2 = xβ3nx1−β3n+1, x3n+3 = xγ3n+1x

1−γ3n+2 for all n ≥ 1. Prove that the sequence

(xn)n≥1 is convergent.

Mihaly Bencze

PP28293. Let ABCD be a convex quadrilateral in which A∡ = C∡ = 90◦.Prove that 4

BD2 = 1(AD+AB)2

+ 1(AD−AB)2

+ 1(CD+CB)2

+ 1(CD−CB)2

.

Mihaly Bencze

PP28294. Let be A ∈ Mn (R) , A = (aij)1≤i,j≤n . Prove that 1n

nPi,j=1

a2ij

!n

≥ det2 (A) .

Mihaly Bencze

PP28295. Let be G = (a, b) and x ∗ y = (a+b)xy−2ab(x+y)+2ab(a+b)2xy−(a+b)(x+y)+c . Determine

all c ∈ R for which (G, ∗) is abelian group. Prove that (G, ∗) ∼=�R∗

+, ·�.

Mihaly Bencze

PP28296. If a, b, c ∈ C∗ and |a| = |b| = |c| then P��a

b − ba

��2 ≤ 6√3.

Mihaly Bencze


P ab(a+c)2(b+c)2

≤ 132Rr

2).P hahb

(ha+hc)2(hb+hc)

2 ≤ s2+r2+4Rr32R 3).

P rarb(ra+rc)

2(rb+rc)2 ≤ 4R+r

16s2r

4).P sin2 A

2sin2 B

2

(sin2 A2+sin2 C

2 )(sin2 B

2+sin2 C

2 )≤ R(2R−r)

2r2


5).P cos2 A

2cos2 B

2

(cos2 A2+cos2 C

2 )(cos2B2+cos2 C

2 )≤ R(4R+r)

2s2

Mihaly Bencze

PP28298. Let be p = 3k + 2, k ∈ N∗ and denote an the sum of decimals of

the number pn. ComputenP

k=1

1a2k

.

Mihaly Bencze

PP28299. Prove that if��xn+1

xn+ xm+1

xm− 2xm+n

xmxn

�� < 2m+n for all n,m ∈ N∗

then (xn)n≥1 is a geometrical progression.

Mihaly Bencze


a�AI2 +AH2

�≥ 8sRr.

Mihaly Bencze


AI4 +AH4�≥ 16s2R2r2

s2−r2−4Rr.

Mihaly Bencze

PP28302. In all acute triangle ABC holdsP

sin (cosA) ≤P sin A2 .

Mihaly Bencze


(arcsin x

x+1 + arcsin 3y3y+1 + arcsin 14z

15z+5 = π

2arctg 1x+5 + arctg 1

y = arctg 2z2

2z2+2z+3

Mihaly Bencze

PP28304. Let ak ∈ Z (k = 1, 2, ..., 2n+ 1) distinct numbers such that2n+1Pk=1

|ak| = n (n+ 1) . Prove that

��2n+1Pk=1

a2n+1k

�� = 0.

Mihaly Bencze


k=1

4

q1k4

+ 14k5

+ 16k6

+ 14k7

+ 1k8

≥ 4√128nn+1 .

Mihaly Bencze



x2

|y+1| =y−4z+1 + ||z|− 1|

y2

|z+1| =z−4x+1 + ||x|− 1|

z2

|x+1| =x−4y+1 + ||y|− 1|

.

Mihaly Bencze


w3a

ha

�λ≤

�sR2r

√s2 − 12Rr

�λfor all

λ ∈ [0, 1] .

Mihaly Bencze

PP28308. Solve in R the equationnP

i=1

�n+1n − xi

�k+

�nP

i=1xi

�k

= n+ 1

when k ≥ 2, k ∈ N is given.

Mihaly Bencze

PP28309. If x ≥ 1 then x2n+1

2n+2 + (2n+1)xn

n+1 ≥ (2n+1)xn+1

n+2 + 1 when n ∈ N.

Mihaly Bencze

PP28310. In all triangle ABC holds Rr − 2 ≥ 3

4

��|√a−

√b|

�√a

�2

. A new

refinement of Euler’s R ≥ 2r inequality.

Mihaly Bencze

PP28311. Let be P (x) =nP

k=0

akxk and ε = cos 2π

n+2 + i sin 2πn+2 . Determine

all ak ∈ R (k = 0, 1, ..., n) such thatn+1Qk=1

P�εk�= (n+ 2)n .

Mihaly Bencze

PP28312. Solve in Q the equationnP

k=1

akk! =

1n! .

Mihaly Bencze


PP28313. If a ∈ (0, 1) ∪ (1,+∞) thennP

k=1

�ak + a−k

�≥ n(n+1)(2n+1)(ln a)2

6 + 2n.

Mihaly Bencze

PP28314. Let be M =�A ∈ M3 (C) |A2 = I3

. Determine all subgroups of

(M3 (C) , ·) from M.

Mihaly Bencze

PP28315. Let ABC be an acute triangle and xn+1 = xn sin2A+ sin2 B

xn,

yn+1 = yn sin2B + sin2 C

yn, zn+1 = zn sin

2C + sin2 Azn

for all n ∈ N,

x0, y0, z0 > 0. Prove that limn→∞

(xn + yn + zn) ≤s(s2+r2−4R2)

3R(s2−(2R+r)2).

Mihaly Bencze

PP28316. Let ABC be a triangle. Determine maxPq

aa sin2 B+b cos2 C

.

Mihaly Bencze


|x+ y|+ 2z = 22x+ |y − 2z| = 16|x+ 3z|+ y = 13

.

Mihaly Bencze


|x+ 2y|+ 3 |z| =√41 + 3

√34

|x|+ |3y − z| =√5 +

√205

|2x+ z|− 2 |y| =√41− 2

√13

.

Mihaly Bencze


s (P

AI)3 ≤ 4�s2 − 3r2 − 6Rr

�2P �cos A

2

�3.

Mihaly Bencze

PP28320. Denote Fk respective Lk the kth Fibonacci respective Lucas

numbers. Prove that n�n2 − 1

� nPk=1

FkLk ≤�n2 − 1

�(Fn+2 − 1) (Ln+2 − 3)+


+3

�nP

k=1

(2k − n− 1)Fk

��nP

k=1

(2k − n− 1)Lk

�.

Mihaly Bencze

PP28321. Determine the gometrical locus of points M in the plane of theconvex quadrilateral ABCD, such that MB = 2MA, MC = 3MA,MD = 4MA.

Mihaly Bencze

PP28322. Determine all a, b, c ∈ N for which�13a + 7b − 2

� �13b + 7c − 2

�(13c + 7a − 2) is divisible by 729.

Mihaly Bencze

PP28323. Determine all a1, a2, a3 > 0 for which

�nP

k=1

ak

�= 1 for all n ≥ 3

where (n+ 1) an+1 = (n− 1) an for all n ≥ 2, and [·] denote the integer part.

Mihaly Bencze

PP28324. Prove that the triangle ABC is equilateral if and only if�ha

wa

�2+

�hb

wb

�2+ 4r

R = 2q

2rR

�ha

wa+ hb

wb

�.

Mihaly Bencze

PP28325. In all triangle ABC holds 3(4R+r)2s2

≥ (4R+r)2+s2

4s2R≥ 6

5R+2r .

Mihaly Bencze


k=1

arctgkk2+2k+2

< 12

�arctg2 (n+ 1)− π2

16

�<

nPk=1

arctg(k+1)k2+1

.

Mihaly Bencze

PP28327. In all triangle ABC holds16(s2−r2−Rr)9(s2+r2+2Rr)

+ Rrs2−3r2−6Rr

≥ 32 .

Mihaly Bencze

PP28328. In all triangle ABC holds s2+r2−8Rr9Rr + r2

4(s2−12Rr)≥ 3

4 .

Mihaly Bencze


PP28329. If a, b, c > 0 and λ > 0 thenP a+(λ+1)b+2λc

(λ+2)a+(2λ+3)b+(3λ+1)c ≤ 32 .

Mihaly Bencze

PP28330. If a, b > 0 then(2a+b)(a2+2b2)

2a2+b2+

(2b+a)(b2+2a2)2b2+a2

≥ 3 (a+ b) .

Mihaly Bencze

PP28331. In all triangle ABC holds R2r ≥ 1

8 + s2+r2

16Rr ≥ 1 (A new refinemento Euler’s R ≥ 2r inequality).

Mihaly Bencze

PP28332. In all triangle ABC holds�4R+rsr

�2 ≥ 2r

�2r − 1

R

�.

Mihaly Bencze

PP28333. If ai > 0 (i = 1, 2, ..., n) andP

cyclic

a1a2 = k then

Pcyclic

�a1 +

1a2

�2≥ (n+k)2

k .

Mihaly Bencze

PP28334. If x > 0 and λ ∈ (−∞, 0] ∪ [1,+∞) then

e−λx +�

272e2x+5ex+2

�λ≥ 23λ−1

�x

ex−1

�λ.

Mihaly Bencze

PP28335. If x ≥ 1 then 2x2 + 16x+ 1x + 1

x+2 + 12x+1 ≥ 5 5

√4 lnx.

Mihaly Bencze

PP28336. If a, b, c > 0 then2P

a2 3pb6 + 15b2c2 (b2 + c2) + c6 ≥P a4 + 3

Pa2b2.

Mihaly Bencze



k=1

�8 ln(k2+k)

(k2+k−1)(k2+k+1)− 27

(k2+k+2)(2k2+2k+3)

�< n

n+1 .

Mihaly Bencze

PP28338. Prove thatn+1Pk=2

1k < n

q1− 1

n√2.

Mihaly Bencze

PP28339. If ak > 0 (k = 1, 2, ..., n) andnP

k=1

ak = n then

Pcyclic

an−11

a1+ n√a2·a3·...·an ≥ n

2 .

Mihaly Bencze

PP28340. If a, b > 0 then

3�a2 + b2

�≥ a

3

qa�2a+ 3

�√3− 1

�b�2

+ b3

qb�2b+ 3

�√3− 1

�a�2.

Mihaly Bencze


k=1

k2 ln kk2+16k+1

≤ n(n−1)(2n+5)192 .

Mihaly Bencze

PP28342. Solve in (1,+∞) the following system:

x12 + 27

4 = 4 lnx2 +27

2(x3+2) +27

4(2x4+1) +1

2x5x22 + 27

4 = 4 lnx3 +27

2(x4+2) +27

4(2x5+1) +1

2x6

−−−−−−−−−−−−−−−−−−−xn

2 + 274 = 4 lnx1 +

272(x2+2) +

274(2x3+1) +

12x4

.

Mihaly Bencze

PP28343. If x ∈�0, π2

�then

sinx3 cos2 x−

√2 cosx+6

√2+ cosx

6+(6−√2) sinx

+√2

3 sin2 x−sinx cosx+6 cosx≥ 3

8 .

Mihaly Bencze


PP28344. If 1 ≤ a ≤ b thenb2−a2

4 − 272 ln b+2

a+2 − 278 ln 2b+1

2a+1 − 12 ln

ba + 43(b−a)

4 ≥ 4 ln bb

aa .

Mihaly Bencze

PP28345. If x ∈�0, π2

�then sin4 x

cos2 x+ cos4 x

2 + 4sin2 x

+ 2 sin2 x cos2 x ≥ 5.

Mihaly Bencze

PP28346. If x ∈�0, π4

�then cos 2x

ln(ctgx) ≥4(2+5 sinx cosx)1+16 sinx cosx .

Mihaly Bencze

PP28347. Denote M the set of all integers in {1, 2, ..., n} which arerelatively prime to n, and

M1 = M ∩�0, kn

�,M2 = M ∩

�kn ,

2kn

�, ...,Mn = M ∩

�(k−1)n

n , ni, where k ≥ 3.

Prove that if cardM ≡ 0 (mod k) then cardM1 = cardM2 = ... = cardMn.

Mihaly Bencze

PP28348. Solve in Z the equation

2�x2 + y2 + z2

�+ xy + yz + zx =

�x+y3 + 1

�3+�y+z

3 + 1�3

+�z+x3 + 1

�3.

Mihaly Bencze

PP28349. If x ∈�0, π2

�then

sin4 x

sin2 x+3√2 cos2 x

+ cos4 x

cos2 x+3√2 sin2 x

+ 4

2+3√sin2 x cos2 x

≥ 32 .

Mihaly Bencze

PP28350. In all triangle ABC holdsP a2

m2a+w2

a≤ 2

�Rr − 1

�.

Mihaly Bencze

PP28351. If a, b > 12 then�

a(a+1)2a−1

�2+

4(a2+b)(a+b2)(a+b−1)2

+�b(b+1)2b−1

�2≥ 6 (a+ b)− 3

2 .

Mihaly Bencze

PP28352. Determine all a, b, c ∈ N for which 10a+b+10b+c+10c+a

a2+b2+c2∈ N.

Mihaly Bencze


PP28353. If x ∈ R then 2�1 + sin2 x cos2 x

�≥ 3

3−sin2 x cos2 x.

Mihaly Bencze

PP28354. Let be ak > 0 (k = 1, 2, ..., n) and denote Fk the kth Fibonaccinumber. Prove that exist ik ∈ N (k = 1, 2, ..., n) such thatnP

k=1

(−1)Fik a2k ≥�

nPk=1

(−1)Fik ak

�2

.

Mihaly Bencze

PP28355. Determine all functions f : R → R for which

(x+ y + z) f (x+ y + z) = f�x2

�+f

�y2�+f

�z2�+2xf (y)+2yf (z)+2zf (x)

for all x, y, z ∈ R.

Mihaly Bencze

PP28356. If xk ∈�0, π2

�(k = 1, 2, ..., n) , then

nPk=1

(sinxk+cosxk)(1−sinxk cosxk)

sin2 xk cos2 xk≤

√n− 1

1

n�

k=1sin2 xk

+ 1n�

k=1cos2 xk

.

Mihaly Bencze

PP28357. If a, b, c > 0 then 2P

a2 (1− b)2 +P

(b− a)2 ≥ 13 (P |a− b|)2 .

Mihaly Bencze

PP28358. If a, b, c > 1 then(a−1)(b−1)(c−1)( 3√

abc−1)3

(√ab−1)

2(√bc−1)

2(√ca−1)

2 ≥ ln a ln b ln c(ln 3√abc)

3

ln2√ab ln2

√bc ln2

√ca.

Mihaly Bencze


1). 2P 1

sin A2

≤ (s−r)2−4R2

s2−(2R+r)2+ 2s

r + 3�2−

√3�

2). (s−r)2−4R2

s2−(2R+r)2≥ 3

�2−

√3�

Mihaly Bencze



ma

wa

�λ≤ 3

�16r

q(R+2r)(s2+2Rr+5r2)

2R

�λ

for all λ ∈ [0, 1] .

Mihaly Bencze

PP28361. Let be x0 = 0, x1 = 1 and xn+1 = (xn + xn−1)qx2n + x2n−1 for all

n ≥ 1. Compute limn→0

xn.

Mihaly Bencze


1

(x41+x4

2)(x1+x2)2 + 1

2(x22+x2

3)3 = 3

16x23x

24

1

(x42+x4

3)(x2+x3)2 + 1

2(x23+x2

4)3 = 3

16x24x

25

−−−−−−−−−−−−−−−−1

(x4n+x4

1)(xn+x1)2 + 1

2(x21+x2

x)3 = 3

16x22x

23

.

Mihaly Bencze


1).Pq

1(a4+b4)(c+b)2

+ 12(a2+b2)3

≤√3

8Rr

2).Pr

1

((s−a)4+(s−b)4)c2+ 1

2((s−a)2+(s−b)2)3 ≤

√3

4r2

3).Pr

1

(r4a+r4b)(ra+rb)

2 + 1

2(r2a+r2b)

3 ≤ (4R+r)√3

4s2r

4).Pr

1

(h4a+h4

b)(ha+hb)2 + 1

2(h2a+h2

b)3 ≤ (s2+r2+4Rr)

√3

16s2r2

Mihaly Bencze


1).P�

1(a4+b4)(a+b)2

+ 12(a2+b2)3

�≤ 3(s2−r2−4Rr)

128s2R2r2

2).P�

1

((s−a)4+(s−b)2)c2+ 1

2((s−a)2+(s−b)2)3

�≤ 3(s2−2r2−8Rr)

16s2r4

3).P�

1

(r4a+r4b)(ra+rb)

2 + 1

2(r2a+r2b)

3

�≤ 3((4R+r)2−2s2)

16s4r2

Mihaly Bencze


PP28365. Prove that there exist infinitely many positive integers n and ksuch that the largest prime divisor of n4 + n2k2 + k4 is equal to the largestprime divisor of (n+ k)4 + (n+ k)2 + 1.

Mihaly Bencze

PP28366. If x, y > 0 then 1(x4+y4)(x+y)2

+ 12(x2+y2)3

≤ 316x3y3

.

Mihaly Bencze


�2(s2−r2−Rr)s2+r2+2Rr

�λ

+�rR

�λ ≤ 2λ · 31−λ for

all λ ∈ [0, 1] .

Mihaly Bencze

PP28368. Determine all function f : R → R for which2f4 (x) + 4 = 3x3 + f

�x2

�+ 4f (x) for all x ∈ R.

Mihaly Bencze

PP28369. Determine the functions f : R → R such that1−x√

4f2(x)+45= 1

(x+1)2

qf(x)+15−3x for all x ∈ R\{5

3}.

Mihaly Bencze

PP28370. Let p ≥ 31 be a prime, andMk = {p+ k, 2p+ k, 3p+ k, ..., (p− 3) p+ k} . Determine all k ∈ N for whichone element of the set Mk is the sum of two integer squares.

Mihaly Bencze

PP28371. If x ∈ [0, 1] then 1−x√4x2+45

≥ 1(x+1)2

qx+15−3x .

Mihaly Bencze

PP28372. If x ∈ R thensin2 x√

45+4 cos2 x+ cos2 x√

45+4 sin2 x≥ 1

(1+sin2 x)2

q1+sin2 x2+3 cos2 x

+ 1(1+cos2 x)2

q1+cos2 x2+3 sin2 x

.

Mihaly Bencze


PP28373. If 0 ≤ a ≤ b ≤ 1 thenbRa

�q5−3xx+1 + 2

√4x+ 5

�dx ≥

12

�b√4b2 + 45− a

√4a2 + 45

�+ 45

4 ln�

2b+√4b2+45

2a+√2a2+45

�.

Mihaly Bencze


q5−3x11+x1

+ 2√5 + 4x2 =

p4x23 + 45q

5−3x21+x2

+ 2√5 + 4x3 =

p4x24 + 45

−−−−−−−−−−−−−−−−q5−3xn

1+xn+ 2

√5 + 4x1 =

p4x22 + 45

.

Mihaly Bencze


k=1

�q5n−3kn+k + 2

q5n+4k

n

�2

≥ 139n2+6n+23n .

Mihaly Bencze

PP28376. In all triangle ABC holds1). 27

�rs

�2+ 2 ≥ 1

3r3� 1

r3a

2). 27Rr2s2

+ 2 ≥ 13r3

� 1

h3a

Mihaly Bencze

PP28377. In all triangle ABC holds1). s2

�s2 + r (4R+ r)

�≥ 2r (4R+ r)2

2). 2 (sr)2�5s2 + r2 + 4Rr

�≥ r

�s2 + r2 + 4Rr

�2

Mihaly Bencze

PP28378. If a, b, c > 0 then 16 (P

a)2 + 18abc ≥ 9P

ab (a+ b) + 36P

ab.

Mihaly Bencze

PP28379. Determine all a, b ∈ N for which 5a

b3+b+1+ 5b

a3+a+1∈ N.

Mihaly Bencze



ma + 2P m2

a

b+c ≥ 4sr(s2−r2−Rr)R(s2+r2+2Rr)

.

Mihaly Bencze

PP28381. If a, b, c > 0 thenP a

3b2+2√3c√a2+b2+c2−bc

≥ 3√3

8√a2+b2+c2

.

Mihaly Bencze


Q �a2 + b2

�≥ 2s2

�s2 + r2 − 6Rr

�2

2).Q�

(s− a)2 + (s− b)2�≥ s2

�s2 + 20Rr

�2

3).Q �

r2a + r2b�≥ 16s4 (2R+ r)2

Mihaly Bencze

PP28383. Let ABCD be a square, and P ∈ BC, Q ∈ CD such thatPAQ∡ = x. Let E and F be the intersections of PQ with AB and AD,respectively. Determine all x such that (AE +AF )2 ≥ 8Area [ABCD] .

Mihaly Bencze


2Q �

cos2 A2 + cos2 B

2

�≥

�Pcos3 A

2 +Q �

cos A2 + cos B

2 − cos C2

��2.

Mihaly Bencze

PP28385. In all triangle ABC holds2Q �

m2a +m2

b

�≥

�Pm3

a +Q

(ma +mb −mc)�2

.

Mihaly Bencze

PP28386. If a, b, c > 0 then 2Q �

a2 + b2�≥ (P

ab (a+ b)− 2abc)2 .

Mihaly Bencze

PP28387. If a, b, c > 0 then�P

a2b− abc�2

+�P

ab2 − abc�2 ≤ 8

27

�Pa2�3

.

Mihaly Bencze

PP28388. Find all functions f : N → N such that m2 + f (mn) + n2 isdivisible by mf (m) +mn+ nf (n) for all m,n ∈ N ∗.

Mihaly Bencze


PP28389. If a, b, c > 0 then�P

a2b− abc�2

+�P

ab2 − abc�2 ≥ 8a2b2c2.

Mihaly Bencze


2P

x2 +P

xy =�x+y

2 + 1�3

+�y+z

2 + 1�3

+�z+x2 + 1

�3.

Mihaly Bencze

PP28391. If x, y, z ≥ 14 then

P ((x+y)2+y+z)((y+z)2+x+y)(x+2y+z−1)2

≥ 4P

x− 34 .

Mihaly Bencze

PP28392. If x, y, z ≥ 13 then

P 1√y−z+(x+y+z)z

≤qP 1

xy .

Mihaly Bencze

PP28393. If x, y, z ≥ 13 then

Pqz

(y−z+(x+y+z)z)(z−x+(x+y+z)) ≤1√xyz .

Mihaly Bencze

PP28394. If x, y, z ≥ 13 thenP

(yP

x+ z − y) (zP

x+ y − z) ≥ (P

xy) (P

x)2 .

Mihaly Bencze

PP28395. If x, y, z ≥ 13 then

P yzy−z+z(x+y+z) ≤ 1.

Mihaly Bencze

PP28396. If x, y, z ≥ 13 then

P z(y+(x+y+z−1)z)(z+(x+y+z−1)x) ≤

(�

x−1)�

x2+((�

x)2−�

x+1)�

xy

xyz(�

x)4.

Mihaly Bencze

PP28397. Prove that

�nQ

k=1

kk�4

≤ exp

�nP

k=1

(k2−1)(k2+16k+1)(k+2)(2k+1)

�.

Mihaly Bencze


PP28398. If 1 ≤ a ≤ b thenb2−a2

4 − 272 ln b+2

a+2 − 278 ln 2b+1

2a+1 − 12 ln

ba + 27

4 (b− a) ≥ 4 ln�

eabb

ebaa

�.

Mihaly Bencze


k=1

�8 ln k(k+1)

(k2+k−1)(k2+k+1)− 27

2k4+4k3+7k2+5k+2

�≤ n

n+1 .

Mihaly Bencze

PP28400. Prove that (n!)4 ≤ exp

�nP

k=1

(k2−1)(k2+16k+1)k(2k2+5k+2)

�.

Mihaly Bencze

PP28401. If xk ∈ (0, 1) (k = 1, 2, ..., n) , then

1).nP

k=1

(1 + xk)2 lnxk ≤ 2

�nP

k=1

x2k − n

�

2).nP

k=1

(1 + xk)�1 + xk + x2k

�lnxk ≤ 2

�nP

k=1

x3k − n

�

Mihaly Bencze

PP28402. If xk ≥ 1 (k = 1, 2, ..., n) thennQ

k=1

xk ≥ exp

2

�

n�

k=1xk−n

�2

n�

k=1x2k−n

.

Mihaly Bencze

PP28403. If xk ∈ (0, 1) (k = 1, 2, ..., n) , thennP

k=1

(1 + xk) lnxk ≤ 2

�nP

k=1

xk − n

�.

Mihaly Bencze


k=1

�1 + 1

k(k+1)

�ln 1

k(k+1) ≤−2n2

n+1 .

Mihaly Bencze


(π +A) ln Aπ < −4π.

Mihaly Bencze



1).P �

1 + sin2 A2

�ln�sin A

2

�< −2− r

2R

2).P �

1 + cos2 A2

�ln�cos A

2

�< −1 + r

2R

Mihaly Bencze

PP28407. If ak > 0 (k = 1, 2, ..., n) then

Pcyclic

aλ+11

(a1+ n√a2a3...an)

λ ≥nλ

�

n�

k=1ak

�λ+1

�

n+(2n−1)n�

k=1ak

�λ for all λ > 0.

Mihaly Bencze


1).P�

rra

+ 1�ln r

ra< −4

2).P�

rha

+ 1�ln r

ha< −4

Mihaly Bencze

PP28409. Determine all functions f : N ∗ → N∗ such that(n− 1)3 < f (n) (f ◦ f) (n) ((f ◦ f ◦ f)n) < n3 + n2 + n+ 1 for all n ∈ N∗.

Mihaly Bencze

PP28410. If λ ∈ [0, 1] thennP

k=1

1k2λ

≤ n�1− 1

n√2

�λ.

Mihaly Bencze

PP28411. Let ABC be a triangle in which B∡ = xC∡, CD = yBC whenextend the side BC be a segment CD. Determine all x, y ∈ R for whichtgBAD∡

2 = a2

4sr .

Mihaly Bencze

PP28412. Determine all functions f : N → N for which

f

nP

k=1

a2k +P

1≤i<j≤naiaj

!=

nPk=1

f2 (ak) + f

P

1≤i<j≤naiaj

!for all

ak ∈ N (k = 1, 2, ..., n) .

Mihaly Bencze


PP28413. Solve in Z the equation (x+yz)2

x+z + (y+zx)2

y+x + (z+xy)2

z+y = x2 + y2 + z2.

Mihaly Bencze


x�y+z

2

�2+ y

�z+x2

�2+ z

�x+y2

�2= x3 + y3 + z3.

Mihaly Bencze

PP28415. If a, b > 0 then ab(a+b)2

+ 2√ab

(√a+

√b)

2 ≤ 14 + 4ab

(a+b)(√a+

√b)

2 .

Mihaly Bencze


4P �

ab + 1 + b

a

�3 ≥ 54P a2+b2

ab ≥ 27(s2+r2+10Rr)2Rr .

Mihaly Bencze

PP28417. If x0 = 1 and xkn+1 + 1 = (xn + 1)k for all n ≥ 0 then determineall k ∈ R for which [xn] = n for all n ≥ 1, where [·] denote the integer part

Mihaly Bencze and Tibor Jakab

PP28418. Solve in Z the equationnP

k=1

x2k = n− 2 +P

1≤i<j≤nxixj .

Mihaly Bencze

PP28419. If ak ∈ [0, 1] (k = 1, 2, ..., n), then

P

1≤i<j≤n

11+aiaj

! P

1≤i<j<k≤n

11+aiajak

!≥ n2(n−1)2(n−2)

72 .

Mihaly Bencze

PP28420. Solve in Z the equationP

cyclic

x1x2...xn−1 = n− 1 +nQ

k=1

xk.

Mihaly Bencze


PP28421. If a, b > 0 then��√3− 1

�a+

�2√3 + 1

�b�q

aa+2b +

��√3− 1

�b+

�2√3 + 1

�a�q

bb+2a ≤

3 (a+ b) .

Mihaly Bencze

PP28422. If a, b, c > 0 then a3(6c−a)2

(8c2+a2)2+ b3(6a−b)2

(8a2+b2)2+ c3(6b−c)2

(8b2+c2)2< a+ b+ c.

Mihaly Bencze

PP28423. If a, b, c > 0 and a+ b+ c = 2 then 27�a2 + b2 + c2

�+64abc ≥ 54.

Mihaly Bencze

PP28424. Compute limt→0

1t

�3t

tR0

√1 + exdx+ 5

t

tR0

eexdx− 3

√2− 5e

�.

Mihaly Bencze

PP28425. If xk > 0 (k = 1, 2, ..., n) , thennP

k=1

x2k√

1+xk≥

�

n�

k=1xk

�2

�

n2+nn�

k=1xk

.

Mihaly Bencze

PP28426. If x, y, z > 0 thenP x2√

1+x+ (

�

x)2√9+3

�

x≥P (x+y)2√

4+2(x+y).

Mihaly Bencze

PP28427. If ai > 0 (i = 1, 2, ..., n) and k ∈ N, k ≥ 2,nP

i=1ai = 1 then

Qcyclic

�1 + ak2a

k3...a

kn

�a1 ≥ 1 +

n

nQp=1

ap

!k

.

Mihaly Bencze

PP28428. If x, y ∈ (0, π2 ) then:

(sin2 x+sin2 y)sin2 + sin2 y ·(cos2 x+cos2 y)cos

2 x+cos2 y

(sinx)2 sin2 x·(sin y)2 sin2 y ·(cosx)2 cos2 x·(cos y)2 cos2 y≤ 4

Daniel Sitaru


PP28429. If a, b, c > 0; a+ b+ c = 3; 0 ≤ x ≤ 1 then:a�ba

�x+ b

�cb

�x+ c

�ac

�x+ b

�ab

�x+ c

�bc

�x+ a

�ca

�x ≤ 6

Daniel Sitaru

PP28430. Find: Ω =P∞

k=1

�P∞n=1n6=k

�1

n2−k2

��

Daniel Sitaru

PP28431. Find: Ω =R �P∞

n=1

�3n sinh3

�x3n

��dx

Daniel Sitaru

PP28432. If x, y, z, t > 0 then:

(xy + yz + zt+ tx)�

1x4 + 1

y4+ 1

z4+ 1

t4

�≥

�1x + 1

y + 1z + 1

t

�2

Daniel Sitaru

PP28433. If a, b, c, d > 0 then:(ab+ bc+ cd+ da)(a4 + b4 + c4 + d4) ≥ abcd(a+ b+ c+ d)2

Daniel Sitaru

PP28434. Let be: Ω(a, b, c) =P∞

n=1an2+bn+c

n! ; a, b, c > 0

Prove that: Ω(a, b, c) + Ω(b, c, a) + Ω(c, a, b) ≥ 3(4e− 1) 3√abc

Daniel Sitaru

PP28435. In ΔABC the following relationship holds:P

CyC(a,b,c) aq�

(b− c)2 + 4r2��(c− a)2 + 4r2

�≥ abc

Daniel Sitaru

PP28436. In acute ΔABC the following relationship holds:2sinA + ssinB + 2sinC + 2cosA + 2cosB + 2cosC > 9

Daniel Sitaru

PP28437. If a1, a2, ..., a8 ≥ 1 then: a41 + a42 + ...+ a48 ≤ (a1a2...a8)4 + 7

Daniel Sitaru


PP28438. Find: Ω = limn→∞ npΩn(a)− (a+ 1)!; a ∈ N;

Ωn(a) =Pn

k=0(k2 − a2 + 1)(a+ k)!; n ∈ N

Daniel Sitaru

PP28439. Let be Ω(a) = limn→∞Pn

k=1

�1

3k sin3(3k sin a)

�

Prove that if a, b, c ∈ [0, π2 ] then: 4�bΩ(a) + cΩ(b) + aΩ(c)

�≤ 3(a2 + b2 + c2)

Daniel Sitaru

PP28440. If 0 < a ≤ b ≤ c then: 1(a−b+c)6

+ 1b6

≤ 1a6

+ 1c6

Daniel Sitaru

PP28441. If a, b > 0, a2 + b2 = 1 then: 1a + 1

b ≥ 2√2.

Daniel Sitaru

PP28442. If x, y > 0; Ω(x, y) =P∞

n=12n2+(2x+2y+5)n+2xy+6x−y3n(n+y)(n+y+1)(n+y+2) then:

Ω(y, x) ≤ 19 3√xy

Daniel Sitaru

PP28443. If a, b, c, d > 0; a3 + b3 + d3 = 1 then: a+b+c+dabcd ≥ 16

Daniel Sitaru

PP28444. If a, b, c, > 0; a3 + b3 + c3 = 1 then: a+b+cabc ≥ 3 3

√9

Daniel Sitaru

PP28445. If a, b, c, d > 0; a2 + b2 + c2 + d2 = 1 then: a+b+c+dabcd ≥ 32

Daniel Sitaru

PP28446. If a, b, c > 0; a2 + b2 + c2 = 1 then: a+b+cabc ≥ 9

Daniel Sitaru

PP28447. If a, b, c > 0; x, y, z, t ∈ R then:(a+b+c)y2

a + (a+b+c)z2

b + (a+b+c)t2

c ≥ x(2y + 2z + 2t− x)

Daniel Sitaru


PP28448. If a, b > 0; x, y, z ∈ R then: (a+b)y2

a + (a+b)z2

b ≥ x(2y + 2z − x)

Daniel Sitaru

PP28449. In ΔABC the following relationship holds:

3

qsinAsinB + 3

qsinBsinC + 3

qsinCsinA − 3

qsinAsinC − 3

qsinBsinA − 3

qsinCsinB < 1

Daniel Sitaru

PP28450. If x ∈�0,√3�then�

1(1+x)2

+ 2x2

(3+2x−x2)

��1

(1+x)2+ 8x2

(3+2x−x2)2

�≥ 9

32 .

Mihaly Bencze

PP28451. If a, b > 0 and 2ab+ a2 = 2a+ b then�2

1+a + 11+b

��2

1+b +1

1+a

�≤ 25

9 .

Mihaly Bencze


k=1

sin π2k(k+1) >

nn+1 .

Mihaly Bencze

PP28453. If xk > 0 (k = 1, 2, ..., n) andnP

k=1

xk ≤ 1 then

nQk=1

�1− 1

xλk

�≥

�2λ − 1

�nfor all λ ≥ 1.

Mihaly Bencze

PP28454. If m,n, k ∈ N ;m,n, k ≥ 2 then�m3 − 1

� �n3 − 1

� �k3 − 1

�≥ 5m2n2k2 + 2mnk − 1.

Mihaly Bencze

PP28455. If m,n, k ∈ N ;m,n, k ≥ 2 then�mk − 1

� �nk − 1

�≥ (2k−1−1)mk−1nk−1

2k−3 + 1.

Mihaly Bencze


PP28456. In all convex quadrilateral ABCD holds1 + 2

PsinA+

Psin2A ≤ 3

Q(1 + sinA) .

Mihaly Bencze


x+ y2 + z3 = 32x2 + y3 + z4 = 74x3 + y4 + z5 = 260

.

Mihaly Bencze

PP28458. If a, b > 0 and ab = 2 then 4a+b +

a+b2ab ≤ 3(a+b)

2 .

Mihaly Bencze

PP28459. Denote Fk and Lk the kth Fibonacci respective Lucas numbers.Prove that:

1).qx2 + xFk + F 2

k +qx2 + xFk+1 + F 2

k+1 ≥q

4x2 + 2xFk+2 + F 2k+2

2).qx2 + xLk + L2

k +qx2 + xLk+1 + L2

k+1 ≥q4x2 + 2xLk+2 + L2

k+2 for

all x ≥ 0

Mihaly Bencze

PP28460. For a positive integer n, we define Fa (n) to be the number of a′ sthat appear as digits after writting the numbers 1, 2, ..., n in their decimalexpansion. Prove that there are finitely many numbers m such thatFa (m) = m for all a a ∈ {2, 3, 4, 5, 6, 7, 8, 9} .

Mihaly Bencze

PP28461. If a0 = 3, a1 = a, a2 = b and an = aan−1 + ban−2 + can−3 for alln ≥ 3 then write an as a polynomial in a, b, c.

Mihaly Bencze

PP28462. In all triangle ABC holds6√14 ≤P

√15 + cos 4A+

P√15− cos 4A ≤ 24.

Mihaly Bencze


PP28463. If a, b, c > 0 then�5−

√5�P

a2 +√5min

�a2, b2, c2

≥ 3

�√5− 1

�Pab.

Mihaly Bencze

PP28464. If a, b, c > 0 then√a2 + ab+ b2 +

√b2 + bc+ c2 ≥

q(a+ c)2 + 2 (a+ c) b+ 4b2.

Mihaly Bencze

PP28465. Denote Fk and Lk the kth Fibonacci respectve Lucas numbers.

Prove thatqx2 + xFk + F 2

k +qx2 + xLk + L2

k +q

x2 + xFk+1 + F 2k+1+

+qx2 + xLk+1 + L2

k+1 ≥q

16x2 + 4x (Fk+2 + Lk+2) + (Fk+2 + Lk+2)2,

for all x ≥ 0.

Mihaly Bencze

PP28466. In all triangle ABC holdsP (ha−ra)(ha+hara+ra)

hara�

ha

√r2a+ra+1+ra

√h2a+ha+1

� ≥ 0.

Mihaly Bencze


cyclic

�ra

qr2b + rb + 1 + rb

pr2a + ra + 1

�≥ 4Rs2

r Qcyclic

(h2a + ha + 1).

Mihaly Bencze

PP28468. Prove that if a, b, c > 1, thenP

loga c ≥P

logab b2. Where

n ∈ N ∗ [1] , a ∈ R\ [0, 1] .

Onose Alina-Iulia

PP28469. Let x, y ∈�0, π2

�. Prove that:

1cosx(sinx+cosx) +

1cos y(sin y+cos y) ≥ 4 tan π

8 .

Onose Alina-Iulia

PP28470. If a, b, c > 0 and a+ b+ c = 27, prove that:ab +

bc +

ca + 3 ≥ 2

3

�√a+

√b+

√c�.

Onose Alina-Iulia


PP28471. If A,B,C are the angles of a triangle, show that the relationshipbelow is true: (cosA+ cosB − cosC)2 + (cosA− cosB + cosC)2+

+(− cosA+ cosB + cosC)2 ≥ 14 + 4 sin A

2 sin B2 sin C

2 .

Maria Mantu and Ioana Mantu

PP28472. Let n be an integer with n ≥ 3 and let ai, i = 1, 2, ..., n, bepositive real numbers such that a1 ≤ a2 ≤ ... ≤ an. Prove that for any realnumber k with k > 1 the following inequality holds:

a1ak2 + a2a

k3 + ...+ an−1a

kn + ana

k1 ≥ ak1a2 + ak2a3 + ...+ akn−1an + akna1

Leonard Giugiuc

PP28473. Let a, b and c be real numbers situated in the interval [1,∞) suchthat a+ b+ c = 1 + 1

a + 1b +

1c . Find the maximum value of the expression

ab+ bc+ ca.

Sladjan Stankovik, Leonard Giugiuc and Diana Trailescu

PP28474. Prove that if k =�√

7 +√3− 4

� �2 +

√3�, then√

x2 + 3 +py2 + 3 + k

√xy + 3 ≥ 2k + 4 for all non negative real numbers x

and y with x+ y = 2. When equality holds?

Leonard Giugiuc, Marius Dragan and Marian Cucoanes

PP28475. In all acute triangle ABC holdsP A(3−A)B+C + 9−π

2 ≥P (A+B)(6−A−B)π+C .

Mihaly Bencze

PP28476. If a, b, c > 0 and λ > 0 thenP a√

a2+λbc+ (P

a)q

�

a�

a3+3λabc≥P (a+b)

√a+b√

a3+b3+2λabc.

Mihaly Bencze

PP28477. In all acute triangle ABC holds3√3

2π + 12

P sinAB+C ≤ 4

3

P cos A2

B+C ≤ 9√3

2π .

Mihaly Bencze



a2x ≥ 4srqP 5−3yz

1+x for all

x, y, z > 0 and x+ y + z = 3.

Mihaly Bencze

PP28479. If ak, bk, ck > 0 (k = 1, 2, ..., n) , then

nPk=1

akbkck + 3

s�nP

k=1

a3k

��nP

k=1

b3k

��nP

k=1

c3k

�≥

≥ 2

n2

nX

k=1

ak

! nX

k=1

bk

! nX

k=1

ck

!.

Mihaly Bencze


a2x ≥ 4sr 10p12P

x4y4z (x+ y), forall x, y, z > 0.

Mihaly Bencze


a2x ≥ 4sr

rP xy

√xy(xy+3z2)

(x+y)z for all

x, y, z > 0.

Mihaly Bencze


a2x ≥ 4sr 8

q818

Q(1 + x2y2) for all

x, y, z > 0 and xyz = 1.

Mihaly Bencze


a2x ≥ 2srq

3√3�

(x+yz)2�

xy for allx, y, z > 0 and x+ y + z = 3.

Mihaly Bencze


a2x ≥ 8srq

15

P x√y√

x+zfor all

x, y, z > 0.

Mihaly Bencze



a2x ≥ 2srq5P x(y+z)

x+1 − 3xyz whenx, y, z > 0 and x+ y + z = 3.

Mihaly Bencze

PP28486. If a, b, c, d > 0 then3�a2 − ab+ b2

�2+ 2abcd+ 3

�c2 − cd+ d2

�2 ≥ 2�a2c2 + b2d2

�.

Mihaly Bencze


a2x ≥ 4srq

12

P x2(y+z)2

y2+z2for all

x, y, z > 0.

Mihaly Bencze


a2x ≥ 4srpP

xλ when x, y, z > 0,x+ y + z = 3 and λ = 2 ln 3−3 ln 2

ln 3−ln 2 .

Mihaly Bencze


a2x ≥ 12sr

√�

x2� x

y

for all x, y, z > 0.

Mihaly Bencze


Pa2x ≥ 8sr

p2 (P

x2y2) (P

x3 + 16xyz) for all x, y, z > 0 and x+ y+ z = 1.

Mihaly Bencze

PP28491. In all triangle ABC holdsPa2x ≥ 4 4√3sr√

26p(x+ y) (y + z) (z + x) for all x, y, z > 0.

Mihaly Bencze


a2x ≥ 4sr

r3√

�

(x+y)�

x2(y+z)for all

x, y, z > 0.

Mihaly Bencze


PP28493. Determine all triangle ABC in whichPa2x ≥ sr

p2 (P

x3 + 9xyz +P

x) for all x, y, z > 0.

Mihaly Bencze

PP28494. In all triangle ABC holdsPa2py2 − yz + z2 ≥ 4sr

px2 + y2 + z2 for all x, y, z > 0.

Mihaly Bencze


a2py2 + yz + z2 ≥ 4sr

Px for all

x, y, z > 0.

Mihaly Bencze


a2x ≥ 4srq

�

a2b2

(�

a2y)(�

a2z)for all

x, y, z > 0.

Mihaly Bencze


a2 ≥ 4srqabc+ 3

pQ(1 + a3).

Mihaly Bencze


a2x ≥ 4sr

rq23

P x√y√

x+27for all

x, y, z > 0.

Mihaly Bencze

PP28499. If a, b, c > 0 and a+ b+ c = 1 thenP �

a2 + b2� �

b2 + c2�≥ 24

�Pa2b2

� 43 .

Mihaly Bencze

PP28500. If a, b, c > 0 and abc = 1 thenP(a+ b)3 (b+ c)3 ≥ 18 (a+ b+ c− 1)2 .

Mihaly Bencze

PP28501. In all triangle ABC holdsP(x+ y) a2 ≥ 4

√3sr 6p4 (x+ y + z − 1) where x, y, z > 0 and xyz = 1.

Mihaly Bencze



a3 ≥ 4srq8s2Rr + 1

2

Pa2 (b− c)2.

Mihaly Bencze

PP28503. If a, b, c > 0 and abc = 1 thenP

(a+ b)3 ≥ 12 (a+ b+ c− 1) .

Mihaly Bencze

PP28504. In all triangle ABC holds (P

a cosA)2 ≤P

m2a.

Mihaly Bencze


cos 2A+ 2 (P

cosA)2 ≤ 3.

Mihaly Bencze


m2a

bc + a2

mbmc

�≥ 25

4 .

Mihaly Bencze

PP28507. In all triangle ABC holdsP �r2a + r2b + r2c

�a �m2

a +m2b +mc

�b ≥ 3s8s3 .

Mihaly Bencze

PP28508. Let ABCD be a convex quadrilateral in which A∡ = C∡ = 90◦.Denote the proiection of A and C to BD with A1 and C1. Prove that

AA1 + CC1 +�2√2− 1

�BD ≤ Perimeter (ABCD) .

Mihaly Bencze

PP28509. In all triangle ABC holdsP a2(a+c)(a+c−b)

b ≥ 24sr

rsr(s2+r2+2Rr)� (c+a)(c+a−b)

b

.

Mihaly Bencze

PP28510. Solve in Z the equation x3+yzy2+z2

+ y3+zxz2+x2 + z3+xy

x2+y2= 0.

Mihaly Bencze



a2xz (x+ z)+b2yx (y + x)+c2zy (z + y) ≥ 2srq2 (P

xz (x+ z))pQ

(x+ y)for all x, y, z > 0.

Mihaly Bencze


a2 + 2n+16 b2 + n(n+1)

2 c2 ≥ 8srn(n+1)

nPk=1

pk3 (k2 + k + 1).

Mihaly Bencze


a2

n+1 + b2

m+1 + c2

k+1 ≥ 4srnmk

nPp=1

mPr=1

kPt=1

q1

pr(p+1)(r+1) +1

rt(r+1)(t+1) +1

pt(p+1)(t+1) .

Mihaly Bencze

PP28514. If A,B ∈ Mn (R) and AB = BA then detrmine all λ ∈ R for

which detnQ

k=0

�A2 + (λ+ k)AB + (λ+ k + 1)B2

�≥ 0.

Mihaly Bencze

PP28515. If A,B ∈ Mn (R) and AB = BA thendet

��A2 + 3AB + 4B2

� �A2 + 4AB + 5B2

��≥ 0.

Mihaly Bencze

PP28516. Let ABC be a triangle in which A ≤ B ≤ C. Prove that

sinB + sinC > (sinA)sinA(1−sinB)sinB(1−sinA) + (sinB)

sinB(1−sinC)sinC(1−sinB) .

Mihaly Bencze

PP28517. In all acute triangle ABC holdsPlog2

�2sinA + 2cosA

�≥ 3 + 1√

2

Psin

�A+ π

4

�.

Mihaly Bencze


k=1

�k4 + 10k3 + 32k2 + 40k + 20

�((k + 1)!)2 =

�n2 + 6n+ 5

�((n+ 2)!)2−20.

Mihaly Bencze


PP28519. If ak > 0 (k = 1, 2, ..., n) andnQ

k=1

ak = 1 thenP

cyclic

a1ea2 ≥ ne.

Mihaly Bencze

PP28520. If a, b, c > 0 then

�a+ba

�2a � b+cb

�2b � c+ac

�2a ≤�2a+b+ca+c

�a+c �2b+c+ab+a

�b+a �2c+a+bc+b

�c+b.

Mihaly Bencze

PP28521. If xk ∈ R (k = 1, 2, ..., n) and xi 6= xj (i 6= j) then

Pcyclic

|x1 + x2| ≤P

cyclic

x1|x1|−x2|x2|x1−x2

≤ 2nP

k=1

|xk| .

Mihaly Bencze


sinA ≤ 2 + 1sinA sinB sinC .

Mihaly Bencze

PP28523. In all triangle ABC holds s2−8Rrr2

+�4R+r

s

�2 ≤ 2+16R2�

1s2

+ 1r2

�.

Mihaly Bencze

PP28524. If 0 ≤ ak ≤ bk (k = 1, 2, ..., n) thennP

k=1

ebk−eakbk−ak

≤ n− 1 +nQ

k=1

ebk−eakbk−ak

.

Mihaly Bencze

PP28525. If a, b, c > 0 and a+ b+ c = 1 thenP 1

a+1 ≤ 3918 + 4

3

P 14−a2

.

Mihaly Bencze

PP28526. If 1 ≤ ak ≤ bk (k = 1, 2, ..., n) then

12

nPk=1

(ak + bk) ≤ n− 1 + 12n

nQk=1

(ak + bk) .

Mihaly Bencze



1). 3P ra

r+ra≤ 13

2 + 4minn

r2a4r2a−r2

;r2b

4r2b−r2

; r2c4r2c−r2

o

2). 3P ha

r+ha≤ 13

2 + 4minn

h2a

4h2a−r2

;h2b

4h2b−r2

; h2c

4h2c−r2

o

Mihaly Bencze

PP28528. If x ∈�0, π2

�then 17 + 112 sin2 x cos2 x ≥ 54

√2 sinx cosx.

Mihaly Bencze

PP28529. In all triangle ABC holds1). 27 + 3

�rs

�2+ 6r(4R+r)

s2≥ r3

P 1r3a

+ 54√2 rs

2). 27 +3(s2+r2+5Rr)

2s2≥ r3

P 1h3a+ 54

s

√Rr

Mihaly Bencze

PP28530. If 0 < x ≤ y < π2 then�q

xy + 1

� �sin y − sin

√xy

�≥ sin y − sinx ≥

�qyx + 1

� �sin

√xy − sinx

�.

Mihaly Bencze

PP28531. If x > 0 thenx6+x3+1

x3 + x2+x+1x + x4+x2+1

x2 ≥ 9�

x2

1+2x2 + 12+x + x

x3+x+1

�.

Mihaly Bencze

PP28532. If a, b, c ∈ [0, 1] then(1+a3)(1+b6)(1+c9)(1−a3)(1−b6)(1−c9)

≥ (1+ab2c3)(1+bc2a3)(1+ca2b3)(1−ab2c3)(1−bc2a3)(1−ca2b3)

.

Mihaly Bencze

PP28533. In all triangle ABC holds1). 324r4

P 1h4a+ 45 ≥ 171r2

P 1h2a

2). 324r4P 1

r4a+ 45 ≥ 171r2

P 1r2a

Mihaly Bencze

PP28534. If ak > 0 (k = 1, 2, ..., n) thenP

cyclic

4√a2a3

2a1+a2+a3≥ 1

4

nPk=1

1√ak.

Mihaly Bencze


PP28535. Prove that exp

�nP

k=1

�√k2 + k +

√k2 + 2k +

√k2 + 3k + 2

��≤

≤ en(3n+5)

2

2n

�(e− 1)3 (e+ 1)

�n

for all n ∈ N∗.

Mihaly Bencze

PP28536. Denote Fk and Lk the kth Fibonacci, respective Lucas numbers.Prove that

1). Fn+2 − 1 ≤nP

k=1

Fλ+1k

+Fλ+1k+1

Fλk+Fλ

k+1

≤ Fn+3 − 2

2). Ln+2 − 3 ≤nP

k=1

Lλ+1k

+Lλ+1k+1

Lλk+Lλ

k+1

≤ Ln+3 − 4 for all λ ≥ 1

Mihaly Bencze


ln2�1 +

√sinA

�≤ s

R .

Mihaly Bencze

PP28538. If f : R → (0,+∞) when 0 < a ≤ b, is continuous and increasing,

then (a+ 3b)abRaf (x) dx

4(a+b)2b2R

(a+b)2f (x) dx ≤ a

bRaf (x) dx

4b2R

(a+b)2f (x) dx.

Mihaly Bencze

PP28539. Solve in R the equation: (arcsin [x])3 (arccos [x])5 = x3�π2 − x

�5where [·] denote the integer part.

Mihaly Bencze and Rovsen Pirguliev


sin3 A2

+ 12Rr

P sin A2

sin B2+sin C

2

+ 24(R+r)r ≥ 216.

Mihaly Bencze

PP28541. In all triangle ABC holds 4s2r ≤ 3R�s2 − r (4R+ r)

�.

Mihaly Bencze


PP28542. In all triangle ABC holds s2λ + 5r2λ + 2 (Rr)λ ≤ 8�RrP m2

a

bc

�λ

for all λ ∈ [0, 1] .

Mihaly Bencze

PP28543. In all triangle ABC holds�P 1sin A

2

�3

≥ 24(R+r)r + 12R

r

P sin A2

sin B2+sin C

2

.

Mihaly Bencze


1).P

b sinA ≤√32

�(s2+r2+4Rr)

2

4sRr − 4s

�

2).P

(s− a) sinA ≤√32

�(4R+r)2−2s2

s

�

3).P

ra sinA ≤√32

�s2−2r(4R+r)

r

�

4).P

sinA sin2 A2 ≤

√32

��s2+r2−8Rr

4Rr

�2+ r

R − 2

�

5).P

cosA cos2 A2 ≤

√32

�(s2+(4R+r)2)

2−16Rs2(4R+r)

16s2R2

�

Mihaly Bencze

PP28545. In all acute triangle ABC holds9−π6 +

P sinAB+C ≥ 1

3

P (A+B)(6−A−B)π+C .

Mihaly Bencze

PP28546. In all triangle ABC holds 2P�

ma

mb

�2≥

�s2 − r2 − 4Rr

�P 1m2

a.

Mihaly Bencze


ctgA ≥ (�

cosA)2

2 sinA sinB sinC .

Mihaly Bencze


cos2A+ (P

cosA)2 ≤ 3.

Mihaly Bencze


PP28549. In all acute triangle ABC holds�Psin3A

� �Pcos3A

�2 ≥ (sinA sinB sinC)3 .

Mihaly Bencze

PP28550. In all acute triangle ABC holds (P

sinA)2 ≤ 2Q sin2 A

cosA .

Mihaly Bencze

PP28551. In all acute triangle ABC holds�2R2

sr

�λ+ (Q

ctgA)λ ≤ 21−λ (P

ctgA)λ for all λ ∈ [0, 1] .

Mihaly Bencze

PP28552. In all acute triangle ABC holds�Psin3A

� �P 1cos3 A

�2 ≥ 8tg3Atg3Btg3C.

Mihaly Bencze

PP28553. In all acute triangle ABC holds�P

cos3A� �P 1

sin3 A

�2≥ 8.

Mihaly Bencze

PP28554. In all acute triangle ABC holds8�P

sin3A�2 �P

cos3A�≥

�Psin2A

�3.

Mihaly Bencze

PP28555. Determine all λ1,λ2 > 0 for which in all acute triangle ABCholds 2

P(sinA)λ1 ≤ 3

P(ctgA)λ2 .

Mihaly Bencze

PP28556. In all triangle ABC holds r2λ +�s2−6Rr−3r2

4

�λ≤ 21−λR2λ for all

λ ∈ [0, 1] .

Mihaly Bencze


(ab)2λ ≤ 31−λ (RP

a)2λ for allλ ∈ [0, 1] .

Mihaly Bencze



b+c ≥ 2√3sr.

Mihaly Bencze

PP28559. If a, b, c > 0 thenP 1

a +P a

b2+c2≥ (

�

a)2�

ab

P 1a+b ≥ 3

√3√

�

a2�

a

P 1a+b ≥ 3

P 1a+b .

Mihaly Bencze


P ab2+c2

≥ 10ss2+r2+2Rr

− s2+r2+4Rr4sRr

2).P ra

r2b+r2c

≥ (4R+r)2((4R+r)2+s2)4s4R

− 1r

3).P sin2 A

2

sin4 B2+sin4 C

2

≥ 8R(2R−r)2(16R2−24Rr+5r2+s2)(s2+r2−8Rr)((2R−r)(s2+r2−8Rr)−2Rr2)

− s2+r2−8Rrr2

4).P cos2 A

2

cos4 B2+cos4 C

2

≥ 4R(4R+r)2(5(4R+r)2+s2)(s2+(4R+r)2)((4R+r)3+s2(2R+r))

− (4R+r)2+s2

s2

Mihaly Bencze


b2+c2≥ 9abc

�

a−4(�

ab)2

4abc�

ab .

Mihaly Bencze

PP28562. In all triangle ABC holdsP a(brc+(c−b)ra−crb)

(bra+crb)(brc+cra)≥ (4R+r)2((4R+r)2+s2)

4s4R.

Mihaly Bencze

PP28563. In all triangle ABC holds 16s2r2 ≤P

a4 ≤ 4s2R2 (A newrefinement of Euler’s R ≥ 2r inequality).

Mihaly Bencze

PP28564. In all triangle ABC holdsP b+c−a

2(a−b)2+c2≥ s(s2+r2+4Rr)

8Rr2(4R+r)− 4R+r

2sr .

Mihaly Bencze

PP28565. In all triangle ABC holds�Pm2

am2bw

2c

�h2ah

2bh

2c ≤

�Pr2ar

2bh

2c

�w2aw

2bw

2c .

Mihaly Bencze


PP28566. In all triangle ABC holds�P

a2ma

� �Pam2

a

�≥ 72s3r3.

Mihaly Bencze


1).Q�

10ab+ 3p4 (a6 + b6)

�≤ 108s2

�s2 + r2 + 2Rr

�2

2).Q�

10rarb +3

q4�r6a + r6b

��≤ 432s4R2

Mihaly Bencze


P 1�

10ab+ 3√

4(a6+b6)≥ 5s2+r2+4Rr

2√3s(s2+r2+2Rr)

2).P 1

�

10rarb+3�

4(r6a+r6b)

≥ (4R+r)2+s2

4s2R√3

Mihaly Bencze

PP28569. In all triangle ABC holdsP tg4 A

2ctg4 C

2

18r2

s2ctgA

2ctgB

2−�

3rsctgC

2+2

≥ 3(s2−2r(4R+r))s2

.

Mihaly Bencze

PP28570. If ak > 0 (k = 1, 2, ..., n) , thenPr

10a1a2 +3

q4�a61 + a62

�≤ 2

√3

nPk=1

ak.

Mihaly Bencze


49�a2

b2+ 2b

a

��b2

a2+ 2a

b

�≥ 225 + 90

�ab +

ba

�+ 9

�ab +

ba

�2.

Mihaly Bencze


� 4�

ctgA2

�

ctgA2ctgB

2

≥ 4

q3�rs

�11qPctgA

2 ctgB2 .

Mihaly Bencze


PP28573. In all triangle ABC holds 2s9r

�PqtgA

2

�2

≥P�

ctgA2+�

ctgB2

�

ctgC2

.

Mihaly Bencze

PP28574. If a, b ≥ 1 then(1+a)(1+b)(1+

√ab)(2+a+b)

(1+√a)(1+

√b)

≥ 2�√

a+√b�+ 2 (a+ b) .

Mihaly Bencze

PP28575. In all triangle ABC holds�P �tg3A2 ctg

2B2 ctg

5C2

� 14

��P �tg3C2 ctg

2B2 ctg

5A2

� 14

�≥ 27.

Mihaly Bencze


3�

ctg2 A2

�

3�

ctg2 B2− 3�

srtgA

2+ 3�

ctg2 C2

� ≥ 3rs

P3

qctgA

2 .

Mihaly Bencze


1). a2+b2

2ab + ab(a+b)2

≥ 54

2). a2+b2

ab + 4ab(a+b)2

≥ 3

Mihaly Bencze

PP28578. If a, b, c > 0 then

9�P

a3� �P 1

a3

�+ 27 (

Pa)

�P 1a

�≥ 108 +

�P a+bc

�3.

Mihaly Bencze

PP28579. If a, b > 0 then 13 + 3a2+2ab+3b2

4(a2+ab+b2)≥ a(a+2b)

(a+b)(2a+b) +b(b+2a)

(a+b)(2b+a) .

Mihaly Bencze

PP28580. In all triangle ABC holdsPs

�

ctgA2

�

ctgB2+�

ctgC2

≥q

3r2s

PqctgA

2 .

Mihaly Bencze


PP28581. If a, b > 0 then 1+3(a6+b6)2a3b3

≥ a(a+2b)2a2+b2

+ 2a2+b2

a(a+2b) +b(b+2a)2b2+a2

+ 2b2+a2

b(b+2a) .

Mihaly Bencze


b+c ≥ 13

��

ab�

a2+

�

a2�

ab

�+ 5

6 ≥ 32 (A

refinement of Nesbit’s inequality).

Mihaly Bencze

PP28583. If a, b > 0 then a3+b3

a2+ab+b2+

2(a2+b2)√3(a2+ab+b2)

≥ a+ b.

Mihaly Bencze

PP28584. If a, b, c > 0 then

6

�2(a4+b4)a3+b3

+ a5+b5

2a2b2

�≥ (2a+b)4

(2a2+b2)(a+2b)+ (a+2b)4

(a2+2b2)(2a+b).

Mihaly Bencze


1).P r2

b

r3a3�

4(729r6+r6b)

≥ 118r3

2).P h2

b

h3a

3�

4(729r6+h6b)

≥ 118r3

Mihaly Bencze

PP28586. In all triangle ABC holds (4R+r)√3

s + k

q3k−

n2−1P �

ctgA2

�n ≥ 6

for all n, k ∈ N, n ≥ 2, k ≥ 6.

Mihaly Bencze

PP28587. If ak > 0 (k = 1, 2, ..., n) then

P a1a2+a3+...+an

+(n−1)n−1

n�

k=1ak

�

(a2+a3+...+an)≥ n+1

n−1 .

Mihaly Bencze

PP28588. If a, b > 0 then 4a2+ab+b2

2a + 4b2+ab+a2

2b + 4aba+b ≥ 4 (a+ b) .

Mihaly Bencze


PP28589. If ak > 0 (k = 1, 2, ..., n) then

P a1a2+a3+...+an

+

n�

k=1

ak�

(a2+a3+...+an)≥ n(n−1)n−1+1

(n−1)n.

Mihaly Bencze


b+c +4abc

�

(a+b) ≥ 2.

Mihaly Bencze

PP28591. If a, b, c > 0, then1).

P 1(a−b+c)2

+ 2P 1

ab ≥ 75(�

a)2

2).P 1

(a−b+c)2+ 1

2

P 1ab ≥ 27

(�

a)2

Mihaly Bencze

PP28592. If a, b, c > 0 then3P

a4 + 9P

a2b2 + 6abcP

a ≥ 6 (P

ab)�P

a2�+ (P |(a− b) (b− c)|)2 .

Mihaly Bencze

PP28593. If ak > 0 (k = 1, 2, ..., n) thenP a21a

22(a21+a22)

a1a2+�

2(a41+a42)+ 1

3

nPk=1

a4k ≥ 2P

a21a22.

Mihaly Bencze

PP28594. If a, b > 0 then a�

2√a2+ab+b2

+ 1a√3

�2+ b

�2√

a2+ab+b2+ 1

b√3

�2≥

≥ 1

(a+ b)2 (a+ 2b) (2a+ b)

�5a2 + 11ab+ 2b2

�2

2a+ b+

�5b2 + 11ab+ 2a2

�2

a+ 2b

!.

Mihaly Bencze

PP28595. If a, b, c > 0 thenP

a3 ≥ 3abc+P

(a− b)2 c.

Mihaly Bencze

PP28596. If 1 < a, b ≤ 103 then

8�

a(a−1)2

+ b(b−1)2

�≥ (10− 2a− b)2 + (10− a− 2b)2 .

Mihaly Bencze


PP28597. If a, b, c > 0 then 1 +�

a2

2�

ab ≤P ab+c ≤ 1

2 +�

a2�

ab (A new

refinement of the inequalityP

a2 ≥P ab).

Mihaly Bencze

PP28598. If a, b, c > 0 and λ > 0 thenP a(b+c)λ+1

(b2+bc+c2)λ≥ 2λ+1

�

ab

(�

a)λ.

Mihaly Bencze

PP28599. If a, b > 0 then�a4 + a3b+ b4

� �a4 + ab3 + b4

�≥

≥�2a2 + b2 + 6 (a− b)2

��a2 + 2b2 + 6 (a− b)2

�a2b2.

Mihaly Bencze

PP28600. If a, b > 0 then(4a2+ab+b2)(a2+ab+4b2)

4ab(a+b)2≤

�12 + 2a2+b2

a(a+2b)

��12 + a2+2b2

b(b+2a)

�.

Mihaly Bencze

PP28601. If a, b > 0 then4a2b2(1+a2)(1+b2)

(1+a)(1+b) +a4(1+a2)

2

(1+a)2+

b4(1+b2)2

(1+b)2≥ 8

�3− 2

√2�ab

�a2 + ab+ b2

�.

Mihaly Bencze

PP28602. If ak > 0 (k = 1, 2, ..., n) , thenP

1≤i<j≤n

a4i+a4jai+aj

≥ (n− 1)nP

k=1

a2k + 3P

1≤i<j≤n(ai − aj)

2 .

Mihaly Bencze

PP28603. If a, b > 0 and λ > 0 then�a2λ+2 + b2λ+2 + aλ+2bλ

� �a2λ+2 + b2λ+2 + aλbλ+2

�≥ 9(2a3+b3)(a3+2b3)a2λb2λ

(2a+b)(a+2b) .

Mihaly Bencze

PP28604. If a, b > 0 then�2p

(a2 + 2) (2a+ 1) +p(b2 + 2) (a+ b+ 1)

�2+

+�2p(b2 + 2) (2b+ 1) +

p(a2 + 2) (a+ b+ 1)

�2≥ 18 (a+ b+ 1)2 .

Mihaly Bencze


PP28605. In all tetrahedron ABCD holds1).

P 3ha+2λrha+λr ≥ 8(λ+6)

λ+4

2).P 3ra+2λr

ra+λr ≥ 8(λ+3)λ+2 for all λ > 0

Mihaly Bencze

PP28606. If x, y, z > 0 thenP x

y+z +Pq

x+y+2z2(x+y) ≥ 9

2 .

Mihaly Bencze

PP28607. If ak > 1 (k = 1, 2, ..., n) and 0 < x ≤ y thennP

k=1

�ayk − axk

�≥ (y − x)

�nP

k=1

ax+y2

k ln ak

�.

Mihaly Bencze

PP28608. Let be P (x) = a0 + a1x+ ...+ amxm, λi > 0 (i = 1, 2, ..., n) ,nQ

i=1λi = 1 then

nPi=1

P k (λi) ≥ nP k (1) , for all k ∈ N .

Mihaly Bencze

PP28609. If ak > 0 (k = 1, 2, ..., n) then

2n(n−1)

�1n

nPk=1

ak

�n2

≥�

nQk=1

ak

� Q1≤i<j≤n

(ai + aj)2

!.

Mihaly Bencze

PP28610. If a, b, c > 1 and 0 < x ≤ y then

ay + by + cy − ax − bx − cx ≥ (y − x)�a

x+y2 ln a+ b

x+y2 ln b+ c

x+y2 ln c

�.

Mihaly Bencze

PP28611. In all triangle ABC holds 4P

ctg2A2 + 45s4

81r4+ 2r(4R+r)

s2≥ 19.

Mihaly Bencze

PP28612. If a, b, c, d > 0 then(P

a)16 ≥ 268

516abcd (a+ b)2 (a+ c)2 (a+ d)2 (b+ c)2 (b+ d)2 (c+ d)2 .

Mihaly Bencze


PP28613. If xk > 0 (k = 1, 2, ..., n) andnP

k=1

xk = n thenP x3

1

x2

√x31+8

≥ n3 .

Mihaly Bencze

PP28614. If xk > 0 (k = 1, 2, ..., n) andnP

k=1

xk = n then

P xλ+11

x2

�

xλ+12 +λ2+2λ

≥ nλ+1 .

Mihaly Bencze

PP28615. If a, b > 0 then6(a2+ab+b2)

a+b +3(a3+b3)

2ab ≥ 514 (a+ b) +

32(a−b)2(a2+ab+b2)ab(a+b)2(2a+b)(a+2b)

.

Mihaly Bencze


1). 4P 1

r4a+ 45

81r4≥ 19((4R+r)2−2s2)

s4r2

2). 4P 1

h4a+ 45

81r4≥ 19

��s2+r2+4Rr

4s2r2

�2− R

s2r3

�

Mihaly Bencze

PP28617. If 0 < a < b < c then (c− a) (b− a) arctg c−b1+cb ≤

(c− b) (b− a) arctg c−a1+ca ≤ (c− b) (c− a) arctg b−a

1+ab .

Mihaly Bencze

PP28618. If a, b, c > 0 then 3P 2a2+bc

b+c ≥ 92

Pa+ 16abc

�

(a−b)2

(�

a)�

(a+b) .

Mihaly Bencze

PP28619. If 0 < a1 ≤ a1 ≤ a2 ≤ ... ≤ an thenQ

1≤i<j≤n

eaj−aaiaj−ai

≥ en−12

n�

k=1ak.

Mihaly Bencze

PP28620. If e ≤ a ≤ b ≤ c then

ln�cb

�c−aln (cb)b−a ≤ ln

�ca

�c−bln (ac)b−a ≤ ln

�ba

�c−bln (ab)c−a . If

0 < a ≤ b ≤ c ≤ e then holds the reverse inequalities.

Mihaly Bencze


PP28621. If 0 < a ≤ b ≤ c then(b− a) (c− a) (c− b) ea+b+c ≤

�eb − ea

�(ec − ea)

�ec − eb

�.

Mihaly Bencze

PP28622. If 0 < a ≤ b < 1 thenb�ln2 b− 2 ln b+ b

�− a

�ln2 a− 2 ln a+ a

�≥

≥ b2 − a2

2

�ln2

�a+ b

2

�− 2 ln

�a+ b

2

�+

a+ b

2

�.

Mihaly Bencze

PP28623. If 1 < a < 1 + 1n then

nQk=1

�ak

k − ak+1

k+1

�< 1

(n+1)(n!)2.

Mihaly Bencze

PP28624. If a > 1 thennQ

k=1

�ak − 1

k+1

�< a

n(n+3)2

(n+1)(n!)2.

Mihaly Bencze

PP28625. Denote Lk the kth Lucas number. Prove that

1).�1 + 1

m

�(Ln+2 − 3) <

nPk=1

Lm+1k+1 −Lm+1

k

Lmk+1−Lm

k<

�1 + 1

m

�(Ln+3 − 4)

2).�1 + 2

m

�(LnLn+1 − 2) <

nPk=1

Lm+2k+1 −Lm+2

k

Lmk+1−Lm

k<

�1 + 2

m

�(Ln+1Ln+2 − 3) for

all m ∈ N∗

Mihaly Bencze

PP28626. Denote Fk the kth Fibonacci number. Prove that

1).�1 + 1

m

�(Fn+2 − 1) <

nPk=1

Fm+1k+1 −Fm+1

k

Fmk+1−Fm

k<

�1 + 1

m

�(Fn+3 − 2)

2).�1 + 2

m

�FnFn+1 <

nPk=1

Fm+2k+1 −Fm+2

k

Fmk+1−Fm

k<

�1 + 2

m

�(Fn+1Fn+2 − 1) for all

m ∈ N∗

Mihaly Bencze

PP28627. In all triangle ABC holds 6qtgA

2 −qtgB

2 +qtgC

2 < 3p

sr .

Mihaly Bencze


PP28628. If ai > 0 (i = 1, 2, ..., n) thennP

i=1ai ≥ k+n−1

snQ

i=1ai

�nP

i=1

k+n−1

qak−1i

�when k ∈ N∗.

Mihaly Bencze

PP28629. If ak ≥ 2 (k = 1, 2, ..., n) and A = 1n

nPk=1

ak, G = n

snQ

k=1

ak,

H = nn�

k=1

1ak

, then A3G3H5 +A3G3 ≤ A3G5H3 +A3H3 ≤ A5G3H3 +G3H3.

Mihaly Bencze


√ra −

√rb +

√rc <

3s√r

2). 6√ha −

√hb +

√hc < 3s

q2R

Mihaly Bencze

PP28631. Compute limn→∞

n

�1−

n+1Rn

e1xdx

�.

Mihaly Bencze

PP28632. If a, b, c > 0 then

3P �

a2 − bc�4

+ 3P

a2 (b+ c)4 ≥��P

a2�2

+P

a2b2�.

Mihaly Bencze

PP28633. Solve in R+ the following system:

4πarctgx1 +�arcsin 2x2

1+x22

�2= π2 + 4arctg2x3

4πarctgx2 +�arcsin 2x3

1+x23

�2= π2 + 4arctg2x4

−−−−−−−−−−−−−−−−−−−−4πarctgxn +

�arcsin 2x1

1+x21

�2= π2 + 4arctg2x2

.

Mihaly Bencze

PP28634. Solve in Z the equation x2+(y+z)2

(x+y)2+z2+ y2+(z+x)2

(y+z)2+x2+ z2+(x+y)2

(z+x)2+y2= 3.

Mihaly Bencze


PP28635. Solve in R the following system: arcsin 2x1

1+x21+ 2arctgx2 =

arcsin 2x2

1+x22+ 2arctgx3 = ... = arcsin 2xn

1+x2n+ 2arctgx1 = π.

Mihaly Bencze

PP28636. Prove that

limx→0

1x

�3

nPk=1

qk2 (k + 1)2 + x− n (n+ 1) (n+ 2)

�= 3n

2(n+1) .

Mihaly Bencze

PP28637. Prove that1R0

x2 arcsin 2xx2+1

dx = π6 + 1

2 − ln 2. Compute

In =1R0

xn arcsin 2xx2+1

dx.

Mihaly Bencze

PP28638. Solve in Z the equation xy2z3

x+2y+3z + yz2x3

y+2z+3x + zx2y3

z+2x+3y = 12 .

Mihaly Bencze

PP28639. If a, b, c, p, q > 0 thenpa+qbp+qR0

ex ln�x2 + x+ c

�dx+

pb+qap+qR0

ex ln�x2 + x+ c

�dx ≤

≤aZ

0

ex ln�x2 + x+ c

�dx+

bZ

0

ex ln�x2 + x+ c

�dx.

Mihaly Bencze

PP28640. In all triangle ABC holds 1 ≤� 1

a2� 1

ab

≤ R2r (A new refinement of

Euler’s R ≥ 2r inequality).

Mihaly Bencze

PP28641. If x ∈�0, π2

�then compute I1 =

Rtg (arccos (sin (arctgx))) dx,

I2 =Rtg (arcsin (cos(arcctgx))) dx.

Mihaly Bencze


PP28642. Prove that limn→∞

1n

nPk=0

n√akbn−k = a−b

ln a−ln b when a > b > 0.

Mihaly Bencze

PP28643. The triangle ABC is equilateral if and only if

2 (cosA)3 |cosB|+ 2 (cosB)3 |cosA| = (1 + cos 2C)�cos2A+ cos2B

�

2 (cosB)3 |cosC|+ 2 (cosC)3 |cosB| = (1 + cos 2A)�cos2B + cos2C

�

2 (cosC)3 |cosA|+ 2 (cosA)3 |cosC| = (1 + cos 2B)�cos2C + cos2A

� .

Mihaly Bencze

PP28644. If x, y, z ∈ R∗+, show that

x2 + 1

2y + 3z+

y2 + 1

2z + 3x+

z2 + 1

2x+ 3y≥ 6

5and

find the values for which the equality is achieved.

Mihaela Berindeanu

PP28645. For a nonnegative integer n let Fn be the nth Fibonacci numberdefined by F0 = 0, F1 = 1 and for all n ≥ 2, Fn = Fn−1 + Fn−2. Show that

1

54F2n

��

4Fn+1 Fn Fn + 2Fn−1

FnFn+1 (2Fn+1 + Fn−1)2 F2n

F 2n+1 + Fn−1Fn+1 F2n F 2

n+2

��

Jose Luis Dıaz-Barrero

PP28646. Let a, b be positive integers. A frog staying at point (a, b) of theplane may jump to point (a− b, b) if a > b or to point (a, b− a) if a < b. Forinstance a valid path for a frog starting at point (17, 12) is the following:(17, 12) → (5, 12) → (5, 7) → (5, 2) → (3, 2) → (1, 2) → (1, 1) Currently, thefrog is at point P (20172017, 123456). It is possible after some jumps to arriveat point (1, 1)?


PP28647. Let a, b, c be three positive numbers such that

ab+ bc+ ca = 3abc. Prove thatq

a+bc(a2+b2)

+q

b+ca(b2+c2)

+q

c+ab(c2+a2)

≤ 3.



PP28648. In all acute triangle ABC holdsQ

(cosA+ cosB cosC)cosA ≥�R+r2R

�R+rR .

Mihaly Bencze


(sinB sinC)cosA ≥�R+r2R

�R+rR .

Mihaly Bencze

PP28650. Prove that 2nP

k=1

(√k2+2k+2−

√k2+1)

2

2k+1 < ln n+1+√n2+2n+2

1+√2

.

Mihaly Bencze

PP28651. ProvenP

k=1

�2k+32k+5

�2k+5(k+3)k+3

(k+1)k+2 ≥ n(n+5)2 .

Mihaly Bencze

PP28652. Denote Lk the kth Lucas number, prove that

nPk=1

�4√Lk +

4pLk+1 +

4pLk+3

�4pLkLk+1Lk+3 ≤ Ln+6 − 13.

Mihaly Bencze

PP28653. In all triangle ABC holds√2sr ≤ min {ab, bc, ca} .

Mihaly Bencze


P a(ha−2r)ha+2r = 2sr(12R+r)

4s2+r2+8Rr

2).P (b+c−a)(ha−2r)

ha+2r =2s(4s2−3r2−40Rr)

4s2+r2+8Rr

Mihaly Bencze

PP28655. Denote Fk the kth Fibonacci number, prove thatnP

k=1

�4√Fk +

4pFk+1 +

4pFk+3

�4pFkFk+1LF ≤ Fn+6 − 8.

Mihaly Bencze


PP28656. If x ≥ 0 then

�1 +

nPk=1

xk

k!

�n+1

≤�1 +

n+1Pk=1

xk

k!

�n

.

Mihaly Bencze

PP28657. If a ∈ (−1, 1) then limn→∞

nQk=0

�1 + a4

k+ a2·4

k+ a3·4

k�= 1

1−a .

Mihaly Bencze

PP28658. Prove that:

1). limn→∞

�1 +

nPk=1

1k!

� 1n

= 1

2). limn→∞

n

�1 +

nPk=1

1k!

� 1n

− 1

!= 1

Mihaly Bencze

PP28659. Prove that

�1 +

nPk=1

1k!

�n+1

<

�1 +

n+1Pk=1

1k!

�n

for all n ∈ N∗.

Mihaly Bencze

PP28660. If a1 = 1 and ak =

r2

q3p4...

√k for k ≥ 2 then compute

limn→∞

1n2

√n

nPk=1

a2k.

Mihaly Bencze

PP28661. Prove that limn→∞

nPk=1

arctg 2525k2+55k+49

= π2 − arctg 8

5 .

Mihaly Bencze

PP28662. Prove that 1!2!...n! < 22n+1−2 for all n ≥ 1.

Mihaly Bencze

PP28663. If a1 = 1 and ak =

s

2

r3

q4p

5...√k for k ≥ 2 then

1).nP

k=1

�akak+1

k(k+1)

�2< 4n

n+1

2).nP

k=1

a2k ≤ n (n+ 1)


3).nP

k=1

1a2ka2k+1

> n4(n+1)

Mihaly Bencze

PP28664. If n ≥ 2, n ∈ N then solve in R the following system:

nnx1 + n− 1 = nx2+1

nnx2 + n− 1 = nx3+1

−−−−−−−−−nnxn + n− 1 = nx1+1

.

Mihaly Bencze


k=1

(k!)−2k < 2n.

Mihaly Bencze

PP28666. If a, b > 1 then 8a�ln2 a− ln a2

�+ 8b

�ln2 b− ln b2

�≥

(3a+ 5b)�ln2

�3a+5b

8

�− ln

�3a+5b

8

�2�+ (5a+ 3b)

�ln2

�5a+3b

8

�− ln

�5a+3b

8

�2�.

Mihaly Bencze


27x+y + 2 = 3y+z+1

27y+z + 2 = 3z+x+1

27z+x + 2 = 3x+y+1.

Mihaly Bencze

PP28668. If x ∈�0, π2

�then compute

Rsinx cosx−x−x2 sinx

x(x+sinx) cosx dx.

Mihaly Bencze

PP28669. If 1 < a ≤ b then

2π ln b

a ln ab+ 2 ln�be

�b � ea

�a<

bRa

lnxdxarctgx < π

2 lnba ln ab+ 2 ln

�be

�b � ea

�a.

Mihaly Bencze


27r3

s3

Pctg4A2 +

�PqctgA

2

�2

≥ 36(4R+r)(s2−2r(4R+r))s3

.

Mihaly Bencze



1).P�

rbra

�2≥ 81r2((4R+r)2−2s2)

s4

2).P�

hb

ha

�2≥ 81r4

��s2+r2+4Rr

4s2r2

�2− R

s2r3

�

Mihaly Bencze


1). 27r3P 1

r4a+�P 1√

ra

�2≥ 36(4R+r)(s2−2r(4R+r))

s4

2). 27r3P 1

h4a+�P 1√

ha

�2≥ 9(s2−r2−4Rr)(s2+r2+4Rr)

2s2r

Mihaly Bencze


1).Pq

1 + rbra

≥q

23 · s

r

2).Pq

1 + hb

ha≥ 2s

r

q2r3R

Mihaly Bencze


ctgA2

qctgA

2 + ctgB2 ≥

�sr

� 32 .

Mihaly Bencze

PP28675. If a, b > 0 then a2+2b2

b(2a+b) +b2+2a2

a(2b+a) +8ab

(a+b)2≥ 4.

Mihaly Bencze

PP28676. If a, b > 0 then2

a2(2a+b)2+ 2

b2(2b+a)2+ 1

(a2+2b2)2+ 1

(b2+2a2)2≥ 1

9

�(2a+b)2

2a6+b6+ (a+2b)2

a6+2b6

�.

Mihaly Bencze

PP28677. In all triangle ABC holdsrs2

P (rctgA2+stgA

2 )ctg2 A

2

tgB2+tgC

2

+P ctgA

2

tgB2+2tgC

2

≥ 4 + rs

P ctg2 A2 (tg

B2−tgA

2 )(tgC2−tgA

2 )tgB

2+tgC

2

.

Mihaly Bencze

PP28678. If a, b, c > 0 and λ > 0 thenP a

(b2+bc+c2)λ≥

�

a

(�

ab)λ.

Mihaly Bencze



1).P�

r(r2a+rbrc)r2a(rb+rc)

+ rbrcr(ra+2rb)

�≥ 4

P (rb−ra)(rc−ra)r2a

2).P�

r(h2a+hbhc)

h2a(hb+hc)

+ hbhc

r(ha+2hb)

�≥ 4

P (hb−ha)(hc−ha)h2a

Mihaly Bencze

PP28680. If ak > 0 (k = 1, 2, ..., n) , thenP �

a21 − a1a2 + a22�r

10a1a2 +3

q4�a61 + a62

�≤ 2

√3

nPk=1

a3k.

Mihaly Bencze


b2+bc+c2≥

�

a�

ab .

Mihaly Bencze

PP28682. Solve in R the equation [xn!] + [x (n+ 1)!] = (n+ 2)! where [·]denote the integer part, and n ∈ N.

Mihaly Bencze

PP28683. Solve in R the equationh

xn+2

i+n

xn+1

o= 1

n where [·] and {·}denote the integer , respective the fractional part and n ∈ N∗.

Mihaly Bencze

PP28684. Solve in R the equation [2017x] + [2018x] = 2019 where [·] denotethe integer part.

Mihaly Bencze

PP28685. The triangle ABC is equilateral if and only if32 +

P�ma

wa

�2+P a+b

c =P (b+c)ma√

bcwa.

Mihaly Bencze

PP28686. Determine all a, b ∈ N and all prime p for which(p+ 2018)a = (p− 2018)b .

Mihaly Bencze


PP28687. Determine all prime p, q, r for which

p3q3r3 + 15p3 + 5q3 + 3r3 = 5p3q3 + q3r3 + 3r3p3 + 120.

Mihaly Bencze


ctg6A2 + 6�sr

�2 ≤ sr

Pctg5A2 .

Mihaly Bencze

PP28689. Let be A =

�a ba−a2−1

b 1− a

�where a, b ∈ C, b 6= 0. Prove that

1). A2n is invertable and�A2n

�−1= A2n−1

for all n ∈ N,n ≥ 2

2). det

�6n+6Pk=0

(−1)k Ak+1

�=

6n+6Pk=0

(−1)k det�Ak+1

�

Mihaly Bencze

PP28690. Solve in Z the equation y8 − x (x+ 1) (x+ 2) (x+ 3) = 136.

Mihaly Bencze

PP28691. In all triangle ABC holds1). r

P 1r6a

+ 6s2r

≤P 1r5a

2). rP 1

h6a+ 3R2

2s4r≤P 1

h5a

Mihaly Bencze

PP28692. If ak > 1 (k = 1, 2, ..., n) , thenP

loga1aλ+12 +aλ+1

3

a22+a23≥ n (λ− 1) for

all λ ≥ 1.

Mihaly Bencze

PP28693. Let ABC be a triangle, denote AD,BE,CF the interiorbisectors, where D ∈ (BC) , E ∈ (CA) , F ∈ (AB) . If one of Area [ABD] ,Area [BCE] , Area [CAF ] is the arithmetical mean of the other two, thenthe triangle ABC is isoscel.

Mihaly Bencze


PP28694. If a, b, c > 0 and a+ b+ c = 3 then(n+1)(

�

a2)2

�

a2b+ 2n

�

ab�

a2≥ n (3n+ 5) for all n ∈ N∗.

Mihaly Bencze

PP28695. If a, b, c > 0 thenP a2c(b+c)√

(a2+b2)(a2+c2)≤P ab.

Mihaly Bencze

PP28696. Solve in Z the equation 1x + 1

x+y + 1x+y+z = 1.

Mihaly Bencze

PP28697. Let (an)n≥1 be an arithmetical progression. Prove that:

1). 1(a1+a2)(a2+a3)

+ 1(a2+a3)(a3+a4)

+ ...+ 1(an+an+1)(an+1+an+2)

=

= n(a1+a2)(an+1+an+2)

2). 1(a1+a2)(a2+a3)(a3+a4)

+ 1(a2+a3)(a3+a4)(a4+a5)

+ ...

+ 1(an+an+1)(an+1+an+2)(an+2+an+3)

= n(a1+a2)(a2+a3)(an+1+an+2)(an+2+an+3)

Mihaly Bencze


2+tg2A≥ 2(R+r)2

12R2+4Rr+r2−s2.

Mihaly Bencze


2+ctg2A≥ 2s2

6R2−4Rr−r2+s2.

Mihaly Bencze


1).Q �

r2a + 9r2�≤ 8s6

27

2).Q �

h2a + 9r2�≤ 8s6

27

Mihaly Bencze

PP28701. In all triangle ABC holdsQ �

1 + 9tg2A2 tg2B2

�≤ 8s2

27r2.

Mihaly Bencze



1).�4R+r4R

�2 ≥ 3�

s4R

�2+ 3

16

�P ��sin2 A2 − sin2 B

2

��2

2).�s2+r2−8Rr

16R2

�2≥ 3r2(2R−r)

32R3 + 316

�P ��sin2 A2 − sin2 C

2

�� sin2 B2

�2

3).�4R+r4R

�2 ≥ 3�

s4R

�2+ 3

16

�P ��cos2 A2 − cos2 B

2

��2

4).�s2+(4R+r)2

16R2

�2≥ 3s2(4R+r)

32R3 + 316

�P ��cos2 A2 − cos2 C

2

�� cos2 B2

�2

Mihaly Bencze

PP28703. In all triangle ABC holds1). s2 ≥ 3r (4R+ r) + 3

16 (P |a− b|)2

2).�s2 + r2 + 4Rr

�2 ≥ 20s2Rr + 316 (P |a− c| b)2

3).�s2+r2+4Rr

2R

�2≥ 6s2r

R + 316 (P |ha − hb|)2

4). (4R+ r)2 ≥ 3s2 + 316 (P |ra − rb|)2

5). s4 ≥ 3s2r (4R+ r) + 316 (P |ra − rc| rb)2 (New refinements of some

classical inequalities).

Mihaly Bencze

PP28704. If λ ∈ (0, 1] thennP

k=1

k12λ ≤ n1−λ (n+ 1)

3λ2 .

Mihaly Bencze

PP28705. If ak > 0 (k = 1, 2, ..., n) and λ ∈ [0, 1] thennP

k=1

�ak+1

a2k+2ak+3

�λ≤ n · 2−3λ

2 .

Mihaly Bencze

PP28706. If 0 < a < b and ab = 4 then computeln bRln a

xe3xdx(ex+2)6

.

Mihaly Bencze


k=1

1k ≥ 4n

(√n+1)

2 .

Mihaly Bencze



k=1

(n−k)(n−k+2)

(V kn )

2 = 1− 1(n!)2

.

Mihaly Bencze


k=1

(n−k)((n−k)2+3(n−k)+3)(V k

n )3 = 1− 1

(n!)3.

Mihaly Bencze

PP28710. ComputeR (x+1)2(x+5)dx

x6+6x5+15x4+20x3+24x2+19x+2where x > 0.

Mihaly Bencze

PP28711. In all tetrahedron ABCD holds

1).P�

rha

�har

< 7

2).P�

r2ra

� 2rar

< 7

Mihaly Bencze


k=1

n−kV kn

= 1− 1n! .

Mihaly Bencze


k=1

�n+1

kn(k+1)

� kn(k+1)n+1

< 2n+ 1.

Mihaly Bencze


1).P�

rra

� rar< 5

2).P�

rha

�har

< 5

Mihaly Bencze

PP28715. Determine all triangles ABC in which (x+ y) a = xb+ yc for allx, y > 0.

Mihaly Bencze


PP28716. If a, b, c, d ∈ N ∗ then�a+ b1+

1b

��c+ d1+

1d

�≤

�a+ ba

1a

��c+ dc

1c

�.

Mihaly Bencze

PP28717. Let be A ∈ M2 (R) where A =

�a bc d

�. Determine all λ ∈ R

and m > 0 such that aλ + bλ + cλ + dλ < m and I2 +A is invertible.

Mihaly Bencze

PP28718. In all triangle ABC holdsP a+b

c5≥

��s2+r2+4Rr

4sRr

�2− 1

Rr

�2

.

Mihaly Bencze


1).P ra+rb

r5c≥

�s2−2r(4R+r)

s2r2

�2

2).P rarb+rbrc

r5cr5a

≥�(4R+r)2−2s2

s4r2

�2

3).P a

(s−a)5≥

�(4r+r)2−2s2

s2r2

�2

4).P c(s−b)

(s−a)5(s−c)5≥

�s2−2r2−8Rr

s2r4

�2

5).P ha+hb

h5c

≥�s2−r2−4Rr

2s2r2

�2

6).P hahb+hbhc

h5ch

5a

≥��

s2+r2+4Rr4s2r2

�2− R

s2r3

�2

Mihaly Bencze

PP28720. Let ABC be a triangle in which A ≤ B ≤ C ≤ π2 . Prove that�

s2+r2+Rr2sr

�2− 4R

r +P A cosA

sin2 A≤ 9(A+C)2

4πAC .

Mihaly Bencze

PP28721. If ai > 0 (i = 1, 2, ..., n) and k ∈ {2, 3, ..., n} , thenQ

cyclic

�a1ak

+ a2ak

+ ...+ak−1

ak+ 1

�ak ≤ k

n�

i=1ai.

Mihaly Bencze


PP28722. If x, y, z ≥ 1 then

P 1x + 2 lnxyz + 9

x+y+z + 6 ln x+y+z3 ≤ 4

P 1x+y + 4

Pln x+y

2 .

If x, y, z ∈ (0, 1) then holds the reverse inequality.

Mihaly Bencze

PP28723. In all acute triangle ABC holdsR+rR + 2 ln 4R2

s2−(2R+r)2+

9(s2−(2R+r)2)s2+r2−4R2 + 6 ln s2−(2R+r)2

3(s2+r2−4R2)≥

≥�s2 + r2 − 4R2

�2+ 4R (R+ r)

�s2 − (2R+ r)2

�

Rr (s2 + r2 + 2Rr)+4 ln

�s2 − (2R+ r)2

�2

32Rr (s2 + r2 + 2Rr).

Mihaly Bencze

PP28724. In all acute triangle ABC holdss2+r2−4R2

s2−(2R+r)2+ 2 ln s2−(2R+r)2

4R2 + 9RR+r + 6 ln R+r

3R ≤

≤ 4R�s2 + 5r2 + 8Rr

�

r (s2 + r2 + 2Rr)+ 4 ln

r�s2 + r2 + 2Rr

�

32R3.

Mihaly Bencze

PP28725. In all triangle ABC holdss2+(4R+r)2

s2+ 4 ln s

4R + 18R4R+r + 6 ln 4R+r

6R ≤

≤8R

�5 (4R+ r)2 + s2

�

(4R+ r)3 + s2 (2R+ r)+ 4 ln

(4R+ r)3 + s2 (2R+ r)

256R3.

Mihaly Bencze

PP28726. In all triangle ABC holds4R+r2R + 4 ln 4R

s + 9s2

s2+(4R+r)2+ 6 ln s2+(4R+r)2

3s2≥

≥ 4X cos2 A

2 cos2 B2

cos2 A2 + cos2 B

2

+ 4 lns4

64R�(4R+ r)3 + s2 (2R+ r)

� .

Mihaly Bencze


PP28727. In all triangle ABC holdss2+r2−8Rr

r2+ 4 ln r

4R + 18R2R−r + 6 ln 2R−r

6R ≤

≤ 8R�16R2 − 24Rr + 5r2 + s2

�

(2R− r) (s2 + r2 − 8Rr)− 2Rr2+ 4 ln

(2R− r)�s2 + r2 − 8Rr

�− 2Rr2

256R3.

Mihaly Bencze

PP28728. In all triangle ABC holds2R−r2R + 4 ln 4R

r + 9r2

s2+r2−8Rr+ 6 ln s2+r2−8Rr

3r2≥

≥ 4X sin2 A

2 sin2 B2

sin2 A2 + sin2 B

2

+ 4 lnr4

64 ((2R− r) (s2 + r − 8Rr)− 2Rr2).

Mihaly Bencze

PP28729. In all triangle ABC holdssR + 2 ln 2R2

sr + 18srs2+r2+4Rr

+ 6 ln s2+r2+4Rr6sr ≥

≥�s2 + r2 + 4Rr

�2 − 8sRr

sR (s2 + r2 + 2Rr)+ 4 ln

r2

8R (s2 + r2 + 2Rr).

Mihaly Bencze

PP28730. In all triangle ABC holdss2+r2+4Rr

2sr + 2 ln sr2R2 + 9R

s + 6 ln s3R ≤ 4R(5s2+r2+4Rr)

s(s2+r2+2Rr)+ 4 ln

s(s2+r2+2Rr)32R3 .

Mihaly Bencze


[nx] cos (a [nx]) dx = n cos(n−1)a−(n−1) cosna−1

4n sin2 a2

, where

n ∈ N∗.

Mihaly Bencze and Ovidiu Furdui


arctg 1([nx])2+3[nx]+3

dx = 1narctg

n−1n+1 where [·] denote

the integer part, and n ∈ N∗.



PP28733. Prove that 12 +

1R0

cos (2a [nx]) dx = sin(2n−1)a2n sin a , where [·] denote

the integer part, and n ∈ N∗.



[nx] sin (a [nx]) dx = n sin(n−1)a−(n−1) sinna

4n sin2 a2

, where [·]denote the integer part, and n ∈ N∗.


PP28735. If a > 1 then1R0

a[nx]dx = an−1n(a−1) , when [·] denote the integer part,

and n ∈ N∗.



sin (2a [nx]) dx = sinna sin(n−1)an sin a , when [·] denote the

integer part, and n ∈ N∗.



[nx]dx

4([nx])4+1= n−1

2(2n2−2n+1), where n ∈ N∗.



ln�[nx]+2[nx]+1

�dx = 1

n ln (n+ 1) , when [·] denote the

integer part, and n ∈ N∗.



[nx] ([nx])!dx = n!−1n , when [·] denote the integer

part, and n ∈ N∗.




([nx])!�([nx])2 + [nx] + 1

�dx = n·n!−1

n , when [·]denote the integer part, and n ∈ N∗.



dx([nx]+1)([nx]+2) =

1n+1 , when [·] denote the integer




[nx]dx([nx]+1)! =

�1− 1

n!

�1n , when [·] denote the integer



PP28743. If 0 ≤ x1 ≤ x2 ≤ ... ≤ xn ≤ 1 then(x2 − x1) arctgx1 + (x3 − x1) arctgx2 + (x4 − x2) arctgx3 + ...+(xn − xn−2) arctgxn−1 + (xn − xn−1) arctgxn ≤ π

2 − ln 2.

Mihaly Bencze and Sitaru Daniel

PP28744. In all tetrahedron ABCD holds1). 1

r2a+ 1

r2b

+ 1r2c

+ 1r2d

+ 8r2

rarbrcrd≥ 1

rarb+ 1

rarc+ 1

rard+ 1

rbrc+ 1

rbrd+ 1

rcrd

2). 1h2a+ 1

h2b

+ 1h2c+ 1

h2d

+ 32r2

hahbhchd≥ 1

hahb+ 1

hahc+ 1

hahd+ 1

hbhc+ 1

hbhd+ 1

hchd

Mihaly Bencze


A (π − 2A) ≤ srπ2

2R2 .

Mihaly Bencze

PP28746. If a, b, c, x, y, z > 0 then�a2 + b2 + c2

� �1x2 + 1

y2+ 1

z2

�+ 2(ab+bc+ca)(x+y+z)

xyz ≥

≥ 13 (a+ b+ c)2

�1x + 1

y + 1z

�2.

Mihaly Bencze


PP28747. If ak > 0 (k = 1, 2, ..., n) , then

P

cyclic

qa1a2a1+a2

!2

≤ n2

nPk=1

ak.

Mihaly Bencze

PP28748. In all acute triangle ABC holdsP sin2 A

A ≥ 4.

Mihaly Bencze


ra

qr2b + rb + 1 + rb

pr2a + ra + 1

�2

r2c ≥≥ 2s2

�s2 − r2 − 4Rr + 2s2r (1 + 3r)

�.

Mihaly Bencze

PP28750. If a, b, c, d, e ∈ [1, 2] then�a2 + b2 + c2 + d2 − 3abcd

� �b2 + c2 + d2 + e2 − 3bcde

�·

·�c2 + d2 + e2 + a2 − 3cdea

� �d2 + e2 + a2 + b2 − 3deab

�·

·�e2 + a2 + b2 + c2 − 3eabc

�≤ 1.

Mihaly Bencze

PP28751. If x, y, z ∈ (0, 1) then

4096�1 + x2

� �1 + y2

� �1 + z2

� �9 + (x+ y + z)2

�3≥

≥ 729�4 + (x+ y)2

�2 �4 + (y + z)2

��4 + (z + x)2

�.

Mihaly Bencze

PP28752. If a, b, c ∈ (0, 1) and λ ≥ 2, then�43

�3λ �1 + aλ

� �1 + bλ

� �1 + cλ

� �3λ + (a+ b+ c)λ

�≥

≥�2λ + (a+ b)λ

��2λ + (b+ c)λ

��2λ + (c+ a)λ

�.

Mihaly Bencze

PP28753. Solve in Z the equation x3 + y3 = z�x2 − y2

�.

Mihaly Bencze



(mamb)λ ≤ 3

�(5R+2r)(s2−r2−4Rr)

12R

�λ

for all λ ∈ [0, 1] .

Mihaly Bencze

PP28755. Denote fk (n) the number of positive integers n such that kdivides

�2nn

�

1). ComputenP

p=1

1fk(p)

2). Compute∞Pp=1

1f2k(n)

when, k ∈ N, k ≥ 2.

Mihaly Bencze

PP28756. If ak ∈ [0, 1] (k = 1, 2, ..., n) , thennP

k=1

ak +nQ

k=1

ak ≥ 4n−1

P1≤i<j≤n

aiaj .

Mihaly Bencze

PP28757. If a, b > 0 and a3b3 − ab (a+ b) + 1 = 0 then�1

2a4+b+ 2

a4+a+b4

��1

2b4+a+ 2

b4+b+a4

�≤ 9

(2a+b)(a+2b) .

Mihaly Bencze

PP28758. If a, b, c > 0 and a+ b+ c = 32 then a2 + b2 + c2 + 32

9 abc ≥ 34 .

Mihaly Bencze

PP28759. Solve in Z the equation�x2+y2

xy+1

�2+

�y2+z2

1+yz

�2+

�z2+x2

1+zx

�2+ 1

xy + 1yz + 1

zx = 6.

Mihaly Bencze

PP28760. In all convex quadrilateral ABCD holdsP sinA+sinB(1+sinA) sinB ≤P 1

sinA .

Mihaly Bencze


PP28761. If x ∈�0, π

2n

�, then

nPk=1

sin kxk2 sin(k+1)x

> nn+1 .

Mihaly Bencze

PP28762. If a, b > 0 and a3b3 − ab (a+ b) + 1 = 0 then

9�2√16a2 + 9 +

√16b2 + 9− 1

��2√16b2 + 9 +

√16a2 + 9− 1

�≤

≤ 196 (2a+ b) (a+ 2b) .

Mihaly Bencze

PP28763. If a, b, c ∈ R and a2 + b2 + c2 ≤ 1 then |a|+ |b|+ |c| ≤ 4√3

3 +3abc.

Mihaly Bencze

PP28764. Denote Fk and Lk the kth Fibonacci respective Lucas numbers.Prove that1). 2FkFk+1

Fk+3+ 2FkFk+3

Fk+1+ 2Fk+1Fk+3

Fk≥ 2Fk+4

2). 2LkLk+1

Lk+3+ 2LkLk+3

Lk+1+ 2Lk+1Lk+3

Lk≥ 2Lk+4

Mihaly Bencze

PP28765. If a, b ≥ 1 then�2ab+1

a + 2a2

b

��2ab+1

b + 2b2

a

�≥ 4 (2a+ b) (a+ 2b) .

Mihaly Bencze

PP28766. If a, b > 0 and 2�a2 + b2

�+ 5ab+ 9 = 9 (a+ b) then

ab (a+ b) + 27a(a+2b) +

27b(b+2a) ≥ 18.

Mihaly Bencze

PP28767. If a, b > 0 then (a+b)2

4ab + a+2√ab

2a+b + b+2√ab

2b+a ≥ 33√a2b

3√

2a(a+b)2+ 3

3√ab2

3√

2b(a+b)2.

Mihaly Bencze

PP28768. If a, b > 0 then�ab

�2+�ba

�2+ 3ab

2(a+b)2≥ 23

4 .

Mihaly Bencze

PP28769. If a, b > 0 then ab +

ba + a(a+2b)

2a2+b2+ b(b+2a)

2b2+a2≥ 4.

Mihaly Bencze


PP28770. If a, b > 0 then a2

b + b2

a + 6a(a+2b)2a+b + 6b(b+2a)

2b+a ≥ 7 (a+ b) .

Mihaly Bencze

PP28771. If a, b > 0 and 2�a4 + b4

�+ 5a2b2 + 9 = 9

�a2 + b2

�then

a3

b + b3

a + (a+ b)2 + 23a

2b (2a+ b) + 23ab

2 (2b+ a) ≥ 6a2b2a+b +

6ab2

2b+a + 6.

Mihaly Bencze

PP28772. If a, b > 0 then ab +

ba + a+2

√ab

2a+b + b+2√ab

2b+a ≥ 4.

Mihaly Bencze

PP28773. If 0 < a1 < a2 < ... < an < an+1 < an+2 and

xn =�a2+a3a1+a2

�a2+a3+�a3+a4a2+a3

�a3+a4+ ...+

�an+1+an+2

an+an+1

�an+1+an+2

then

compute limn→∞

xn

n .

Mihaly Bencze

PP28774. If a, b > 0 then a3�b√a+1√ab

�+ b3

�a√b+1√ab

�+

√a+

√b

a3b3≥ 2

�ab+ a+b

ab

�.

Mihaly Bencze

PP28775. Determine all a, b ∈ N for which

2 (a+ b) = 3 +�√

a�√

2a��

+h√

b�√

2b�i

where [·] denote the integer.

Mihaly Bencze

PP28776. If a1 > 0 and (an − 1) (a1 + a2 + ...+ an−1) = 1 for all n ≥ 2

then compute∞Pk=1

1a21+a22+...+a2

k

.

Mihaly Bencze

PP28777. Let f : (0,+∞) → (0,+∞) a decreasing function for which exist

k > 0 such that limx→∞

f(x+k)f(x) = 1. Prove that lim

n→∞1n

nPp=1

f(x+k+p)f(x) = 1.

Mihaly Bencze

PP28778. If 3xn + 5xn = n+ 1 for all n ≥ 1, then compute limn→∞

xnxn+1

(lnn)2.

Mihaly Bencze


PP28779. Let be (xn)n≥1 a sequence for which the sequence (yn)n≥1 isconvergent, where yn = axn+2 + bxn+1 + cxn for all n ≥ 1. Determine alla, b, c ∈ R for which (xn)n≥1 is convergent.

Mihaly Bencze

PP28780. Prove that1R

−1

�√cos πx

2+�

cos 3πx5

�

dx

1+e−x ≤ 2.

Mihaly Bencze


(x+ 1) esinπxdx ≥ 4e2π−1.

Mihaly Bencze

PP28782. Prove that 41R12

ln�1 + x2

�dx ≤ ln 2 +

1R−1

ln�1 + x2

�dx.

Mihaly Bencze

PP28783. Let be f : R → R where f (0) = 0 and f (x) =nQ

k=1

cos�

kekx−e−kx

�

for all x 6= 0. Prove that f is antiderivable.

Mihaly Bencze


(xy)6 + 2√3 = 9 (yz)2

(yz)6 + 2√3 = 9 (zx)2

(zx)6 + 2√3 = 9 (xy)2

.

Mihaly Bencze

PP28785. Determine the function f : C → C wheref (z) = az3 + bz2 + cz + d (a, b, c, d ∈ C) such that |f (z)| = f (|z|) for allz ∈ C.

Mihaly Bencze


PP28786. If xk > 1 (k = 1, 2, ..., n) andnP

k=1

xn−1k = n (n− 1)n−1 then

Pcyclic

logx1

�xn−12 + xn−1

3 + ...+ xn−1n

�≥ n2.

Mihaly Bencze


b ≥ 2(2s2−3r2−12Rr)s .

Mihaly Bencze

PP28788. In all triangle ABC holdsP r2a

rb≥ 5(4R+r)2−12s2

4R+r .

Mihaly Bencze


tgA2 tg

B2

rtgA

2

tgC2

+P

tgA2

qtgB

2 tgC2 + 4r

s

PqtgA

2 tgB2 ≥ 12r

s�

�

tgA2tgB

2

+ 2.

Mihaly Bencze

PP28790. In all triangle ABC holds 2s

P√ab ≥ 11− s2+r2

2Rr .

Mihaly Bencze

PP28791. In all triangle ABC holdsP a2+b2

ac ≥ 2(5s2−3r2−12Rr)s2+r2+4Rr

.

Mihaly Bencze

PP28792. Solve the following system

[x] y2z + x {y} z = 108, 18[y] z2x+ y {z}x = 143, 635[z]x2y + z {x} y = 76, 534

,

where [·] and {·} denote the integer, respective the fractional part.

Mihaly Bencze

PP28793. In all triangle ABC holdsP a+b

b+c ≥ 2 + 6 3√2Rr3√s2+r2+2Rr

− 1s

P√ab.

Mihaly Bencze

PP28794. In all triangle ABC holdsP ra+rb

rb+rc≥ 2 + 6 3

pr4R − 2

�√rarb

4R+r .

Mihaly Bencze


PP28795. In all triangle ABC holds4(s2−r2−Rr)s2+r2+2Rr

+ s2+r2+4Rr2(s2−r2−4Rr)

≥ 4.

Mihaly Bencze

PP28796. If ak ∈ Z (k = 1, 2, ..., n) such thatnP

k=1

ak = 0 and�

nPk=1

ak�x2k−1

�= 0 for all x ∈ R, then a1 = a2 = ... = an = 0, where [·] and

{·} denote the integer, respective fractional part.



ab

�2+P�

rarb

�2+ 15r

2R + 60Rrs2+r2+2Rr

≥ 272 .

Mihaly Bencze


n

��π4

�2 − n1R0

xnarctgxdx1+x2n

�.

Mihaly Bencze


2+tg2 B

2

1+√3(tgA

2+tgB

2 )≥ 6

√3r+2s

4R+r+3s .

Mihaly Bencze


k=1

�k 10√2018

< 17n(n+1)

40 where {·} denote the

fractional part.

Mihaly Bencze

PP28801. If r > 0 and λ ∈ [0, 1] thennP

k=1

({rk})λ ≤ n�n+12 max ({r} , 1− {r})

�λwhere {·} denote the fractional

part.

Mihaly Bencze

PP28802. Determine all ak ∈ N∗ (k = 1, 2, ..., n) for whichna1a2

o+n

a2a3

o+ ...+

nana1

o= 1 where {·} denote the fractional part.

Mihaly Bencze



k=1

�n+

hn

1k + 1

k

i� 1k ∈ R\Q where [·] denote the

integer part.

Mihaly Bencze

PP28804. Denote Fk and Lk the kth Fibonacci respective Lucas numbers.Compute

1).∞Pk=1

12Fk

2).∞Pk=1

12Lk

Mihaly Bencze


k=1

�log2

2n

k

�= 2n − 1 where [·] denote the integer

part.

Mihaly Bencze

PP28806. Determine all increasing functions f : N → N such that

n�

k=1

f(xk)

1+�

1≤i<j≤n

f(xi+xj)∈ N∗ for all xk ∈ N (k = 1, 2, ..., n) .

Mihaly Bencze


q(ax+ by + cz)2 + λ (a2 + b2 + c2) ≥ 4Area [ABC]

qP xyab + λ

P a2

b2for all

x, y, z,λ > 0 (A generalization of Weitzenbock’s inequality).

Mihaly Bencze


ax+ by + cz + λ�a2 + b2 + c2

�≥ 4Area [ABC]

�qP xyab + λ

qP a2

b2

�for all

x, y, z,λ > 0 (A generalization of Weitzenbock’s inequality).

Mihaly Bencze



ax+ by + cz + a2 + b2 + c2 ≥ 4Area [ABC]

�qP xyab +

qP a2

b2

�for all

x, y, z > 0 (A generalization of Weitzenbock’s inequality).

Mihaly Bencze

PP28810. In all triangle ABC holdsp(ax+ by + cz) (a2 + b2 + c2) ≥ 4Area [ABC] 4

r�P xyab

� �P a2

b2

�for all

x, y, z > 0 (A generalization of Weitzenbock’s inequality).

Mihaly Bencze

PP28811. Let ABC be a triangle, denote M the area of triangle formed by

ma,mb,mc. Prove that s2 − r2 − 4Rr ≥ 8M3

rP�

ma

mb

�2.

Mihaly Bencze

PP28812. Let ABC be a triangle, denote T the area of triangle formed by

cos A2 , cos

B2 , cos

C2 . Prove that 4R+ r ≥ 8T

sP�

cos A2

cos B2

�2

.

Mihaly Bencze


ax+ by + cz ≥ 4Area [ABC]q

xyab +

yzbc + zx

ca for all x, y, z > 0. (A

generalization of Weitzenbock’s inequality).

Mihaly Bencze


aλ + bλ + cλ ≥ 4Area [ABC]

qP(ab)λ−2 for all λ > 0 (A generalization of

Weitzenbock’s inequality).

Mihaly Bencze

PP28815. In all triangle ABC holds 3 + 2P

sin (4n+1)A2 sin A

2 ≥ 0 for alln ∈ N.

Mihaly Bencze



an + bn + cn ≥ 4Area [ABC]qP

(ab)n−2 for all n ∈ N,n ≥ 2. (A

generalization of Weitzenbock’s inequality).

Mihaly Bencze


1).P�

cosA+2 cosC2−cosB

�2cos4 A

2 ≤ (4R+r)2−s2

2R2

2).P�

cosA+2 cosC2−cosB

�2sin4 A

2 ≤ 8R2+r2−s2

2R2

Mihaly Bencze

PP28818. In all triangle ABC holdsP 3(cosA+2 cosC)2+16 cosB

(4+cos2 B)rarc≤ 4(4R+r)

s2r.

Mihaly Bencze

PP28819. In all triangle ABC holdsP 3(cosA+2 cosC)2+16 cosB

(4+cos2 B)(sinA+sinB)≤ 4R(5s2+r2+4Rr)

s(s2+r2+2Rr).

Mihaly Bencze

PP28820. In all triangle ABC holdsP (2−cosA)2

cosB+cosC ≥ 27(R+r)8R .

Mihaly Bencze

PP28821. In all triangle ABC holdsP (2−cosA)2

cosB+2 cosC ≥ 3(R+r)2R .

Mihaly Bencze


a2 + b2 + c2 ≥ 4Area [ABC]qP a2

b2≥ 4

√3Area [ABC] (A new refinement of

Weitzenbock’s inequality).

Mihaly Bencze


ab(s−a)(s−b) ≤�2R−r2sr2

�2.

Mihaly Bencze


PP28824. In all triangle ABC holdsP |(b−c)(c−a)|

ab ≤ (s2−r2−4Rr)(s2−3r2−12Rr)4s2r2

.

Mihaly Bencze


ab(b+c)(c+a) ≤�

s2−r2−Rrsr(s2+r2+2Rr)

�2.

Mihaly Bencze


ab(s−c) ≤(4R+r)2

4s3r2.

Mihaly Bencze

PP28827. In all triangle ABC holdsnP

k=1

(P

cos (2k + 1)A) ≤ 3n2 .

Mihaly Bencze

PP28828. In all triangle ABC holds 12 +P a2+b2

mamb≥ 20s2

�

mamb.

Mihaly Bencze

PP28829. In all triangle ABC holdsP√

bc cos2 A2 ≥ 3s

2 − 12

P�√a−

√b�2

.

Mihaly Bencze

PP28830. In all triangle ABC holdsP (a+b)(b+c)

ab ≤�s2+r2+4Rr

2sr

�2.

Mihaly Bencze


(a+b)(a+c) +P�

ab+c

�2≥ 8s2

5s2+r2+4Rr.

Mihaly Bencze

PP28832. In all triangle ABC holdsP 1√

(s−a)(a+b)(a+c)≤ s2−r2−Rr

2r√s(s2+r2+2Rr)

.

Mihaly Bencze


(1− cosA) ≤ s2−(2R+r)2

4R2

�4R2

s2−(2R+r)2− 1

�3.

Mihaly Bencze



1).Q �

1− sin A2

�≤ r

4R

�4Rr − 1

�3

2).Q �

1− cos A2

�≤ s

4R

�4Rs − 1

�3

Mihaly Bencze


1).Q �

1 + sin A2

� 1

sin A2 ≥ 4

3

�1 + r

4R

� 4Rr

2).Q �

1 + cos A2

� 1

cos A2 ≥ 4

3

�1 + s

4R

� 4Rs

Mihaly Bencze


(1− sinA) ≤ sr2R2

�2R2

sr − 1�3

.

Mihaly Bencze


(1 + sinA)1

sinA ≥ 43

�1 + sr

2R

� 2R2

sr .

Mihaly Bencze


Q(1 + cosA)

1cosA ≥ 4

3

�1 + s2−(2R+r)2

4R2

� 4R2

s2−(2R+r)2 .

Mihaly Bencze

PP28839. If x > 0 then2(shx)

23

�

1+(shx)23

�

ch2x+ (shx)

43

1+sh4x+ 2(shx)

23

1+shx ≤ 3.

Mihaly Bencze

PP28840. Solve the following system:

h2x

x+{y}

i+h

2yy+{x}

i= 2h

x2

2x+{y}

i+h

y2

2y+{x}

i= 2

, when

[·] and {·} denote the integer, respective the fractional part.

Mihaly Bencze

PP28841. Compute limx→0

x�12 − arctgx−arcsinx

x3

�.

Mihaly Bencze


PP28842. If an = n (a1 + a2 + ...+ an−1) for all n ≥ 1 then determine alla1 ∈ N and all n ∈ N for which an have the last 500 decimal equals.

Mihaly Bencze

PP28843. If xk > 0 (k = 1, 2, ..., n) thenQ

cyclic

x21+x2

2x1+x2

≥nQ

k=1

xk.

Mihaly Bencze


Q �a2 + b2

�≥ 8s2Rr

�s2 + r2 + 2Rr

�

2).Q �

r2a + r2b�≥ 4s4Rr

3).Q �

sin4 A2 + sin4 B

2

�≥ r2((2R−r)(s2+r2−8Rr)−2Rr2)

512R5

4).Q �

cos4 A2 + cos4 B

2

�≥ s2((4R+r)3+s2(2R+r))

512R5

Mihaly Bencze

PP28845. Determine all n, k ∈ N for which fromzn1 − zk1 = zn2 − zk2 = zn3 − zk3 = zn4 − zk4 holds that two of there are equals,z1, z2, z3, z4 ∈ C.

Mihaly Bencze

PP28846. In all triangle ABC holdsPqsinA sinB

sin2 A−sinA sinB+sin2 B≥ s2+12Rr+3r2

s2−4Rr−r2.

Mihaly Bencze


3x21−4x1 + x22 = 5x3 + 3x4

3x22−4x2 + x23 = 5x4 + 3x5

−−−−−−−−−−3x

21−4xn + x21 = 5x2 + 3x3

.

Mihaly Bencze

PP28848. If x1 > 0 and xn+1 =a1xn+b1[xn]+c1{xn}a2xn+b2[xn]+c2{xn} for all n ≥ 1 then prove

that the given sequence is convergent, and compute its limit, whereai, bi, ci > 0 (i = 1, 2) .

Mihaly Bencze


PP28849. In all triangle ABC holdsP s+rctgA

2�

tgA2+ctgA

2

≥ 6√sr.

Mihaly Bencze


P r+rara

√rra+rbrc

≥ 6s

2).P r+ha

ha

√rha+hbhc

≥ 3√2Rs

Mihaly Bencze

PP28851. If ak > 0 (k = 1, 2, ..., n) , thenP

cyclic

√19a21+2a1a2+4a22

4a2+a3≥ n.

Mihaly Bencze and Daniel Sitaru


1).Pp

4 + sin 2A+ 15 sin2A ≥ 1 + 4s+rR

2).P√

4 + sin 2A+ 15 cos2A ≥ 4 + 4r+sR

Mihaly Bencze


1R0

n√2016 + x ln (2017 + x) dx.

Mihaly Bencze

PP28854. If ak > 0 (k = 1, 2, ..., n) , thenP

cyclic

p76a21 + 230a1a2 + 178a22 ≥ 21

nPk=1

ak.

Mihaly Bencze


tgA2 + tgB

2

�7 �tgC

2

�5 ≥ 128r2

9s2.

Mihaly Bencze


1).P�

1r − 1

ra

�71r2a

≥ 1289s4r5

2).P�

1r − 1

ha

�71h2a≥ 32R2

9s4r7

Mihaly Bencze



r5a(r5ar5c+r10b

+s2rr4br3c)

≤ (4R+r)2−2s2

(s2r)5.

Mihaly Bencze


h5a

�

h5ah

5c+h10

b+ 2s2r

Rh4bh3c

� ≤R3

�

(s2+r2+4Rr)2−16s2Rr

�

128s10r10.

Mihaly Bencze


k2−1k8+5k4+1

≤ ζ(3)12 , where ζ denote the Riemann zeta

function.

Mihaly Bencze


k=2

k3

k8+5k4+1≤ 3n2−n−2

48n(n+1) .

Mihaly Bencze

PP28861. If xk > 0 (k = 1, 2, ..., n) then compute maxnQ

k=1

xk

(k+xk)

Mihaly Bencze

PP28862. If ai > 0 (i = 1, 2, ..., n) , k ∈ N∗ then

Pcyclic

(a1+a2+...+an−1)k

(a1a2...an−1)k

n−1≥

(n−1)k+1

�

n�

i=1ai

�k+1

(�

n−1√a1a2...an−1)

k .

Mihaly Bencze

PP28863. If 1 < a < b thenbRa

x3dxx8+5x4+1

≤ 124 ln

(b−1)(a+1)(b+1)(a−1) .

Mihaly Bencze

PP28864. In all triangle ABC holds1). 16R ≤ 5r + s2

r

2). 4s√r(4R+r)

≤ 9s√r

(4R+r)√4R+r

+ (4R+r)√4R+r

s√r

Mihaly Bencze


PP28865. If an =nP

k=1

1√n2+k

then nn+1 ≤

nPk=1

a2k

k2≤ 2− 1

n .

Mihaly Bencze

PP28866. Determine all continuous functions f : [0, 1] → [0,+∞) for which

af (x) + f�x2

�+ f

�1− x2

�≤ a

1R0

f (x) dx, where a > 0.

Mihaly Bencze


maxnP 8h2

a+395h4

a+34h2a+80

;P 8r2a+39

5r4a+34r2a+80

o≤ 1

2r .

Mihaly Bencze


maxnP 8h2

a+395h4

a+34h2a+80

;P 8r2a+39

5r4a+34r2a+80

o≤ 1

2r .

Mihaly Bencze


5x41 + 34x22 + 80 > 16x33 + 78x45x42 + 34x23 + 80 > 16x34 + 78x5−−−−−−−−−−−−−−5x4n + 34x21 + 80 > 16x32 + 78x3

.

Mihaly Bencze


k=1

8k2+39(k+1)(5k4+34k2+80)

< n2n+1 .

Mihaly Bencze

PP28871. Determine all n ∈ N for which for all i, j ∈ {0, 1, ..., n}�ni

�divide�

nj

�or

�nj

�divide

�ni

�.

Tibor Jakab



h3x2+1

3

i+h3y2+2

3

i=h2z2+1

2

ih3y2+1

3

i+h3z2+2

3

i=h2x2+1

2

ih3z2+1

3

i+h3x2+2

3

i=h2y2+1

2

i , when [·] denote the integer part.

Mihaly Bencze

PP28873. If xk ≥ 1 (k = 1, 2, ..., n) then

(x1 + x3)√x2 − 1 + (x2 + x4)

√x3 − 1 + ...+ (xn + x2)

√x1 − 1 ≤

≤ x1x2 + x2x3 + ...+ xnx1.

Mihaly Bencze

PP28874. Determine all ak > 0 (k = 1, 2, ..., n) if a1 = 1 and

6n (n+ 1)nP

k=1

kak = (2n+ 1) a2na2n+1.

Mihaly Bencze

PP28875. Computeh1k

2

i+ k

h2k

3

i+ k2

h3k

4

i+ ...+ kn−1

hnk

n+1

iwhen [·]

denote the integer part.

Mihaly Bencze

PP28876. If x, y, z ≥ 1 thenP 1+x2

(2+x2+y2)(2+x2+z2)≤

�

(1+xy)2

4(1+x2)(1+y2)(1+z2).

Mihaly Bencze

PP28877. If xk ≥ 1 thenP

cyclic

(1+x21)(1+x2

2)1+x1x2

≤ n+nP

k=1

x2k.

Mihaly Bencze

PP28878. If xk > 0 (k = 1, 2, ..., n) andnP

k=1

x2k = n then

Pcyclic

(1+x21)(1+x2

2)1+x1x2

≥ 2n.

Mihaly Bencze


PP28879. If xk > 0 (k = 1, 2, ..., n) andnP

k=1

x2k = n then

Pcyclic

(1+x21)(1+x2

2)...(1+x2r)

1+x1x2...xr≥ n2r−1, when r ∈ {2, 3, ..., n} .

Mihaly Bencze

PP28880. If bn =nP

k=1

2k−1h

k2

k+1

ithen compute

∞Pn=2

1bn.

Mihaly Bencze

PP28881. Solve in R the equation

h1

2017−x

i+h

11−2017x

i= 1

2017−[x] +1

1−[2017x] when [·] denote the integer part.

Mihaly Bencze

PP28882. Determine all a, b > 0 for which

�rha2

2

i+h(b+1)2

2

i�+

�rhb2

2

i+h(a+1)2

2

i�≥ 1 when [·] and {·} denote the

integer respective the fractional part.

Mihaly Bencze

PP28883. Let (bn)n≥1 be a geometrical progression for which b1 > 0, q > 1

when q is the ratio. Prove thatnP

k=1

bk+2−bk+1

(2k−1)(b2k−2−b1)≤ n

2n+1 .

Mihaly Bencze

PP28884. If ak > 0 (k = 1, 2, ..., n) such thatnP

k=1

a4k = 1 then

Pcyclic

√a1

1+a2 4√n≥ 1

2n .

Mihaly Bencze

PP28885. If a1 = 1 then determine all ak > 0 (k = 1, 2, ..., n) for whicha211 +

a222 + ...+ a2n

n =a31+a32+...+a3na1+a2+...+an

.

Mihaly Bencze


PP28886. If ak ≥ 1 (k = 1, 2, ..., n) andnP

k=1

ak = n+ 1 then

Pcyclic

pa1a2...an−1 (an − 1) ≤

r Pcyclic

a1a2...an−1.

Mihaly Bencze

PP28887. If an =�√

1�+�√

3�+�√

5�+ ...+

h√4n2 + 4n− 1

i, when [·]

denote the integer part, then compute∞Pk=1

1ak.

Mihaly Bencze

PP28888. If a1 =16 and an+1 =

n+1n+3

�an + 1

2

�for all n ≥ 1 then compute

nPk=1

[ak], when [·] denote the integer part.

Mihaly Bencze

PP28889. If Ak =2k+1Pp=0

hpk2 + 2p

iwhen [·] denote the integer part, then

compute∞Pk=1

1A2

k

.

Mihaly Bencze

PP28890. Determine all arithmetical progression (an)n≥1 and (bn)n≥1 forwhich (anbn)n≥1 is arithmetical progression too.

Mihaly Bencze

PP28891. Determine all geometrical progression (an)n≥1 and (bn)n≥1 forwhich (an + bn)n≥1 is geometrical progression too.

Mihaly Bencze

PP28892. If a, b, c > 0 thennP

k=1

�a

a2+k2(k+1)2+ b

b2+k2(k+1)2+ c

c2+k2(k+1)2

�≤ 3n

2(n+1) .

Mihaly Bencze



P aa2+rbrc

≤ 12r

2).P a

a2+hbhc≤ 1

2r

Mihaly Bencze


{x}+�x+ 1

n

+�x+ 1

n2

=

�n3x

+ 1

n + 1n2 when n ∈ N∗.

Mihaly Bencze


max��tgA

2 − tgB2

�� ;��tgB

2 − tgC2

�� ;��tgC

2 − tgA2

�� ≤ (4R+r)2

3s2.

Mihaly Bencze

PP28896. In all triangle ABC holdsP a

a2+(sin B2sin C

2 )4 ≤ 4R(2R−r)

r2.

Mihaly Bencze


a2+(cos B2cos C

2 )4 ≤ 4R(4R+r)

s2.

Mihaly Bencze


a2+(s−b)4(s−c)4≤ s2−2r2−8Rr

2s2r4.

Mihaly Bencze

PP28899. Denote B1 = {L1} , B2 = {L2, L3} , B3 = {L4, L5, L6, } ..., whereLk denote the kth Lucas number. Determine the sum of elements of the setBn.

Mihaly Bencze

PP28900. If a ∈ Q then the equation [x] {x} = a have an infinitely manysolutions in Q.

Mihaly Bencze


PP28901. If ak > 0 (k = 1, 2, ..., n) thenP

cyclic

a1a21+a2a3...an

≤n�

k=1

√ak

2

�

n�

k=1ak

.

Mihaly Bencze

PP28902. Denote A1 = {F1} , A2 = {F2, F3} , A3 = {F4, F5, F6} , ..., whenFk denote the kth Fibonacci number. Determine the sum of elements of setAn.

Mihaly Bencze

PP28903. If a1 = a+ 12 , a ∈ N∗ and an+1 = a2n − an + 3

4 for all n ≥ 1 thenprove that the number an+1 + an+2 is composite for all n ≥ 1.

Mihaly Bencze

PP28904. Let (an)n≥1 be an arithmetical progression with a1 > 0, r > 0and 4a1a2 < r, where r is the ratio. Denote

Sn =q1 + r2

a1a2a3+q

1 + r2

a2a3a4+ ...+

q1 + r2

anan+1an+2. Prove that

[Sn] = n for all n ≥ 3, when [·] denote the integer part.

Mihaly Bencze

PP28905. If a1 = 2, a2 = 6 and4an+2 = (n+ 3)

�4 +

√4an+1 + 1 +

√4an + 1

�for all n ≥ 1, then

nPk=1

1akak+2

= 118 − 1

3(n+1)(n+2)(n+3) .

Mihaly Bencze

PP28906. If ak ∈ [0, 1] (k = 1, 2, ..., n) , thenP

cyclic

a1a2+a33+3n−2

≤ 13 .

Mihaly Bencze

PP28907. Determine all p prime for which p2 + p+ 1 is prime andp2

n+ p2

n−1+ 1 is divisible by p2 + p+ 1 for all n ≥ 1.

Mihaly Bencze


PP28908. How many distinct divisors have the numbersnQ

k=1

�p2

k+ p2

k−1�

when p is a prime.

Mihaly Bencze

PP28909. If Sk =3n+1Pp=1

��p3

��kwhen [·] denote the integer part, then

compute S1, S2, S3, S4.

Mihaly Bencze

PP28910. Prove that3n+1Pk=1

��k3

��2 ≥ n2(3n+1)4 , where [·] denote the integer

part.

Mihaly Bencze

PP28911. If Sn =3n+1Pk=1

�k3

�when [·] denote the integer part, then

nPp=1

Sp =n(n+1)2

2 .

Mihaly Bencze

PP28912. Prove that 172n+ 172

n−1+ 1 is divisible by 307 for all n ≥ 1.

Mihaly Bencze


y+z + y2017

z+x + z2017

x+y = 32 .

Mihaly Bencze


1).Pq

(s−a)(s−b)ac ≤ 3

2

2).Pr

cos A2cos B

2

(cos A2+cos B

2 )(cosB2+cos C

2 )≤ 3

2

Mihaly Bencze

PP28915. In all triangle ABC holdsP sin A

2

sin B2+(sin C

2 )3+7

≤ 13 .

Mihaly Bencze


PP28916. If dk =∞Pn=1

1(Ln)

k when Ln denote the nth Lucas number, then

compute∞Pk=1

(−1)kdkk! .

Mihaly Bencze


y+z + y3

z+x + z3

x+y = −6.

Mihaly Bencze

PP28918. Let be xk =∞Pn=1

1pkn

where pn denote the nth prime. Compute

∞Pk=1

(−1)kxk

k! .

Mihaly Bencze


y+z + y2

z+x + z2

x+y = −3.

Mihaly Bencze

PP28920. Compute∞R0

∞R0

∞R0

ln(x+y+z)dxdydz(ex+1)(ey+1)(ez+1) .

Mihaly Bencze

PP28921. If ck =∞Pn=1

1(Fn)

k where Fn denote the nth Fibonacci number,

then compute∞Pk=1

(−1)kCk

k! .

Mihaly Bencze

PP28922. Let be

1 + x(1−x)2

+ x2

(1−x)2(1−x2)2+ x3

(1−x)2(1−x2)2(1−x3)2+ ... =

∞�

n=0anxn

(1−x)2(1−x2)2(1−x3)2....

Compute∞Pn=0

1a20+a21+...+a2n

.

Mihaly Bencze


PP28923. Let be ak =∞Pn=1

1nk . Compute

∞Pk=1

(−1)kakk! .

Mihaly Bencze

PP28924. Let be bk =∞Pn=1

1(2n+1)k

. Compute∞Pk=1

(−1)kbkk! .

Mihaly Bencze


arctg�6−

√5

3√5

�3+

√5

2

�n− 6+

√5

3√5

�3−

√5

2

�n�= π

4 .

Mihaly Bencze

PP28926. Prove that

��

1 1 0 0 0 ... 013 1 2 0 0 ... 017

13 1 3 0 ... 0

− − − − − − −1

2n−11

2n−1−1... ... ... ... 1

��

= n!1·3·7·...·(2n−1) .

Mihaly Bencze

PP28927. Prove that∞Pn=1

1n�

k=1(−1)n+k(nk)kn

= e− 1.

Mihaly Bencze

PP28928. Compute

1).∞Pk=1

Γ(sin2 kx)k!

2).∞Pk=1

Γ(cos2 kx)k! , where Γ denote the gamma function.

Mihaly Bencze

PP28929. Let A = (aij)1≤i,j≤n when aij =hij

i+hji

iwhere [·] denote the

integer part. Compute detA.

Mihaly Bencze


PP28930. Compute Ik =∞R0

e−x (lnx)k dx.

Mihaly Bencze

PP28931. Compute Vn =1R0

1R0

...1R0

x21+2x2

2+...+nx2n

nx1+(n−1)x2+...+xndx1dx2...dxn.

Mihaly Bencze

PP28932. Compute∞Pn=1

1p1+p2+...+pn

where pk denote the kth prime.

Mihaly Bencze

PP28933. Let be A = (aij)1≤i,j≤n where aij = iεj + jεi where εn = 1(ε 6= 1) . Compute detA.

Mihaly Bencze

PP28934. Let be A = (aij)1≤i,j≤n where aij =�i+ji

��i+jj

�. Compute detA.

Mihaly Bencze

PP28935. Compute

1).∞Pk=1

x2k sin(2k−1)x(2k+1)!

2).∞Pk=1

x2k cos(2k−1)x(2k+1)!

Mihaly Bencze

PP28936. Compute∞Pk=1

xk+yk

(x+y)k.

Mihaly Bencze

PP28937. Prove that

0 <

bRa

3xx2+x+1

dx

!2

−

bRa

3x sinxdxx2+x+1

!2

−

bRa

3x cosxdxx2+x+1

!2

≤ 112 .

Mihaly Bencze


PP28938. Compute1). 1− cosx

2 +B1cos 2x2! + ...+ (−1)nBn

cos 2nx(2n)! + ...

2). 1− sinx2 +B1

sin 2x2! + ...+ (−1)nBn

sin 2x(2n)! + ... when Bn denote the nth

Bernoulli number.

Mihaly Bencze

PP28939. Denote F and L the set of Fibonacci, respective Lucas numbers,and be x1 = 2, x2 = 8, xn+2 − 4xn+1 + xn = 0.

1). Compute∞Pn=1

1n2+y2n

where yn = {xn|n ∈ N} ∩ F

2). Compute∞Pn=1

1n2+z2n

where zn = {xn|n ∈ N} ∩ L

Mihaly Bencze


arctg(7+4

√3)

k−2+(7−4√3)

k

3 = π2 .

Mihaly Bencze

PP28941. Solve in Z the equation y2 = 1 + 2x+ 3x2 + ...+ (n+ 1)xn.

Mihaly Bencze


1).P

sinAesinA ≥P cos A2 cos B−C

2 ecosA2cos B−C

2

2).P

cosAecosA ≥P sin A2 cos B−C

2 esinA2cos B−C

2

Mihaly Bencze

PP28943. If xk > 0 (k = 1, 2, ..., n) , thennP

k=1

xkexk ≥ P

cyclic

x1+x22 e

x1+x22 .

Mihaly Bencze

PP28944. In all triangle ABC holdsP 1sin2 A cos2 A

≤ 4 + 1sin2 A sin2 B sin2 C

+ 1cos2 A cos2 B cos2 C

.

Mihaly Bencze

PP28945. Prove that 1k + 2k + ...+ nk ≤ n− 1 + (n!)k , where k ∈ N .

Mihaly Bencze


PP28946. Prove that 2n ≤ n+ (n!)n+1

(1!2!...n!)2for all n ∈ N∗.

Mihaly Bencze

PP28947. If xk, yk ≥ 1 (k = 1, 2, ..., n) then�n− 1 +

nQk=1

x2k

��n− 1 +

nQk=1

y2k

�≥

�nP

k=1

xkyk

�2

.

Mihaly Bencze

PP28948. If ak > 0 (k = 1, 2, ..., n) , thennP

k=1

1ak+1 ≤ n− 1 + 1

1+n�

k=1ak

.

Mihaly Bencze

PP28949. If xk ≥ 1 (k = 1, 2, ..., n) , then

nx1 + (n− 1)x2 + ...+ 2xn−1 + xn ≤ n(n−1)2 + x1 + x1x2 + ...+ x1x2...xn.

Mihaly Bencze


P a2−ab+b2

(a2+b2)(2(a2+b2)−ab)≤ 1

12Rr

2).P r2a−rarb+r2

b

(r2a+r2b)(2(r2a+r2

b)−rarb)≤ 4R+r

6s2r

Mihaly Bencze


k=1

k2+k+1(2k2+2k+1)(3k2+3k+2)

≤ n6(n+1) .

Mihaly Bencze


1).P sin4 A

2−sin2 A

2sin2 B

2+sin4 B

2

(sin4 A2+sin4 B

2 )(2(sin4 A

2+sin4 B

2 )−sin2 A2sin2 B

2 )≤ 4R(2R−r)

3r2

2).P cos4 A

2−cos2 A

2cos2 B

2+cos4 B

2

(cos4 A2+cos4 B

2 )(2(cos4A2+cos4 B

2 )−cos2 A2cos2 B

2 )≤ 4R(4R+r)

6s2

Mihaly Bencze


PP28953. If xk ≥ 1 (k = 1, 2, ..., n) , thennQ

k=1

(x1 + x2 + ...+ xk) ≤nQ

k=1

(k − 1 + x1x2...xn) .

Mihaly Bencze

PP28954. If xk > 0 (k = 1, 2, ..., n) , then 1 +nP

k=1

sh2xk ≤nQ

k=1

ch2xk.

Mihaly Bencze


k=1

�1k2

+ 1(k+1)2

+ (2k+1)4

2k2(k+1)2(2k2+2k+1)

�≥ 6n

n+1 .

Mihaly Bencze

PP28956. If xk > 0 (k = 1, 2, ..., n) , then

13

nPk=1

x2k ≥ Pcyclic

(x21−x1x2+x2

2)x1x2

(x21+x2

2)(2(x21+x2

2)−x1x2).

Mihaly Bencze


1).�

1ha

+ 1hc

�hd�

1hb

+ 1hd

�hc�

rhc

+ rhd

�hc+hd ≤�

1hc

� 1hd

�1hd

� 1hc

2).�

1ra

+ 1rc

�rd�

1rb

+ 1rd

�rc �r2rc

+ r2rd

�rc+rd ≤�

1rc

�rd�

1rd

�rc

Mihaly Bencze

PP28958. If a, b, c > 0 thenP a+b

(b+c)(c+a) ≥9

2(a+b+c) .

Mihaly Bencze

PP28959. If a, b, c > 0 and λ > 0 thenP a+λb+c

(b+λc+a)(c+λa+b) ≥9

(λ+2)(a+b+c) .

Mihaly Bencze

PP28960. If a, b, c > 0 then�1 + a

c

�2c �1 + b

a

�2a �1 + c

b

�2b ≤�2a+b+ca+c

�a+c �2b+c+ab+a

�b+a �2c+a+bc+b

�c+b.

Mihaly Bencze



1).�1 + rc

ra

��1 + 1

xrb

�xrc � r(xrc+1)rc(xr+1)

�xrc≤ 1

2).�1 + hc

ha

��1 + 1

xhb

�xhc�r(xhc+1)hc(xr+1)

�xhc

≤ 1 for all x > 0

Mihaly Bencze

PP28962. If a, b, c > 0 then�1 + a

c

�c �a+2b+ca+b+c

�a+b+c≤

�2(a+b+c)a+b+2c

�a+b+2c

and his permutations.

Mihaly Bencze

PP28963. If n ∈ N ∗ then�1 + 1

n

�n �n+2n+3

�n+3≤ 4

�2n+32n+5

�2n+5.

Mihaly Bencze

PP28964. If a, b > 0 and x ∈ R then�1 + a

sin2 x

�sin2 x �1 + b

cos2 x

�cos2 x ≤ a+ b+ 1.

Mihaly Bencze

PP28965. Prove that

1).�1 + Fk

Fk+2

�Fk+2�1 +

Fk+1

Fk+3

�Fk+3 ≤�Fk+2+Fk+4

Fk+1+Fk+3

�Fk+4

2).�1 + Lk

Lk+2

�Lk+2�1 +

Lk+1

Lk+3

�Lk+3 ≤�Lk+2+Lk+4

Lk+1+Lk+3

�Lk+4

for all k ∈ N∗

Mihaly Bencze

PP28966. In all tangential quadrilateral ABCD with radius R andsemiperimeter s holds

�1 + tgC

2 ctgA2

�ctgC2�1 + tgB

2 ctgD2

�ctgD2 ≤

�s

R(ctgC2+ctgB

2 )

�ctgC2+ctgD

2

.

Mihaly Bencze



1).P�

(2+ctg2 A2 ) ln(2+ctg2 A

2 )ln(1+ctg2 A

2 )

� 1

2+ctg2 A2 ≥ 18R

8R−r

2).P�

(2+tg2 A2 ) ln(2+tg2 A

2 )ln(1+ctg2 A

2 )

� 1

2+tg2 A2 ≥ 18R

10R+r

Mihaly Bencze


1).P�

(2+tg2A) ln(2+tg2A)ln(1+tg2A)

� 12+tg2A

≥ 18R2

9R2−s2+(2R+r)2

2).P�

(2+ctg2A) ln(2+ctg2A)ln(1+ctg2A)

� 12+ctg2A

≥ 18R2

6R2+s2−r(4R+r)

Mihaly Bencze

PP28969. In all triangle ABC holds 1R√Rr

≤P sinA

ra√

a(s−a)≤ 1

2r√Rr

(A

refinement of Euler’s R ≥ 2r inequality).

Mihaly Bencze

PP28970. In all triangle ABC holds s4r2

≥P m2a

s−a ≥ s2Rr (A refinement of

Euler’s R ≥ 2r inequality).

Mihaly Bencze

PP28971. In all triangle ABC holds 3R4sr2

≥P 1(b+c)(s−a) ≥

32sr (A


Mihaly Bencze

PP28972. In all triange ABC holds 9R2

r2(s2+(4r+r)2)≥P 1

m2a≥ 18R

r(s2+(4R+r)2)(A refinement of Euler’s R ≥ 2r inequality).

Mihaly Bencze


PP28973. In all triangle ABC holds 4√RrP

ra

qs−aa ≥ 4

√3Area [ABC]

(Weitzenbock’s type inequality).

Mihaly Bencze

PP28974. In all triangle ABC holds 2sr2

R+r

P √b2+bc+c2

s−a ≥ 4√3Area [ABC]

(Weitzenbock’s type inequality).

Mihaly Bencze

PP28975. In all triangle ABC holds 13r2

≥P 1

wawb≥ 2

3Rr (A refinement ofEuler’s R ≥ 2r inequality).

Mihaly Bencze


2P a2 sin A

2

cosA cos B−C2

≥ 4√3Area [ABC] (Weitzenbock’s type inequality).

Mihaly Bencze


k=1

e−xk+1−e−xk

x dx = nγn+1 where γ is the

Euler-Mascheroni constant.

Mihaly Bencze


k=1

e−x2k+1−e−x2k−1

x dx = nγ2n+1 where γ is the


Mihaly Bencze


AB∞R0

e−xA−e−xB

x dx = 0.

Mihaly Bencze


cos2A sin2A∞R0

e−xcos2 A−e−xsin

2 A

x dx =((2R+r)2−R2−s2)γ

R2 , where γ is the


Mihaly Bencze


PP28981. If xk > 0 (k = 1, 2, ..., n) , thennP

k=1

∞R0

e−tch2x−e−tsh

2t

t dt ≥ n2−γn�

k=1

sh2xkch2xk

where γ is the Euler-Mascheroni

constant.

Mihaly Bencze

PP28982. Prove that, if 0 < a ≤ b thenbRa

�∞R0

e−xch2t−e−xsh

2t

x dx

�dt = 2γsh2(a−b)

sh2ash2b where γ is the Euler-Mascheroni

constant

Mihaly Bencze


1γ

P ∞R0

e−xsin2 A−e−xcos

2 A

x dx =�

s2+r2−4R2

s2−(2R+r)2

�2− 8R(R+r)

s2−(2R+r)2−�s2+r2+Rr

2sr

�2+ 4R

r

where γ is the Euler-Mascheroni constant.

Mihaly Bencze

PP28984. Compute∞P

k1=1

∞Pk2=1

∞Pk3=1

1(k21+k22)(k

22+k23)(k

23+k21)

.

Mihaly Bencze


sinx(1−(sinx)2n+2)cos2 x

dx where n ∈ N.

Mihaly Bencze


cosx(1−(cosx)2n+2)dxsin2 x

where n ∈ N.

Mihaly Bencze

PP28987. Compute

π2R0

tgx ln�1+sin 2x1−sin 2x

�dx.

Mihaly Bencze



Hn

ζ (2k)−

nPp=1

1p2k

!, where ζ denote the

Riemann zeta function and Hn =nP

i=1

1i and k ∈ N∗.

Mihaly Bencze

PP28989. Compute

π2R0

π2R0

ln(sinx) ln(cos y) ln(1−cosx) ln(1−sin y)dxdysinx cos y .

Mihaly Bencze

PP28990. Let be f (x) =nP

k=1

(k (k + 1))x − n(n+1)(n+2)3 . Compute

limn→∞

1n3

n(n+1)(n+2)3R0

f−1 (x) dx.

Mihaly Bencze

PP28991. Prove that if 0 < a ≤ b < 1 thenbRa

(√x+

√1−x) ln(x−x2)1+

√2x

dx =√2 ln bb(1−a)1−a

aa(1−b)1−b .

Mihaly Bencze

PP28992. Prove that if 0 < a ≤ b < 1 thenbRa

(√x+

√1−x)(sinπx+sinπ(1−x))dx

1+√2x

= 4√2

π sin (b−a)π2 cos π(b+a−1)

2 .

Mihaly Bencze

PP28993. If xk > 0 (k = 1, 2, ..., n) andnP

k=1

xk = n thennQ

k=1

(1+xk)1+xk

xxkk

≤ 4n.

Mihaly Bencze

PP28994. If x, y, z > 0 then 64(x+1)x+1(y+1)y+1(z+1)z+1(x+y+z+3)x+y+z+3

xxyyzz(x+y+z)x+y+z ≤

≤ 27(x+y+2)x+y+2(y+z+2)y+z+2(z+x+2)z+x+2

(x+y)x+y(y+z)y+z(z+x)z+x .

Mihaly Bencze



1). (a+1)a+1(b+1)b+1(c+1)c+1

aabbcc≤

�1 + 2s

3

�3 �1 + 3

2s

�2s

2). (ra+1)ra+1(rb+1)rb+1(rc+1)rc+1

rraa rrbbrrcc

≤�1 + 4R+r

3

�3 �1 + 3

4R+r

�4R+r

Mihaly Bencze

PP28996. If xk > 0 (k = 1, 2, ..., n) then

nQk=1

(xk + 1)�1 + 1

xk

�xk ≤�1 + 1

n

nPk=1

xk

�n1 + n

n�

k=1

xk

n�

k=1

xk

.

Mihaly Bencze

PP28997. If x, y, z > 0 and x+ y + z = 3 then4096(x+1)x+1(y+1)y+1(z+1)z+1

xxyyzz ≤ (5−x)5−x(5−y)5−y(5−z)5−z

(3−x)3−x(3−y)3−y(3−z)3−z .

Mihaly Bencze

PP28998. If x ∈�0, π2

�then 3 sin2 x cos2 x+ (tgx)6 cos

2 x + (ctgx)6 sin2 x ≥ 1.

Mihaly Bencze

PP28999. Denote Fk and Lk the kth Fibonacci respective Lucas numbers.Prove that1). (Fk+2)

Fk+2 ≥ (Fk)Fk (Fk+1)

Fk+1

1). (Lk+2)Lk+2 ≥ (Lk)

Lk (Lk+1)Lk+1 for all k ∈ N∗

Mihaly Bencze

PP29000. In all triangle ABC holds (n+ 1)P 1

cos A2

≥ (2n+ 1)√3 + 4R+ r

for all n ∈ N∗.

Mihaly Bencze

PP29001. In all triangle ABC holds (n+ 1)P

r4a ≥ s2r (12R+ (9n+ 7) r)for all n ∈ N∗.

Mihaly Bencze


PP29002. If a, b, c ∈ (0, 1) ∪ (1,+∞) thenP (an+1−1)(2a−b−c)

(a−1)(b+c) ≥ 0.

Mihaly Bencze

PP29003. If a, b, c > 0 thennP

k=2

�P a(b+c)2+(k2+k−4)a2

�≥ 3(n−1)

2(n+1) .

Mihaly Bencze

PP29004. If a, b, c > 0 then 3√3P

a2 ≥�a2 + b2 − c2

�+ 4

Pab.

Mihaly Bencze

PP29005. If a, b, c > 0 then 9�P

a2�2 ≥ 3

P �a2 + b2 − c2

�2+ 8 (

Pab)2 .

Mihaly Bencze


b+c−a ≥ 2(s2−r2−4Rr)s2+r2+4Rr

.

Mihaly Bencze

PP29007. In all triangle ABC holdsP (s−b)(s−c)

b2c2(s−a)≥

√3

8R3 .

Mihaly Bencze

PP29008. Determine all λ ≥ 0 for which in all triangle ABC holdsP a3

b+c−a ≥≥ 4√3sr + λ

P(a− b)2 .

Mihaly Bencze

PP29009. In all triangle ABC holds1

2sr

Pa3 (s− b) (s− c) ≥ 4

√3Area [ABC] (Weitzenbock’s type inequality).

Mihaly Bencze


1). rλ2P�

1ra

� 3λ2+ 6

sλ≤ r

λ−22P�

1ra

� 3λ−22

2). rλ2P�

1ha

� 3λ2+ 6

sλ≤ r

λ−22P�

1ra

� 3λ−22

for all λ > 0.

Mihaly Bencze



k=2

�(k+1) ln(k+1)

ln k

� 1k+1 ≥ 2

n+1 .

Mihaly Bencze

PP29012. If 1 < a ≤ b then ln b+1a+1 +

bRa

�(x+1) ln(x+1)

lnx

� 1x+1

dx ≥ b− a.

Mihaly Bencze

PP29013. If xk > 1 (k = 1, 2, ..., n) then

nPk=1

�(x2

k+1) ln(x2

k+1)

2 lnxk

� 1

x2k+1

≥

�

n�

k=1xk

�2

n+n�

k=1

x2k

.

Mihaly Bencze

PP29014. If xk > 1 (k = 1, 2, ..., n) , then

nPk=1

�(xk+1) ln(xk+1)

lnxk

� 2xk+1 ≥

�

n�

k=1xk

�2

n+2n�

k=1xk+

n�

k=1x2k

.

Mihaly Bencze

PP29015. If xk > 0 (k = 1, 2, ..., n) , thennP

k=1

sh2xk�

cos

�

π

sh2xk+ch2xk

�

≥ 1n

�nP

k=1

shxk

�2

+ 1n

�nP

k=1

chxk

�2

.

Mihaly Bencze


k=1

1cos π

2(2k+1)≥ n(4n+5)

4(n+1) .

Mihaly Bencze


1).Qr

cos�π2

�a−ba+b

��≤ 16Rr

s2+r2+2Rr

2).Qq

cos�π2

�a−bc

��≤ 2r

R 3).Qr

cos�π2

�ra−rbra+rb

��≤ 2r

R

Mihaly Bencze


PP29018. If xk ∈�0, π2

�(k = 1, 2, ..., n) , then

nPk=1

1�

cos(π2cos 2xk)

≥ n

sin

�

2n

n�

k=1xk

� .

Mihaly Bencze


2 sin (x+ y) cos (x− y) ≥qcos

�π2 cos 2x

�+qcos

�π2 cos 2y

�.

Mihaly Bencze

PP29020. Prove that sin 2x ≥qcos

�π2 cos 2x

�for all x ∈ R.

Mihaly Bencze

PP29021. If a, b, c > 0 thenQr

cos�π2

�a−ba+b

��≤ 8abc

(a+b)(b+c)(c+a) .

Mihaly Bencze

PP29022. In all triangle ABC holdsPq

cos�π2 cos 2A

�≤ 2sr

R2 .

Mihaly Bencze

PP29023. If a, b, c > 0 thenQ

cos�π2

�a−c

a+2b+c

��≤

�8(a+b)(b+c)(c+a)

(a+2b+c)(b+2c+a)(c+2a+b)

�2.

Mihaly Bencze


a2 cos(π2 (

b−cb+c ))

≥ s(s2+r2−6Rr)16sR2r2

≥ 12Rr .

Mihaly Bencze


1).P

(a+ b) cos�π2

�a−ba+b

��≤

2�

(s2+r2+4Rr)2+8s2Rr

�

sRr

2).P

(ra + rb) cos�π2

�ra−rbra+rb

��≤ s2+r(4R+r)

R

3).P

(sinA+ sinB) cos�π2

�sinA−sinBsinA+sinB

��≤ (s2+r2+4Rr)

2−8s2Rr

sR(s2+r2+2Rr)


4).P

a cos�π2

�b−ca

��≤ r(s2+(4R+r)2)

sR

Mihaly Bencze


k=1

k+1k2

cos�π2

�k−1k+1

��≤ 4n

n+1 .

Mihaly Bencze


k=1

rk cos

�π2

�k−1k+1

��≤ 2n

n+1 .

Mihaly Bencze

PP29028. If x, y, z ∈ [0, 1] then(z − y + 1) sinx+ (x− z + 1) sin y + (y − x+ 1) sin z ≤≤ sin ((x− y) z + y) + sin ((y − z)x+ z) + sin ((z − x) y + x) .

Mihaly Bencze


1). s2−r(4R+r)2R2 cosx+

�1− s2−(2R+r)2

2R2

�cos y ≤P cos

�x sin2A+ y cos2A

�

2). s2−r(4R+r)2R2 sinx+

�1− s2−(2R+r)2

2R2

�sin y ≤

Psin

�x sin2A+ y cos2A

�for

all x, y ∈�0, π2

�

Mihaly Bencze

PP29030. If x, y ∈�0, π2

�and λ ∈ [0, 1] then

λ sinx+(1−λ) sin ycos(λx+(1−λ)y) ≤ sin(λx+(1−λ)y)

λ cosx+(1−λ) cos y .

Mihaly Bencze

PP29031. If x, y, z ∈ [0, 1] then(z − y + 1) cosx+ (x− z + 1) cos y + (y − x+ 1) cos z ≤≤ cos ((x− y) z + y) + cos ((y − z)x+ z) + cos ((z − x) y + x) .

Mihaly Bencze

PP29032. If x, y ∈�0, π2

�then

nPk=1

cos

�x+(k2+k−1)y

k(k+1)

�≥ n

n+1 (cosx+ n cos y) .

Mihaly Bencze


PP29033. If x, y ∈�0, π2

�then

nPk=1

sin

�x+(k2+k−1)y

k(k+1)

�≥ n

n+1 (sinx+ n sin y) .

Mihaly Bencze

PP29034. If λ ∈ [0, 1] and x, y ∈�0, π2

�then

λ (1− λ) sin (x+ y) + λ2 sinx cosx+ (1− λ)2 sin y cos y ≤≤ 1

2 sin (2λx+ 2 (1− λ) y) .

Mihaly Bencze

PP29035. If λ ∈ [0, 1] and x1, y1, x2, y2 ∈�0, π2

�then

λ cos x1+x22 cos x1−x2

2 + (1− λ) cos y1+y22 cos y1−y2

2 ≤≤ cos λ(x1+x2)+(1−λ)(y1+y2)

2 cos λ(x1−x2)+(1−λ)(y1−y2)2 .

Mihaly Bencze

PP29036. If x, y ∈�0, π2

�and t ∈ R then

(cosx− cos y)2 sin2 t cos2 t+ cosx cos y ≤≤ cos

�sin2 tx+ cos2 ty

�cos

�cos2 tx+ sin2 ty

�.

Mihaly Bencze

PP29037. If t ∈ R and x, y ∈�0, π2

�then

sinx+ sin y ≤ sin�x sin2 t+ y cos2 t

�+ sin

�x cos2 t+ y sin2 t

�.

Mihaly Bencze

PP29038. If t ∈ R and x, y ∈�0, π2

�then

cosx+ cos y ≤ cos�x sin2 t+ y cos2 t

�+ cos

�x cos2 t+ y sin2 t

�.

Mihaly Bencze


1). 2 (cosx+ cos y) ≤P cos�

rrax+

�1− r

ra

�y�+P

cos��

1− rra

�x+ r

ray�

2). 2 (cosx+ cos y) ≤P cos�

rhax+

�1− r

ha

�y�+P

cos��

1− rha

�x+ r

hay�

for all x,y∈�0, π2

�.

Mihaly Bencze


PP29040. If ak > 0 (k = 1, 2, ..., n) , then�a1a2

+ 1�a2 �a2

a3+ 1

�a3· ... ·

�ana1

+ 1�a1

≤ 2a1+a2+...+an .

Mihaly Bencze

PP29041. If x, y ∈ R then�1 + sin2 x

sin2 y

�sin2 y �1 + cos2 x

cos2 y

�cos2 y≤ 2.

Mihaly Bencze

PP29042. In triangle ABC holds A ≥ B ≥ C. Prove thatc sinA+ b sinB + a sinC ≤ 2s2

3R .

Mihaly Bencze

PP29043. In all triangle ABC holdsP ra(λs−a)

s−a ≥ (4R+r)(4(λ−1)R+(λ+2)r)3r for

all λ ≥ 1.

Mihaly Bencze

PP29044. In all acute triangle ABC holds�s2 − r2 − 4Rr

� �s2 − (2R+ r)2

�+ 45

�s2 − (2R+ r)2

�2+ 16s2r2 ≤

≤ 2�s2 − r2 − 4Rr

�2.

Mihaly Bencze

PP29045. If x, y, z > 0 then in triangle ABC holds

xOI + yGI + zGO ≤√

x2+y2+z2

3

�18R2 − 26Rr + 7r2 − s2

�.

Mihaly Bencze

PP29046. Let A1B1C1 be the Morley triangle of ABC. Denote Rm and Rthe circumradii of triangles A1B1C1 and ABC. Prove thatRm

R ≤ 8√3

3

Qsin A+B

6 ≤ 8√3

3

Qsin π+A

12 ≤ 8√3

3

Qsin 3π−A

24 ≤ 8√3

3

Qsin 5π+A

48 ≤... ≤ 8

√3

3

�sin π

9

�3.

Mihaly Bencze and Mehmet Sahin

PP29047. In all triangle ABC holds nra+krb+prcn+k+p ≥ nrama+krbmb+prcmc

nma+kmb+pmcfor all

n, k, p ∈ N∗.

Mihaly Bencze


PP29048. Let ABC be an acute triangle, denote A1B1C1 his Morleytriangle, denote Rm and R the circumradii of triangles A1B1C1 and ABC.

Prove that 64√3ABC

81π3 < Rm

R < 8√3ABC81 .

Mihaly Bencze


P(ma +mb)

3 > 27s4

�s2 − 3r2 − 6Rr

�

2).P

(ma +mb)2 (mb +mc)

2 > 8116

��s2 + r2 + 4Rr

�2 − 16s2Rr�

3).P 1

(ma+mb)2 < 4

9

��s2+r2+4Rr

4sRr

�2− 1

Rr

�

4).P 1

(ma+mb)2(mb+mc)

2 <81(s2−r2−4Rr)

128s2R2r2

Mihaly Bencze

PP29050. In all triangle ABC holdsP ra

rb+rc≥ (4R+r)2

2s2.

Mihaly Bencze

PP29051. Let Ra, Rb, Rc be the circumradius of triangles BOC, COA,

AOB, in triangle ABC. Prove thatP Ra

R+2Ra= 1

4 +�4R+r2s

�2.

Mihaly Bencze


1).P�

1 + rra

� rrb ≤ 4

�1 + R

r

�

2).P�

1 + 1ha

� rhb ≤ 13

4 + r(4R+r)4s2

Mihaly Bencze


rarb+rc

�λ≤ 1

3λ−1

�R3r − 1

6

�λfor all

λ ∈ [0, 1] .

Mihaly Bencze

PP29054. In all acute triangle ABC holdsPq

(1 + sinA)cosA (1 + cosA)sinA ≤ 3 + srR2 .

Mihaly Bencze


PP29055. In all triangle ABC holdsAI ·BI · CI = 2R2 (ha − 2r) (hb − 2r) (hc − 2r) .

Mihaly Bencze


(1 + sinA)sinB ≤ 3 + s2+r2+4Rr4R2 .

Mihaly Bencze

PP29057. In all acute triangle ABC holdsP(1 + cosA)cosB ≤ 3 + s2+r2−4R2

R2 .

Mihaly Bencze

PP29058. In all acute triangle ABC holdsP(1 + sinA)sinA (1 + cosA)cosA ≤ 6 + 2sr

R2 .

Mihaly Bencze


2�1 + cos B

2 cos C2 − cos2 A

2

�+ r

2R ≤ 2P

cos A2 .

Mihaly Bencze

PP29060. In all triangle ABC holdsP ma(mb+mc)

mb+mc−ma≥ 2

Pma.

Mihaly Bencze


m2am

2b ≥ 9s2r2 + 3

2

�P ��a2 − b2��2 .

Mihaly Bencze

PP29062. If ak > 0 (k = 1, 2, ..., n) , thennQ

k=1

a2k+ak+1

ak(ak+1) ≥ 1 + 1n�

k=1ak

+ 1n�

k=1(ak+1)

.

Mihaly Bencze

PP29063. In all triangle ABC holdsP (cos A

2+cos B

2−cos C

2 )3

cos C2

≥ 4R+r2R .

Mihaly Bencze


PP29064. If ak > 0 (k = 1, 2, ..., n) , thenP

cyclic

a21√2a2a3+

√a22+a23

≥ 12√2

nPk=1

ak.

Mihaly Bencze

PP29065. If ak > 0 (k = 1, 2, ..., n) , thenPcyclic

a21

(a2+a3)2�√

2a2a3+√

a22+a23

� ≥ n2

8√2

n�

k=1ak

.

Mihaly Bencze


k=1

1k(

√2k+

√k2+1)

≥ n√2(n+1)

.

Mihaly Bencze

PP29067. If ak > 0 (k = 1, 2, ..., n) thenP

cyclic

1√2a1a2+

√a21+a22

≥ n2

2n�

k=1ak

.

Mihaly Bencze


1).Q�√

ab+q

a2+b2

2

�≤ 2s

�s2 + r2 + 2Rr

�

2).Q�p

(s− a) (s− b) +

q(s−a)2+(s−b)2

2

�≤ 4sRr

3).Q�

√rarb +

qr2a+r2

b

2

�≤ 4s2R

4).Q�√

hahb +

qh2a+h2

b

2

�≤ s2r(s2+r2+2Rr)

R2

5).Q�

sin A2 sin B

2 +q

12

�sin4 A

2 + sin4 B2

��≤ (2R−r)(s2+r2−8Rr)−2Rr2

32R3

6).Q�

cos A2 cos B

2 +q

12

�cos4 A

2 + cos4 B2

��≤ (4R+r)3+s2(2R+r)

32R3

Mihaly Bencze

PP29069. Denote Fk and Lk the kth Fibonacci respective Lucas numbers.Prove that

1).nP

k=1

�Fk −

√FkLk + Lk

�2 ≥ Fn+2+Ln+2

2 − 2


2).nP

k=1

�F 2k − FkLk + L2

k

�2 ≥ FnFn+1+LnLn+1

2 − 1

Mihaly Bencze

PP29070. If x ∈�0, π2

�then

1). sinx+ cosx ≥ 1√2+√sinx cosx

2). 1 ≥ sinx cosxq

12 − sin2 x cos2 x

Mihaly Bencze

PP29071. If ak > 0 (k = 1, 2, ..., n) , thenP

1≤i<j≤n

(2a2i−3aiaj+2a2j)(ai+aj)

a4i+a4j≤ (n− 1)

nPk=1

1ak.

Mihaly Bencze

PP29072. If ak > 0 (k = 1, 2, ..., n) , then determine all λ > 0 for whichnP

k=1

ak ≥ λ n

snQ

k=1

ak + (n− λ)

s1n

nPk=1

a2k.

Mihaly Bencze


dx(1+x4)(1+xe)(1+xπ)

.

Mihaly Bencze

PP29074. If A =

�λ+ 1 λ2 + λ+ 1−1 −λ

�where λ ∈ C then

A− I2 −A2018 = O2.

Mihaly Bencze


1n3

P1≤i<j<k≤n

�(n3+i3)(n3+j3)(n3+k3)

n6+i3+j3+k3

� 13

.

Mihaly Bencze


Innn when In =

nR0

�xn + xn+1

�arctgxdx.

Mihaly Bencze


PP29077. Determine all a, b ∈ R for whichP

cyclic

xa1+xb

2x2+x3+...+xn

≥ 2nn+1 .

Mihaly Bencze


k=1

hqn− 1

k(k+1) +qn+ 1

k(k+1) +√ni= n

�√9n− 2

�when [·] denote the

integer part.

Mihaly Bencze

PP29079. Solve the equationnP

k=1

hx

(3k−1)(3k+1)(2k+2)

i= n when [·] denote

the integer part.

Mihaly Bencze

PP29080. Solve the equationnP

k=1

hx2k−1+x2k+1

(2k−1)(2k+1)

i= n+ 1 when [·] denote the

integer part.

Mihaly Bencze

PP29081. If x, y, z > −1 thenP x4+4x3+16x2+24x+16

x3+4x2+6x+4≥ 12.

Mihaly Bencze

PP29082. Let be M(a, b, c) a common point of the plane x+ y + z = λ1,and the sphere x2 + y2 + z2 = λ2

2 not lying on the cartesian axis. Determineall λ1,λ2 ∈ R for which

P aba+b does not depend of the position of point M.

Jose Luis Diaz-Barrero and Mihaly Bencze

PP29083. Prove thatp−1Pk=0

(−1)k cos�kπ+1

p

�∈ R\Q for all p prime numbers.

Mihaly Bencze


1).Q

abc ≤�

4s3

9(s2+r2+4Rr)

�9(s2+r2+4Rr)

2).Q

(s− a)(s−b)(s−c) ≤�

s3

9r(4R+r)

�9r(4R+r)

Mihaly Bencze


PP29085. Solve in Z the equation2 (3x+ 4y + z)3 + 8 + 4yz + 3x+ z = 4y2 + z2.

Mihaly Bencze and Stanciu Neculai


(n!)2nQ

k=1

��1 + 1

k

� 1k+1 − 1

�.

Mihaly Bencze


k=1

arctg 2k+1(k3+3k+1)2

arctg 2k2+6k+3k4+6k3+11k2+6k−1

=

(arctg3)2 − (arctg (n+ 1) (n+ 3))2 .

Mihaly Bencze


p18 + 3x− y2 +

√9− z2 +

p9− 6x+ y2 +

√9z − 3x2 ≤ 12p

18 + 3y − z2 +√9− x2 +

p9− 6y + z2 +

p9x− 3y2 ≤ 12√

18 + 3z − x2 +p9− y2 +

√9− 6z + x2 +

p9y − 3z2 ≤ 12

.

Mihaly Bencze


2tg2 B

2

tgA2tgB

2+3(tg2 B

2tg2 C

2+tg2 C

2tg2 A

2 )≥ 1

3 .

Mihaly Bencze

PP29090. In all triangle ABC holdsP r2ar

2c

ra(r2br2c+3r(r2b+r2c)ra)≥ 1

3r .

Mihaly Bencze

PP29091. Prove thatrQ

p=1

nPk=1

(kr − p+ 1)! can be perfect r power.

Mihaly Bencze

PP29092. Prove that

�nP

k=1

((3k − 2)!)2��

nPk=1

((3k − 1)!)2��

nPk=1

(3k!)2�

can be a perfect cube.

Mihaly Bencze


PP29093. Determine all A ∈ M2 (C) for which

An +An−1 =

�2 (2n− 1) 20170 2

�.

Mihaly Bencze

PP29094. If ak > 0 (k = 1, 2, ..., n) andnP

k=1

ak ≥ n then

Pcyclic

a21a1+a22+a23+...+a2n

≥ n.

Mihaly Bencze

PP29095. If A =

2+√3

2

√32

√32√

32

2+√3

2

√32√

32

√32

2+√3

2

then

An = I3 +(3

√3+2)

n−2n

3·2n

1 1 11 1 11 1 1

.

Mihaly Bencze

PP29096. If x ∈�0, π2

�and (cosx)a = sinx, (sinx)b = cosx then compute

a2 + b2.

Mihaly Bencze

PP29097. In all triangle ABC holdsP wa

ma≥ 4√

6(s2−r2−4Rr)

P bcb+c cosA.

Mihaly Bencze

PP29098. Prove that (n+ 1) (n+ 2) ... (kn) is divisible by kn and is notdivisible by kn+1, when n, k ∈ N∗.

Mihaly Bencze

PP29099. If xk > 0 (k = 1, 2, ..., n) andnP

k=1

xk = n, thenP

cyclic

x1√x2 ≤ n.

Mihaly Bencze



�nP

k=1

xk�+ 1

�

n�

k=1xk

� =nP

k=1

{x}k + 1n�

k=1

{x}x

when [·] and {·} denote the integer, respective the fractional part.

Mihaly Bencze

PP29101. Determine all a ∈ C for whichnP

k=0

�a2n+ aak + 1

�a

= 1a2

�a3n+ a2

a2n+ a

�+ 1

a2

�a2n+ aan+ 1

�+ 1

a

�a2n+ aan+ 1

�a2

.

Mihaly Bencze

PP29102. Determine all a ∈ C for which�anPk=0

(−1)k�ank

��a2n−kan

�ak�a

=�ann

� 12a2.

Mihaly Bencze

PP29103. Solve in Z the equation xa1a2...an − yb1b2...bm = a1a2...anb1b2...bm,where ai, bj ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} (i = 1, 2, ..., n; j = 1, 2, ...,m) .

Mihaly Bencze

PP29104. Solve the following system

{x1}+�x22

= x3+[x4]

2017

{x2}+�x23

= x4+[x5]

2017−−−−−−−−−−{xn}+

�x21

= x2+[x3]

2017

, when [·]

and {·} denote the integer part, respective the fractional part.

Mihaly Bencze

PP29105. Prove that�2n2r

� rPk=0

(−1)k�nk

��n

2r−k

�=

�2nn

� rPk=0

(−1)k�2rk

��2n−2rn−k

�.

Mihaly Bencze

PP29106. Determine all a ∈ C for whichnP

k=0

�a2nak

�a= 1

a2

�a3na2n

�+ 1

a2

�a2nan

�+ 1

a

�a2nan

�a.

Mihaly Bencze


PP29107. Prove that

�nP

k=0

�2n2k

�2��

n−1Pk=0

�2n

2k+1

�2�

= 14

�4n2n

�2 − 14

�2nn

�2.

Mihaly Bencze

PP29108. Prove that

�rP

k=0

�n2k

��n

2r−2k

��r−1Pk=0

�n

2k+1

��n

2r−2k−1

��= 1

4

�2n2r

�2 − 14

�nr

�2when r ≥ 1.

Mihaly Bencze

PP29109. Prove that4nPk=0

(−1)k�4nk

�3=

�6n2n

� 4nPk=0

(−1)k�4nk

�2.

Mihaly Bencze

PP29110. Determine all a ∈ C for whichanPk=0

(−1)k�ank

�a+1= (−1)n

�ann

��(a+1)n

n

�.

Mihaly Bencze

PP29111. Compute

�nP

k=0

(−1)k�nk

�3�2

−2nPk=0

(−1)k�2nk

�3.

Mihaly Bencze

PP29112. Let A,B ∈ Mn (C) such that AB = BA and A2013 = In,B2017 = In. Compute rank (A+B) .

Mihaly Bencze

PP29113. Prove that n2(n+1) <

nPk=1

√2k(2k+1)−

√(2k−1)(2k+2)

k < 1+ 12 + ...+ 1

n .

Mihaly Bencze

PP29114. If A,B ∈ M2 (Q) such that AB = BA, detA = −λ,

det�A+

√λB

�= 0, then compute det

�A2 +B2 −AB

�, where λ > 0.

Mihaly Bencze


PP29115. Determine A ∈ M3 (R) for which Tr (A) = 9 and

A4 =

�3897 4968621 792

�.

Mihaly Bencze


x1 − x22 + x3 = x2 − x23 + x4 = ... = x2n − x21 + x2 = 1.

Mihaly Bencze

PP29117. If x1 = 1 and (1 + xn)xn+1 = xn + 2 for all n ≥ 1 then�xn−1 −

√2� �

xn+1 −√2� �

xn +√2�2

=

=�xn−1 +

√2� �

xn+1 +√2� �

xn −√2�2

for all n ≥ 2.

Mihaly Bencze


sin�π2

�√4n2 + 2n+ 3 +

√4n2 + 10n+ 9

��·

· cos�π2

�√4n2 + 2n+ 3−

√4n2 + 10n+ 9

��

Mihaly Bencze

PP29119. Solve the equation A4 − 4A3 + 5A2 =

�40 8020 40

�and compute

An, where n ∈ N .

Mihaly Bencze

PP29120. Determine all A ∈ M2 (C) for which A2 = 4A− 3I2 and computeAn, where n ∈ N .

Mihaly Bencze

PP29121. If√nxn + n+ 1−

√nxn + 1 ≤

√n2 ≤ √

nxn + n−√nxn for all

n ≥ 1 then compute∞Pn=1

�916 − xn

�2.

Mihaly Bencze



xlg(5x2)1 = 2 · 51−log2 x3

xlg(5x3)2 = 2 · 51−log2 x4

−−−−−−−−−−xlg(5x1)n = 2 · 51−log2 x2

.

Mihaly Bencze


k=1

(−1)k−1 (nk)k2

≥ 1 + 1√2+ ...+ 1

n√n.

Mihaly Bencze

PP29124. Prove that

1).nP

k=0

��nk

�Lk

� 1k+1 ≥ n+1

√L2n

2).nP

k=0

��nk

�2kLk

� 1k+1 ≥ n+1

√L3n when Ln denote the kth Lucas number.

Mihaly Bencze

PP29125. Prove that

1).nP

k=1

(Lk)1k ≥ n

√Ln+3 − 3

2).nP

k=1

�L2k

� 1k ≥ n

√LnLn+1 − 2 where Lk denote the kth Lucas number.

Mihaly Bencze

PP29126. Prove that

1).nP

k=0

��nk

�Fk

� 1k+1 ≥ n+1

√F2n

2).nP

k=0

��nk

�2kFk

� 1k+1 ≥ n+1

√F3n when Fk denote the kth Fibonacci number.

Mihaly Bencze

PP29127. Prove that:

1).nP

k=1

(F2k+1)1k ≥ n

√F2n+2

2).nP

k=1

(F2k)1k ≥ n

√F2n+1 − 1 when Fk denote the kth Fibonacci number.

Mihaly Bencze


PP29128. Prove that�n0

�+q�

n1

�+ 3

q�n2

�+ ...+ n+1

q�nn

�≥ 2.

Mihaly Bencze

PP29129. Prove that

1). F1 +√F2 +

3√F3 + ...+ n

√Fn ≥ n

√Fn+2 − 1

2). F 21 +

pF 22 + 3

pF 23 + ...+ n

pF 2n ≥ n

√FnFn+1 when Fk denote the kth

Fibonacci number.

Mihaly Bencze

PP29130. Determine the function f : R → R for which

nPk=1

xk ≤nP

k=1

5xkf (xk) ≤�

nPk=1

xk

�5

n�

k=1xk

for all xk ∈ R (k = 1, 2, ..., n) .

Mihaly Bencze

PP29131. If xk ∈�18 , 1

�(k = 1, 2, ..., n) then

Pcyclic

logx1

�x24 − 1

64

�≥ 2n.

Mihaly Bencze


x1 (x2x3...xn)k = a1

a2

x2 (x1x3...xn)k = a2

a3−−−−−−−−−xn (x1x2...xn−1)

k = ana1

where ap ∈ R (p = 1, 2, ..., n) and k ∈ N.

Mihaly Bencze

PP29133. If an =�1 + 1

n

�nthen solve in R the following system:

x1√x2x3...xn = a1

x2√x1x3...xn = a2

−−−−−−−−xn

√x1x2...xn−1 = an

.

Mihaly Bencze



3x+ y + z − 28xy = 7�x2 + 4y2

�

x− y + 2z + 4xz = 2�x2 + z2

�

4x− 4y + z + 12xy = 4x2 + 9y2.

Mihaly Bencze

PP29135. In all triangle ABC holds −6− 2√3 ≤P sinA−3

cosA+2 ≤ −6 + 2√3.

Mihaly Bencze

PP29136. Solve in R the following system:�√x1 − 1 +

√2− x2

� �3√x3 − 1 + 3

√2− x4

�= ...

=�√

xn − 1 +√2− x1

� �3√x2 − 1 + 3

√2− x3

�= 1.

Mihaly Bencze

PP29137. Solve in R the following system: 3|x1+1| − 2 |3x2 − 1|− 3x3 =

= 3|x2+1| − 2 |3x3 − 1|− 3x4 = ... = 3|xn+1| − 2 |3x1 − 1|− 3x2 = 2.

Mihaly Bencze


|z1 − |z2 + 1|| = |z3 + |z4 − 1|||z2 − |z3 + 1|| = |z4 + |z5 − 1||−−−−−−−−−−−−−|zn − |z1 + 1|| = |z2 + |z3 − 1||

.

Mihaly Bencze

PP29139. If x ∈�0, π2

�then�

2sinx + 3− cosx + 4− sinx + 6cosx� �

2cosx + 3− sinx + 4− cosx + 6sinx�≥ 16.

Mihaly Bencze


�x21 + 4

�52x

22−10x3+7 = x4�

x22 + 4�52x

23−10x4+7 = x5

−−−−−−−−−−−�x2n + 4

�52x

21−10x2+7 = x3

.

Mihaly Bencze



2x1 + 3x2 + 5x3 + 7x4 = 15x5 + 22x2 + 3x3 + 5x4 + 7x5 = 15x6 + 2−−−−−−−−−−−−−−−2xn + 3x1 + 5x2 + 7x3 = 15x4 + 2

.

Mihaly Bencze


x1x2x3 = x34 + 3x5 − x6 − x7 − x8x2x3x4 = x35 + 3x6 − x7 − x8 − x9−−−−−−−−−−−−−−−xnx1x2 = x33 + 3x4 − x5 − x6 − x7

.

Mihaly Bencze

PP29143. If εn = 1 (ε 6= 1) then

(z − 1)2 + (z − ε)2 +�z − ε2

�2+ ...+

�z − εn−1

�2= nz2 for all z ∈ C.

Mihaly Bencze


1 + log5 x1 = log3�4 +

√5x2

�

1 + log5 x2 = log3�4 +

√5x3

�

−−−−−−−−−−−−−1 + log5 xn = log3

�4 +

√5x1

�.

Mihaly Bencze


√1 + 2x1 +

3√1 + 3x2 +

4√1 + 4x3 +

5√1 + 5x4 = 4 + 4x5√

1 + 2x2 +3√1 + 3x3 +

4√1 + 4x4 +

5√1 + 5x5 = 4 + 4x6

−−−−−−−−−−−−−−−−−−−−−−−−−√1 + 2xn + 3

√1 + 3x1 +

4√1 + 4x2 +

5√1 + 5x3 = 4 + 4x4

.

Mihaly Bencze

PP29146. In all triangle ABC holdsP (a+b)2

3√a2

b2�

3√a2+3

3√c2� ≥ 3.

Mihaly Bencze



4√sinx1 + 2

√cosx2 = log2

81+x2

3

4√sinx2 + 2

√cosx3 = log2

81+x2

4

−−−−−−−−−−−−−4√sinxn + 2

√cosx1 = log2

81+x2

2

.

Mihaly Bencze

PP29148. Determine the functions f : R → R where f (x) = ax2 + bx+ c

such that

f ◦ f ◦ ... ◦ f| {z }

n−time

(x) +

f ◦ f ◦ ... ◦ f| {z }

n−time

(−x) = 0 for all x ∈ R.

Mihaly Bencze


x21 +3px62 + 7x33 =

px44 + 8x5

x22 +3px63 + 7x34 =

px45 + 8x6

−−−−−−−−−−−−−−x2n + 3

px61 + 7x32 =

px43 + 8x4

.

Mihaly Bencze


2sin 3x1 = sin3 x2 + 8sinx3

2sin 3x2 = sin3 x3 + 8sinx4

−−−−−−−−−−2sin 3xn = sin3 x1 + 8sinx2

.

Mihaly Bencze

PP29151. We consider the sequence (an)n≥1 such that a1 = 0, a2 = 1 andall an, an+1, an+2 for n odd are in arithmetical progression and for n even are

in geometrical progression. Prove thatnP

k=2

k2

a2k−1a2k= n−1

n .

Mihaly Bencze

PP29152. If ak =h√

k2 − k + 1 +√k2 + k + 1

iwhen [·] denote the integer

part, then 9 (n+ 1)2�

nPk=1

akak+1

�2

= 4

�nP

k=1

a3k

��n+1Pk=1

a3k

�.

Mihaly Bencze



9�tg2A2 + tg2B2

�+ ctg2C2 ≥ 9.

Mihaly Bencze


x1 (x1 + x2 + ...+ xn)k = 13

x2 (x1 + x2 + ...+ xn)k = 23

−−−−−−−−−−−−xn (x1 + x2 + ...+ xn)

k = n3

when k ∈ N.

Mihaly Bencze

PP29155. If a, b, c ≥ 0, a2 + b2 + c2 = 1, thenP a2

1+2014bc ≥ 32017 .

Mihaly Bencze

PP29156. Computeb−1Pk=1

nk2ab

owhere a, b ∈ N∗ and {·} denote the

fractional part.

Mihaly Bencze

PP29157. Prove that the number n4 + 49n2 + 2027 can be the sum of twoprime number, for all n ∈ N.

Mihaly Bencze

PP29158. If {a1, a2, ..., an} =�13, 23, ..., n3

then solve the following

system:

x1

�nP

k=1

xk

�= a1

x2

�nP

k=1

xk

�= a2

−−−−−−−−xn

�nP

k=1

xk

�= an

.

Mihaly Bencze

PP29159. Let be a0 = x and 5an+1 = a2n − 10an + 50 for all n ≥ 0. Provethat exist infinitely many x ∈ R\Q for which an ∈ N.

Mihaly Bencze


PP29160. Let AkBkCk triangles for which BkAkCk∡ ≥ 90 (k = 1, 2, ..., n)and denote hk the altitude from Ak to side BkCk (k = 1, 2, ..., n) . Prove thatnQ

k=1

1hk

≥nQ

k=1

1AkBk

+nQ

k=1

1AkCk

.

Mihaly Bencze

PP29161. If ma (x, y) =x+y2 , mg (x, y) =

√xy, mh = 2xy

x+y ,

mp (x, y) =q

x2+y2

2 then solve in R the following system:

mh (x1, x2) +mp (x2, x3) = mg (x3, x4) +ma (x4, x5)mh (x2, x3) +mp (x3, x4) = mg (x4, x5) +ma (x5, x6)−−−−−−−−−−−−−−−−−−−−−−−mh (xn, x1) +mp (x1, x2) = mg (x2, x3) +ma (x3, x4)

.

Mihaly Bencze

PP29162. Determine all x ∈ R\Q for which x (x+ 1) and x2 (x+ 2017) areinteger numbers.

Mihaly Bencze

PP29163. If ma (x, y) =x+y2 ,mg (x, y) =

√xy,mh (x, y) =

2xyx+y ,

mp (x, y) =q

x2+y2

2 then solve in R the following system:

mh (x1, x2) +ma (x2, x3) = mg (x3, x4) +mp (x4, x5)mh (x2, x3) +ma (x3, x4) = mg (x4, x5) +mp (x5, x6)−−−−−−−−−−−−−−−−−−−−−−−mh (xn, x1) +ma (x1, x2) = mg (x2, x3) +mp (x3, x4)

.

Mihaly Bencze

PP29164. If x ∈ R then2 +

h�1 +

h1

sin2 x

i�cos2 x

i+��1 +

�1

cos2 x

��sin2 x

�=h

1sin2 x

i+�

1cos2 x

�when

[·] denote the integer part.

Mihaly Bencze

PP29165. If a is odd and natural number not a perfect cube, then for alln,m ∈ N∗ holds {m (a+ 3

√a)} 6= {n (a− 3

√a)} where {·} denote the

fractional part.

Mihaly Bencze


PP29166. Let (an)n≥1 be an arithmetical progression formed by naturalnumbers, and M = {k ∈ N∗|aak = ak+1 + ak + ak−1} . Prove that if2019 ∈ M and M 6= ∅ then M = N∗.

Mihaly Bencze

PP29167. In all triangle ABC holds s2+(4R+r)2

4sR ≥√3 + 3s

2(4R+r) .

Mihaly Bencze

PP29168. If

yn =(Lk)

n+1n +(Lk+1)

n+1n

Lk+2+

(Lk+1)n+1n +(Lk+2)

n+1n

Lk+3+

(Lk+2)n+1n +(Lk)

n+1n

Lk+Lk+2, where Lk

denote the kth Lucas number, then yn−1 + yn+1 ≤ 2yn for all n≥ 2 and k ≥ 1.

Mihaly Bencze

PP29169. Determine all ak > 0 (k = 1, 2, ..., n) if a1 = 1 anda5113

+a5223

+ ...+ a5nn3 = 2n+1

6 anan+1.

Mihaly Bencze

PP29170. If xn =Fn+1k

+Fn+1k+1

Fnk+Fn

k+1+

Fn+1k+1 +Fn+1

k+2

Fnk+1+Fn

k+2+

Fn+1k+2 +Fn+1

k

Fnk+2+Fn

kwhen Fk denote the

kth Fibonacci number, then xn−1 + xn+1 ≤ 2xn for all n ≥ 2 and k ≥ 1.

Mihaly Bencze

PP29171. Prove that (an)n≥1 is an arithmetical progression if and only if2 (a2 + a3 + ...+ an−1) = (n− 2) (a1 + an) for all n ≥ 3.

Mihaly Bencze


k=1

{kx} = 1 when {·} denote the

fractional part.

Mihaly Bencze


k=1

�k2x

= 1 when {·} denote the

fractional part.

Mihaly Bencze


PP29174. If xk > 0 (k = 1, 2, ..., n) such thatnP

k=1

1xk

≤ n thennP

k=1

x2k ≥ n.

Mihaly Bencze


k=3

p25 + (k − 3) (k − 1) (k + 2) (k + 4) = 2n3+6n2−39n+36

6 .

Mihaly Bencze

PP29176. Prove thathpn (n+ 1) +

pn (n+ 2) +

p(n+ 1) (n+ 2)

i= 3n+ 2 for all n ∈ N∗,

where [·] denote the integer part.

Mihaly Bencze

PP29177. If ak > 0 (k = 1, 2, ..., n) , a1 = 1 and1

a1+a2+ 1

a2+a3+ ...+ 1

an−1+an= an − 1 for all n ≥ 1 then

2�

1an+1

− 1�< a1 + a2 + ...+ an < 2

an.

Mihaly Bencze

PP29178.Prove thatnP

k=1

h(k+1)2

k+2

i h(k+2)2

k+3

i= n(n+1)(n+2)

3 where [·] denote the

integer part.

Mihaly Bencze

PP29179. If r, n,m, p ∈ N ∗ then

r+1Pk=1

hkn+1

kn+kn−1+...+k+1

i hkm+1

km+km−1+...+k+1

i hkp+1

kp+kp−1+...+k+1

i= r2(r+1)2

4 , where

[·] denote the integer part.

Mihaly Bencze

PP29180. If x ∈ R then1+27tgx

3(1+3tgx+9ctg4x)+ 1+27ctgx

3(1+3ctgx+9tg4x)+

81(tg3x+tg3y)1+81(tgx+ctgx) ≥ 2.

Mihaly Bencze


PP29181. Determine all r, s ∈ R\Q for which�rs

�n+�sr

�n ∈ N for alln ∈ N.

Mihaly Bencze

PP29182. Prove that

π2R0

2n√cosxdx2n√sinx+ 2n√cosx

= π4 when n ∈ N∗.

Mihaly Bencze

PP29183. Determine all n ∈ N for whichnP

k=1

k5nk is divisible by 5.

Mihaly Bencze

PP29184. Solve in R the equationp3 (2x+ 3) (3y + 5) +

p6 (x+ 2) (y + 1) +

p2 (x+ 1) (3y + 1) =

=p

5 (2x+ 3) (9y + 8).

Mihaly Bencze

PP29185. If x, y > 0 and x2 + y2 ≤ 1 then1√

x2+2√2y+3

+ 1√y2+2

√2x+3

+ 1√9+xy

≥ 1.

Mihaly Bencze

PP29186. Let f : R → R be a continuous convex function andλ ∈ [0, 1] , a, b > 0. Prove that

λa

aR0

f (x) dx+ 1−λb

bR0

f (x) dx ≥ 1λa+(1−λ)b

λa+(1−λ)bR0

f (x) dx.

Mihaly Bencze

PP29187. Let f : R → R be a continuous function and

λk > 0, ak > 0 (k = 1, 2, ..., n) such thatnP

k=1

λk = 1. Prove that

nPk=1

λk

ak

akR0

f (x) dx ≥ 1n�

k=1

λkak

n�

k=1λkakR0

f (x) dx.

Mihaly Bencze


PP29188. Determine all k ∈ N∗ for which (k − 1)! is divisible by k+1.

Mihaly Bencze

PP29189. If P0 = 0, P1 = 1 and Pn = 2Pn−1 + Pn−2 for all n ≥ 2, then

compute limn→∞

nPk=1

arctg2(Pk+Pk+1)Pk+1

PkP2k+1Pk+2−1

· arctg�

2P 2k+1

1+PkP2k+1Pk+2

�.

Mihaly Bencze and Jose Luis Dıaz-Barrero

PP29190. Determine all p ∈ N∗ such that limn→∞

nPk=1

12n+pk−1 = ln

√2.

Mihaly Bencze


P aa+2b < 4

3 − 23Rr6s2

2).P −a+b+c

a−b+3c < 43 − 23r2

3s

3).P ra

ra+2rb< 4

3 − 23s2r3(4R+r)3

Mihaly Bencze

PP29192. If a1 = 1 and an+1 (1 + nan) = an for all n ≥ 1, and

λ = limn→∞

1n4

�1a1

+ 2a2

+ ...+ nan

�, then compute n

�λ− 1

n4

nPk=1

kak

�.

Mihaly Bencze

PP29193. Determine all A,B,C ∈ M3 (C) for whichdet

�A3 +B3 + C3

�= 1, AB = C,BC = A,CA = B and A2 +B2 + C2 = I3.

Mihaly Bencze

PP29194. Determine sinAk the subunitary root of the equation2 (ak + 2)x2 − 2 (ak + 1) (ak + 1)x− ak = 0, where k ∈ {1, 2, 3}. Determineall a1, a2, a3 ∈ R for which A1 +A2 +A3 = π.

Mihaly Bencze

PP29195. Compute

1).1R0

arctgxx+1 dx 2).

1R0

arctgx·arctg(x+1)(x+1)(x+2) dx

Mihaly Bencze



3y+5z + y3

3z+5x + z3

3x+5y = 32 .

Mihaly Bencze

PP29197. If xn = 1k

k + 2k

k2+ ...+ nk

kn when k ∈ N, k ≥ 2, then compute [xn],where [·] denote the integer part.

Mihaly Bencze and Nicolae Papacu

PP29198. If xn > 0, n ∈ N∗ andnP

k=1

xk+1

k2xk< n

n+1 , then the sequence (xn)n≥1

is convergent.

Mihaly Bencze


loga x+ loga+1 (y + 1) = 2 loga+2 (z + 2)loga y + loga+1 (z + 1) = 2 loga+2 (x+ 2)loga z + loga+1 (x+ 1) = 2 loga+2 (y + 2)

.

Mihaly Bencze

PP29200. Let ABC be a triangle, then determine all x, y > 0 for which16P �

cos A2

�x+y �sin B

2

�x �sin C

2

�y ≥ 3sR .

Mihaly Bencze

PP29201. Determine all a, b, c ∈ R for which the equationsx2 + (a+ c)x+ b2 − c = 0, x2 + (b+ a)x+ c2 − a = 0,x2 + (c+ b)x+ a2 − b = 0 have real roots.

Mihaly Bencze


k=1

tg2 π3(2k+1) ≤

n(16n2+48n+17)3 .

Mihaly Bencze


k=1

ln(1+√k−1)

k2(k+1)< n

2(n+1) .

Mihaly Bencze


PP29204. Determine all a ∈ N, a ≥ 2 for which an − 1 is divisible bya(n+ 1) for an infinity of n ∈ N.

Mihaly Bencze

PP29205. Determine all a, b, c ∈ Z for which a+b+c−abc1+ab+bc+ca ∈ Z.

Mihaly Bencze


k=1

sin πk(k+1)n ≥ n

n+1 sinπn

Mihaly Bencze

PP29207. If zk ∈ C∗ (k = 1, 2, ..., n) such that|z1| = |z2| = ... = |zn| = |z1 + z2 + ...+ zn| then compute

card

��z1z2

�t+�z2z3

�t+ ...+

�znz1

�t|t ∈ N

�.

Mihaly Bencze

PP29208. If k ∈ N ∗ then solve in R the equation

[kx] +�

1kx

= {(k + 1)x}+

h1

(k+1)x

iwhere [·] , respective {·} denote the

integer respective the fractional part.

Mihaly Bencze

PP29209. If ak ∈ Q (k = 1, 2, ..., 2n) such thatan = 1

n (a1 + a3 + ...+ a2n−1) and an+1 =1n (a2 + a4 + ...+ a2n) , then

compute2nPk=1

1a2k

.

Mihaly Bencze

PP29210. Determine all prime p and all k ∈ N for which (p+ 1)k+1 − pk+1

is a perfect k power.

Mihaly Bencze

PP29211. Solve in Z the equation (x− y) (1 + xy) =�1− x2

� �1− y2

�.

Mihaly Bencze


PP29212. Let ABC be a triangle such that x cosA+ y cosB = 1 andy sinA+ x sinC = (x+ y) sinA. Determine all x, y > 0 for which ABC isequilateral.

Mihaly Bencze

PP29213. In all triangle ABC holds a2x+ b2y + c2z ≤ 4p

x2 + y2 + z2srfor all x, y, z > 0.

Mihaly Bencze


a2 ≥ sr

�4√

�

xy�

x

�for all x, y, z > 0.

Mihaly Bencze

PP29215. In all triangle ABC holds�a2xy + b2yz + c2zx

�2 �a2x+ b2y + c2z

�2 ≥≥ 256xyz (x+ y + z) (xy + yz + zx) (sr)4 for all x,y,z¿0.

Mihaly Bencze


a2eb+c + b2ec+a + c2ea+b ≥ 4√3sre

2(a+b+c)3 .

Mihaly Bencze


a2b� �P

a2c�≥ 16 (

Pab)2 (sr)2 .

Mihaly Bencze


a2bλ� �P

a2cλ�≥ 16

�P(ab)λ

�2(sr)2 for all λ ≥ 0.

Mihaly Bencze


Pa ≥ 4s

pr2R

2).P

a2 ≥ 4√3sr

3).P

a3 ≥ 4√s2 + r2 + 4Rrsr

4).P

a2 (s− b) ≥ 4pr (4R+ r)sr

5).P

a2rb ≥ 4s2r

6).P �

a sin A2

�2 ≥ srR

√s2 + r2 − 8Rr


7).P �

a cos A2

�2 ≥ srR

qs2 + (4R+ r)2

Mihaly Bencze


a2 sinA ≥ 2srR

√s2 + r2 + 4Rr.

Mihaly Bencze


a2 cosA ≥ 2srR

√s2 + r2 − 4R2.

Mihaly Bencze

PP29222. In all triangle ABC holds a2ea + b2eb + c2ec ≥ 4√3sre

a+b+c3 .

Mihaly Bencze


r2 + r2a + r2b + r2c ≥ 16R2 − 31−λ�a

2λ + b

2λ + c

2λ

�λfor all λ ∈ [0, 1] .

Mihaly Bencze


Pa2tgA

2 ≥ 4sr

2).P

a2ma ≥ 4s2rq

2rR

Mihaly Bencze

PP29225. In all triangle ABC holds 36r2 ≤ (P

HA) (P

IA) ≤ 6R (2R− r)(A new refinement of Euler’s R ≥ 2r inequality).

Mihaly Bencze


HAλ +P

IAλ ≤ 3

�Rλ +

�2(2R−r)

3

�λ�

for all λ ∈ [0, 1] .

Mihaly Bencze

PP29227. Let ABC be a triangle and denote F the area of the triangleformed by sides ma,mb,mc. Prove that

F ≤ 5(s2−r2−4Rr)2−3(s2+r2+4Rr)

2+48s2Rr

16sr .

Mihaly Bencze


PP29228. In all acute triangle ABC holdsP �

cosAsinB sinC

�λ ≤ 3�23

�λfor all

λ ∈ [0, 1] .

Mihaly Bencze

PP29229. Let ABC be a triangle and denote F the area of the triangleformed by sides cos A

2 , cosB2 , cos

C2 . Prove that

F ≤ 116sr

P �b2 + c2 − a2

�cos2 A

2 .

Mihaly Bencze

PP29230. Consider the triangles AkBkCk with sides ak, bk, ck and areasFk = Area [AkBkCk] (k = 1, 2, ..., n) , then

16

�nQ

k=1

Fk

� 2n

+ 2

�nQ

k=1

ak

� 4n

+ 2

�nQ

k=1

bk

� 4n

+ 2

�nQ

k=1

ck

� 4n

≤

≤nQ

k=1

�a2k + b2k + c2k

� 2n for all n ∈ N∗, n ≥ 2 (A generalization of Padoe’s

inequality).

Mihaly Bencze


1).P

a2u ≤ 31−u�8R2 + 4sr

3√3

�u

2).��

8R2�v

+�

4sr3√3

�v� 1v ≥ 2

1v−1

�Pa2�for all u ∈ [0, 1] and v ≥ 1.

Mihaly Bencze


aλ ≥ 31−λ (2s)λ ≥ 31−λ2

�4√3Area [ABC]

�λ2 for all λ ≥ 1.

Mihaly Bencze

PP29233. In all triangle ABC holds s2λ + 5r2λ ≥ 6�8Rr3

�λfor all λ ≥ 0.

Mihaly Bencze

PP29234. In all triangle ABC holds 2Rλ +�3√3− 4

�rλ ≥

�3√3− 2

�1−λsλ

for all λ ≥ 0.

Mihaly Bencze


PP29235. In all triangle ABC holdsxaλ

b+c +ybλ

c+a + zcλ

a+b ≥ (xa+yb+zc)λ

(x+y+z)λ−2((y+z)a+(z+x)b+(x+y)c)for all x, y, z > 0 and

λ > 0.

Mihaly Bencze

PP29236. In all triangle ABC holds wλa + wλ

b +mλc ≤ 31−

λ2 sλ for all

λ ∈ [0, 1] .

Mihaly Bencze


hλa≥ 3

(3r)λ≥ 3

�3

qR

2s2r2

�λ

for all

λ > 0.

Mihaly Bencze


3 (3r)λ ≤ 3�

3√s2r

�λ≤P

�a cos A

2

�λ ≤ 31−λ�2�s2 − r2 − 4Rr

��λ2�4R+r2R

�λ2 ≤

≤ 3Rλ�4R+r2R

�λ2 ≤ 3

�3R2

�λfor all λ ∈ [0, 1] .

Mihaly Bencze


2 +P √

s−a+√s−b√

s−a≤ 2

s2r

q2 (s2 −m2

a)�s2 −m2

b

�(s2 −m2

c).

Mihaly Bencze


s+ 3Pp

(s− a) (s− b) ≤ 2s

Pq(s2 −m2

a)�s2 −m2

b

�.

Mihaly Bencze


(cos (B − C))λ ≤ 31−λ�P ha

ma

�λfor all λ ∈ [0, 1] .

Mihaly Bencze

PP29242. In all triangle ABC holds√2Pp

s (s− a) ≤Pps2 −m2

a.

Mihaly Bencze


PP29243. In all triangle ABC holds 18 ≤ s2+r2+4RrRr ≤ 9R

r (A newrefinement of Euler’s R ≥ 2r inequality).

Mihaly Bencze


(ab)2λ ≥ 3�8s2Rr

3

�λ≥ 3

�163 Area [ABC]

�λfor all λ > 0.

Mihaly Bencze


3�2s2r2

R

�λ3 ≤Phλa ≤Pmλ

a ≤ 31−λ�2s−

�6√3− 9

�r�λ ≤

≤ 31−λ (4R+ r)λ ≤ 31−λ�9R2

�λfor all λ ∈ [0, 1] .

Mihaly Bencze

PP29246. In all triangle ABC holds�2s3

�λ ≤�

(a sin A2 )

λ

�

(sin A2 )

λ ≤ 12 max

naλ + (b+ c)λ , bλ + (c+ a)λ , cλ + (a+ b)λ

ofor

all λ ≥ 0.

Mihaly Bencze

PP29247. In all triangle ABC holds7R2 ≤ (4R+ r)2 + r2 − 2s2 = r2 +

Pr2a ≤ 16R2 − 4s2

3 .

Mihaly Bencze


(a cosA)λ ≤ 31−λ�2srR

�λ ≤ 31−λ�

2s2

3√3R

�λ≤ 31−λsλ for all λ ∈ [0, 1] .

Mihaly Bencze

PP29249. In all triangle ABC holds Rλ + rλ ≤ 21−λmax�hλa , h

λb , h

λc

for all

λ ∈ [0, 1] (A generalization of Erdos inequality).

Mihaly Bencze

PP29250. In all triangle ABC holds a2b2c2 ≥ 4s2

27R2a2b2c2 ≥

�4Area[ABC]√

3

�3.

Mihaly Bencze


PP29251. In all acute triangle ABC holds s2−(2R+r)2

4R2 =Q

cosA ≤ r2

2R2 ≤ 18 .

Mihaly Bencze

PP29252. In all triangle ABC holds1). 512s2Rr ≤ 27

�s2 + r2 + 2Rr

�2

2). 4 (4R+ r)3 r ≤ 27s2R2

3). 16�s2 + r2 + 4Rr

�3 ≤ 27s2�s2 + r2 + 2Rr

�2

4). 512 (2R− r)3 r2 ≤ 27�(2R− r)

�s2 + r2 − 8Rr

�− 2Rr2

�2

5). 512 (4R+ r)3 s2 ≤ 27�(4R+ r)3 + s2 (2R+ r)

�2

Mihaly Bencze


(sin A2 )

2n ≥P 1

(sin A2sin B

2 )n ≥ 3

�4Rr

� 2n3 ≥ 3 · 4n for all n ∈ N.

Mihaly Bencze

PP29254. In all triangle ABC holds 18 ≥Q sin A+B

4 ≥ 14

pr2R . (A


Mihaly Bencze

PP29255. If x, y, z ∈�0, π4

�then

tgxtgytgztg3�x+y+z

3

�≤ tg2

�x+y2

�tg2

�y+z2

�tg2

�z+x2

�.

Mihaly Bencze


(sin A2 )

n ≥ 2P 1

(sin A+B4 )

n − 3 · 2n ≥ 3 · 2n for all n ∈ N.

Mihaly Bencze


8x2−12x+1√x2−2x

dx ≥ 4 (b− a) (b+ a− 1) .

Mihaly Bencze


8x3−28x2+25x−2(x−2)

√x−2

dx ≥ 2 (b− a) (b+ a− 2) .

Mihaly Bencze



Pha cosA ≤ s2−r2−4Rr

2R

2).P

ra cosA ≤ s2−2r(4R+r)2r

3).P

sin2 A2 cosA ≤ 1

2

�s2+r2−8Rr

4Rr

�2+ r

2R − 1

4).P

cos2 A2 cosA ≤ (s2+(4R+r)2)

2−16Rs2(4R+r)

32s2R2

Mihaly Bencze


a�b2 + c2

�≥ 9abc+

Pa3.

Mihaly Bencze


am2a = s

2

�s2 + 5r2 + 2Rr

�.

Mihaly Bencze

PP29262. In all triangle ABC holdsP a2λ

mbmc≥ 4λ91−λ (

Pmamb)

λ−1 for allλ > 0.

Mihaly Bencze


a3

ra

�λ≤ 3

�4sR3

�λfor all λ ∈ [0, 1] .

Mihaly Bencze

PP29264. Let ABC be a triangle in which A∡ = 90◦ and x, y > 0. Provethat xyha ≤ (xb+ yc)min {x, y}++1

2

�x2 + y2 −min

�x2, y2

− 2min {x, y}

px2 + y2

�a.

Mihaly Bencze


(sinA)λ12 (cosB)

λ22 ≤

≤r

32−λ1−λ2

�2 cos A

2 − 3√3

2

�λ1 �2P

sin A2 − 3

2

�λ2 ≤ 31+λ14

2λ1+λ2

2

for all

λ1,λ2 ∈ (0, 1) .

Mihaly Bencze

PP29266. In all triangle ABC holdsP aλ+1

ra≤ abc

r

�12s

Paλ−1

�for all λ ≥ 0.

Mihaly Bencze


PP29267. In all triangle ABC holds 4 ≤�

s2+r2−4R2

s2−(2R+r)2

�2− 8R(R+r)

s2−(2R+r)2≤

�Rr

�2

(A new refinment of Euler’s R ≥ 2r inequality).

Mihaly Bencze

PP29268. Solve in Q the equation(11x + 12x + 13x + 14x) (31y + 33y + 35y + 37y + 39y + 41y) = 1320x+y.

Mihaly Bencze

PP29269. Solve in Q the equation(3x + 4x + 5x) (11y + 12y + 13y + 14y) (31z + 33z + 35z + 37z + 39z + 41z) == 7920x+y+z.

Mihaly Bencze

PP29270. If a, b, c > 0 then 2abc+ (P

a)�P

a2�≥ 11

27 (P

a)3 .

Mihaly Bencze

PP29271. Solve in Q the equation(1x + 2x + ...+ 24x) (15y + 16y + ...+ 34y) = 70x+y.

Mihaly Bencze

PP29272. If x ≥ 0 then�524288x16 + 2048

�16 ≥ 256(2048x16+256x15+4480x13+17472x11+22880x9++11440x7 + 2184x5 + 140x3 + 2x− 8)16.

Mihaly Bencze

PP29273. If x ≥ 0 then 9�3x3 + 3x2 + 2x+ 1

�3+9

�9x3 + 6x2 + 3x+ 1

�3 ≥≥

�3x3 + 6x2 + 5x+ 1

�3+�9x3 + 15x2 + 6x+ 1

�3.

Mihaly Bencze

PP29274. If x, y ≥ 0 then 4�3x3y + 9y4

�3+ 4y12 ≥

�10x4 + 9xy3

�3.

Mihaly Bencze

PP29275. If x ≥ 0 then�272x8 + 17

�8 ≥ 17�16x8 + 16x7 + 56x5 + 28x3 + 2x− 1

�8.

Mihaly Bencze



�√x2+1−

√y2+1

�2

(x2+1)(y2+1)≥ π

4 − 12e

1−π4 .

Mihaly Bencze

PP29277. If a ∈ C then�a4 + 2ai

�3+�1 + a3i

�3+�3a2 + 5ai− 5

�3+�4a2 − 4ai+ 6

�3+

+�5a2 − 5ai− 3

�3=

�1 + 2a3i

�3+�a4 − ai

�3+�6a2 − 4ai+ 4

�3.

Mihaly Bencze


k!(n+k)! =

1(n−1)(n−1)! for all n ≥ 2.

Mihaly Bencze

PP29279. Prove that

� ∞Pk=0

cos kxk!

�2

= e2 cosx +

� ∞Pk=0

sin kxk!

�� ∞Pk=0

(−1)k sin kxk!

�, for all x ∈ R.

Mihaly Bencze

PP29280. Prove that for all n ∈ N holds 2he2+12e (2n)!

i= [e (2n)!] +

h(2n)!e

i

when [·] denote the integer part.

Mihaly Bencze

PP29281. Prove that [e (2n+ 2)!]−h(2n+2)!

e

i= 4 (n+ 1)

he2−12e (2n+ 1)!

i

for all n ∈ N where [·] denote the integer part.

Mihaly Bencze


e(sinπ2x)

2n+(cos π

2x)

2n

dx ≤

≤ e2(2n−1)!!

(2n)!! + maxx,y∈[0,1]

qe(sin

π2x)

2n+(cos π

2x)

2n

−q

e(sinπ2y)

2n+(cos π

2y)

2n

!2

.

Mihaly Bencze



e(sinπ2x)

2n+1+(cos π

2x)

2n+1

dx ≤ e4(2n)!!

π(2n+1)!!+

+ maxx,y∈[0,1]

qe(sin

π2x)

2n+1+(cos π

2x)

2n+1

−qe(sin

π2y)

2n+1+(cos π

2y)

2n+1

!2

, for all

n ∈ N.

Mihaly Bencze

PP29284. Prove that

1).1R0

ecosπxdx ≤ maxx,y∈[0,1]

�√ecosπx −

√ecosπy

�2+ 1

2).1R0

esinπxdx ≤ maxx,y∈[0,1]

�√esinπx −

√esinπy

�2+ e

2π

Mihaly Bencze

PP29285. Solve in Z the equation x+2yy+3z + y+2z

z+3x + z+2xx+3y = 9

4 .

Mihaly Bencze


�x+ 1

x

�xdx ≤ 2

√e+ max

x,y∈[0,1]

��x+ 1

x

�x2 −

�y + 1

y

� y2

�2

.

Mihaly Bencze

PP29287. If a ∈ [0, 1] then ea−1a ≤

√ea + max

x,y∈[0,1]

�√eax −

√eay

�2.

Mihaly Bencze

PP29288. Prove that

1). πn (n+ 1) ≤nP

k=1

�k!�ek

�k�2≤ e2n(n+1)

2

2). 2π2n(n+1)(2n+1)3 ≤

nPk=1

�k!�ek

�k�4≤ e4n(n+1)(2n+1)

6

3). 2π3n2 (n+ 1)2 ≤nP

k=1

�k!�ek

�k�6≤ e6n2(n+1)2

4

Mihaly Bencze


PP29289. Prove that2(

√n+1−1)e <

nPk=1

�ke

�k · 1k! <

√nπ .

Mihaly Bencze


k=0

12k+1(n−k)!

=√e

2n+1n!Γ�n+ 1, 12

�.

Mihaly Bencze

PP29291. If x ∈�0, π2

�then

�sin2 x

� �9−sinx3−sinx

� 3sin x

+�cos2 x

� �9−cosx3−cosx

� 3cos x ≥ 82− 18 (sinx+ cosx) .

Mihaly Bencze


k=1

kq

n2 (n+ 1)2 − k2 (k + 1)2 < πn2(n+1)2

8 .

Mihaly Bencze

PP29293. Solve in Z the equation x+y2

z2+x+ y+z2

x2+y+ z+x2

x+z2= 3.

Mihaly Bencze

PP29294. Solve in Z the equationx(y2+z2)

y+z +y(z2+x2)

z+x +z(x2+y2)

x+y = 3t2.

Mihaly Bencze

PP29295. If xk ≥ 3 (k = 1, 2, ..., n) , then

nQk=1

(3xk−1)xk−2

(xk−1)xk≥

�

3n

n�

k=1xk−1

�

n�

k=1xk−2n

�

1n

n�

k=1xk−1

�

n�

k=1xk

.

Mihaly Bencze

PP29296. If xk ≥ 3 (k = 1, 2, ..., n) , thennP

k=1

�3xk−1xk−1

�xk ≥ 1n

�3

nPk=1

xk − n

�2

.

Mihaly Bencze


PP29297. If a, b, c ≥ 3 then(3a−1)a−2

(a−1)a· (3b−1)b−2

(b−1)b· (3c−1)c−2

(c−1)c· (a+b+c−1)a+b+c−6

(a+b+c3

−1)a+b+c ≥

≥�

3(a+b)2

−1�a+b−4

( a+b2

−1)a+b ·

�

3(b+c)2

−1�b+c−4

( b+c2

−1)b+c ·

�

3(c+a)2

−1�c+a−4

( c+a2

−1)c+a .

Mihaly Bencze

PP29298. If a, b > 0 and 6a = b�2a2 + 2 + ab

�then 9

8 ≤ 2a2+b2

a(a+2b) ≤65 .

Mihaly Bencze

PP29299. If a, b > 0 and 2 (2a+ b) = a�ab+ 2b2 + a2

�then

2a2�2 + a2

�≥ (1 + ab)

�2 + a2 − b2

�.

Mihaly Bencze

PP29300. Determine all triangles ABC such that2s ≥

�2√3− 1

�R+ 2

�√3 + 1

�r.

Mihaly Bencze

PP29301. If a, b > 0 then�a4 + 2a3b+ b4

� �a4 + 2ab3 + b4

�≥ 4a3b3 (a+ b)2 .

Mihaly Bencze


k=3

�3k−1k−1

�k≥ 3n(n+1)(2n−1)

2 + n− 29.

Mihaly Bencze

PP29303. In all triangle ABC holds�10R2 −P a2

�Pa3 cosA = 4R2

�Pa2 − 6R2

�Pa cos3A.

Mihaly Bencze

PP29304. If a, b, c ∈ (0, 1] and abc = 1, then 5P

a+ 11+

�

ab ≥ 16.

Mihaly Bencze

PP29305. Determine all a, b ∈ N for which (a+ 1)b + a2 − a+ 1 is aninteger power of 3.

Mihaly Bencze



a3 ≥P a3 cos (B − C) ≥ 3abc.

Mihaly Bencze

PP29307. Determine all functions f : R → R for which2f (xy + yz + zx) = f2 (x+ y + z)+ f2 (x)+ f2 (y)+ f2 (z) for all x, y, z ∈ R.

Mihaly Bencze

PP29308. If x, y, z > 0 then

3P

(x+ y)2 ≥�P

x+pP

x2�2

+ 2 (P

x)pP

x2.

Mihaly Bencze

PP29309. If a, b, c > 0 then ln (1 + a) ln (1 + b) ln (1 + c) ln3�1 + 3

√abc

�≤

≤ ln2�1 +

√ab�ln2

�1 +

√bc�ln2 (1 +

√ca) .

Mihaly Bencze

PP29310. Find all function f : N → N such that for all a, b, c ∈ N holds(f (a) + b+ c) (a+ f (b) + c) (a+ b+ f (c)) == (f (a+ b) + c) (f (b+ c) + a) (f (c+ a) + b) .

Mihaly Bencze


k=1

1k3(k+1)3

≥ 16(√n+1−1)

4

n3(n+1)2.

Mihaly Bencze

PP29312. If a, b > 0 then�a√b+ b

√a+ a

qa+b2

��√a+

√b�√

a+ b+ a√2�+

+

�a√b+ b

√a+ b

qa+b2

��√a+

√b�√

a+ b+ b√2�≤

≤�5a2 + 8ab+ 5b2

�√a+ b.

Mihaly Bencze

PP29313. In all triangle ABC holds arctg π3−ABCπ(π2−

�

AB)≥ π3

4(2π2−�

AB).

Mihaly Bencze


PP29314. If a0 = 1, b0 = 2, c0 = 3 and an+1 = an + 1bn

+ 1cn,

bn+1 = bn + 1cn

+ 1an, cn+1 = cn + 1

an+ 1

bnfor all n ≥ 0, then compute

max {an, bn, cn} .

Mihaly Bencze

PP29315. In all acute triangle ABC holds�2rR

�2s ≥Q (cos (B − C))a .

Mihaly Bencze

PP29316. Let ABC be a triangle. Determine all x, y > 0 for which2r ≤P 1

wa≤ x

r + yR .

Mihaly Bencze

PP29317. In all acute triangle ABC holds�3√3 +

PctgA

�(P

tgA) ≥ 36.

Mihaly Bencze

PP29318. If a, b, c > 0 thenPq

a+b+2ca+b =

qQ a+b+2ca+b + 4

�

a�

√(a+b)(a+b+2c)

.

Mihaly Bencze

PP29319. If ak ≥ 1 (k = 1, 2, ..., n) and λ ≥ 0, then

(n+ λ+ 1)

�nP

k=1

ak

� Pcyclic

1λa1+a2+a3+...+an

!≤ n2 + λ

P1≤i<j≤n

|ai − aj | .

Mihaly Bencze

PP29320. If x, y, z > 0 thenPq

x+yz ≥

√2 +

q(x+y)(y+z)(z+x)

xyz .

Mihaly Bencze

PP29321. If x, y, z > 0 then

Pqx+yz ≤

q(x+y)(y+z)(z+x)

xyz +√2(x+y+z)

4√xyz(

√x+

√y+

√z).

Mihaly Bencze


PP29322. If xk > 0 (k = 1, 2, ..., n) andnP

k=1

xk = n, then

Pcyclic

x31

3�

4(x62+1)

≥ n2 .

Mihaly Bencze

PP29323. If ak > 0 (k = 1, 2, ..., n) and λ ∈ (0, 1] such thatnP

k=1

1ak+λ ≥ 1

then (n− λ− 1)nP

k=1

a2k + λ (n− λ)nP

k=1

ak ≥ 2P

1≤i<j≤naiaj .

Mihaly Bencze

PP29324. If x, y, z > 0 then 2�P

x2�2

+ 12P

xy ≥ 9P

xy2 + 9P

x.

Mihaly Bencze

PP29325. If a, b, c > 0 and abc+P

ab ≤ 4 thenP

a ≥P ab.

Mihaly Bencze

PP29326. Denote Fk and Lk the kth Fibonacci, respective Lucas numbers.Prove that

1).nP

k=1

F 6k+1

(3F 2k−4Fk+3)

2 ≥ 14 (3FnFn+1 − 4Fn+2 + 3n+ 4)

2).nP

k=1

L6k+1

(3L2k−4Lk+3)

2 ≥ 14 (3LnLn+1 − 4Ln+2 + 3n+ 6)

Mihaly Bencze

PP29327. If x ∈�0, π2

�then

1). 1+sin6 x

(3 sin2 x−4 sinx+3)2 + 1+cos6 x

(3 cos2 x−4 cosx+3)2≥ 9− 4 (sinx+ cosx)

2). 1+sin12 x

(3 sin4 x−4 sin2 x+3)2 + 1+cos12 x

(3 cos4 −4 cos2 x+3)2≥ 5− 6 sin2 x cos2 x

Mihaly Bencze

PP29328. If xk > 0 (k = 1, 2, ..., n) thenP 3

√1+x6

1

3x22−4x2+3

≥ n3√4

.

Mihaly Bencze



x2dx(3x2−4x+3)3

≥ 112arctg

b3−a3

1+(ab)3.

Mihaly Bencze


x5dx(3x2−4x+3)3

≥ 124 ln

b6+1a6+1

.

Mihaly Bencze


k=1

k6+1(3k2−4k+3)2

≥ n(2n2−n+3)8 .

Mihaly Bencze

PP29332. If xk > 0 (k = 1, 2, ..., n) , then

4nnP

k=1

x6k+1

3x2k−4xk+3

≥�3

nPk=1

x2k − 4nP

k=1

xk + 3n

�2

.

Mihaly Bencze

PP29333. If xk > 0 (k = 1, 2, ..., n) , thenP

cyclic

x61+1

(3x21−4x1+3)(3x2

2−4x2+3)2 ≥ n

4 .

Mihaly Bencze

PP29334. If x ∈�0, π2

�, then

1). 32 + 16�sin6 x+ cos6 x

�≥ (9− 4 (sinx+ cosx))3

2). 32 + 16�sin12 x+ cos12 x

�≥

�5− 6 sin2 x cos2 x

�3

Mihaly Bencze

PP29335. In all triangle ABC holdsP 64 sin6 A+9�

4 sin2 A− 8√3sinA+3

�2 ≥ s2−r(4R+r)2R2 − 2s√

3R+ 9

4 .

Mihaly Bencze

PP29336. In all acute triangle ABC holdsP 64 cos6 A+1

(12 cos2 A−8 cosA+3)2≥ 13

4 − 2rR − 3(s2−(2R+r)2)

2R2 .

Mihaly Bencze


PP29337. If xk > 0 (k = 1, 2, ..., n) then

nPk=1

�xk +

4(x6k+1)

(3x2k−4xk+3)

2

��x2k + xk + 1

�≥ 3

n3

�nP

k=1

xk

�4

+ 3n

�nP

k=1

xk

�2

+ 3n.

Mihaly Bencze

PP29338. If xk > 0 (k = 1, 2, ..., n) , thenP x2

1

3�

4(x62+1)

≥

�

n�

k=1xk

�2

3n�

k=1x2k−4

n�

k=1xk+3n

.

Mihaly Bencze

PP29339. Solve in (0,+∞) the following system

4(x61+1)

(3x21−4x1+3)

2 = 3x22 − 4x3 + 3

4(x62+1)

(3x22−4x2+3)

2 = 3x23 − 4x4 + 3

−−−−−−−−−−−−−4(x6

n+1)(3x2

n−4x1+3)2= 3x21 − 4x2 + 3

.

Mihaly Bencze

PP29340. If x ∈�0, π2

�then

4 (sinx+ cosx) + 3

q4�1 + sin6 x

�+ 3p4 (1 + cos6 x) ≤ 9.

Mihaly Bencze

PP29341. If x ∈ R then

6 sin2 x cos2 x+ 3

q4�1 + sin12 x

�+ 3p4 (1 + cos12 x) ≤ 7.

Mihaly Bencze


4�x61 + 1

�=

�3x22 − 4x3 + 3

�3

4�x62 + 1

�=

�3x23 − 4x4 + 3

�3−−−−−−−−−−−−−4�x6n + 1

�=

�3x21 − 4x2 + 3

�3

Mihaly Bencze


PP29343. If xk > 0 (k = 1, 2, ..., n) , then

4n2

�n+

nPk=1

x6k

�≥

�3

nPk=1

x2k − 4nP

k=1

xk + 3n

�3

.

Mihaly Bencze

PP29344. If ak > 0 (k = 1, 2, ..., n) and (a1 + a2) (a2 + a3) ... (an + a1) = 2n

thenQ

cyclic

�1 +

qa1a2(a1+a2)

2

�≤ 2n − 1 +

nQk=1

ak.

Mihaly Bencze


1).P 1

ra3√

4(729r6+r6a)≥ 1

18r3

2).P 1

ha3√

4(729r6+h6a)

≥ 118r3

Mihaly Bencze

PP29346. In all triangle ABC holdsP (tgA

2tgB

2 )3

�

1+729(tgA2tgB

2 )6≥ 1

9 3√2.

Mihaly Bencze

PP29347. If x ∈�0, π2

�and λ =

√33−sin2 2x−1

2 then�√2 +

qλ�λ+ sin2 x

�sinx

��√2 +

pλ (λ+ cos2 x) cosx

�≤

≤ 2√2(7+λ sin2 x cos2 x)√

2+sinx cosx.

Mihaly Bencze

PP29348. If a, b, c > 0 andP

a+ abc = 4 thenP 1

a+1 ≥ n2−32(n−1) for all

n ∈ N,n ≥ 3.

Mihaly Bencze

PP29349. If a, b > 0 and 2a+ b+ a2b = 4 then a (3b+ 1) ≤ b+ 1.

Mihaly Bencze


PP29350. If ak > 0 (k = 1, 2, ..., n) , then

Pcyclic

a1a2+a3+...+an

+

n�

k=1

ak�

cyclic

(a1+a2+...+an−1)≥ n(n−1)n−1+1

(n−1)n.

Mihaly Bencze

PP29351. If a, b > 0 and a+ ab+ b = 12a then 1

3a+4(a3+b3)+ 1

8a3+3b≤ 6

5 .

Mihaly Bencze

PP29352. If a, b > 0 then�4a2+ab+b2

a + aba+b

��4b2+ab+a2

b + aba+b

�≥ 169

16 (a+ b)2 .

Mihaly Bencze

PP29353. If x ∈ R then3�1 + sin4 x

� �1 + cos4 x

�≥ 2

�2 + sin2 x

� �2 + cos2 x

�.

Mihaly Bencze

PP29354. If a, b > 0, then16

�a3 + 3

� �b3 + 3

�≥ (3 + a) (3 + b) (2 + a+ b)2 .

Mihaly Bencze

PP29355. If a, b, c, d > 0 thenP

cyclic

k

qa3+3

b+c+d+1 ≥ 4 for all k ∈ N∗.

Mihaly Bencze

PP29356. If ak > 0 (k = 1, 2, ..., n) , thenP

cyclic

�an−11 +n−1

a2+a3+...+an+1

�λ

≥ n for all

λ ≥ 0.

Mihaly Bencze

PP29357. If x ∈ R then�√32 + cos2 8x− 1

�sin2 x cos2 x

�2 sin4 x+

�√32 + cos2 2x− 1

�cos2 x

�·

·�2 cos4 x+

�√32 + cos2 2x− 1

�sin2 x

��33− 2

√32 + cos2 2x

�≤ 256.

Mihaly Bencze



1).P

(a− b+ c)√b ≤P (a− b+ c)

32

2).P �

cos A2 − cos B

2 + cos C2

�qcos B

2 ≤P�cos A

2 − cos B2 + cos C

2

� 32

3).P

(ma −mb +mc)√mb ≤

P(ma −mb +mc)

32

Mihaly Bencze

PP29359. If a, b > 0 and a (a+ b)2 = 4 then a3b�a2 + b2

�≤ 2.

Mihaly Bencze

PP29360. Let ABCDA1B1C1D1 be a cuboid. Prove that

AB2 ·B1C +BC2 ·B1A+BB21 ·AC ≤

√2�AB3 +BC3 +BB3

1

�.

Mihaly Bencze

PP29361. If ak > 0 (k = 1, 2, ..., n) thenP

cyclic

a21pa22 + a23 ≤

√2

nPk=1

a3k.

Mihaly Bencze

PP29362. If ak > 0 (k = 1, 2, ..., n) thenP

cyclic

(a1+a2)2(a1+a3)

2

a1(a21+a2a3)≥ 8

nPk=1

ak.

Mihaly Bencze

PP29363. If ak > 0 (k = 1, 2, ..., n) thenQ

cyclic

�a21 + a2a3

�≤

�

cyclic

(a1+a2)3

4nn�

k=1ak

.

Mihaly Bencze


1).Q �

a2 + bc�≤ s2(s2+r2+2Rr)

3

32Rr

2).Q �

r2a + rbrc�≤ s4R3

r

3).Q �

sin4 A2 + sin2 B

2 sin2 C2

�≤ ((2R−r)(s2+r2−8Rr)−2Rr2)

3

1048576R8(2R−r)

4).Q �

cos4 A2 + cos2 B

2 cos2 C2

�≤ ((4R+r)3+s2(2R+r))

3

1048576R8(4R+r)

Mihaly Bencze


PP29365. If x ∈ R thensin6 x cos6 x

sin8 x+cos6 x+cos2 x+1+ 8 cos6 x

cos8 x+sin2 x+9+ 8 sin6 x

sin6 x+16≤ sin8 x+cos8 x+6 sin2 x cos2 x+16

6 .

Mihaly Bencze

PP29366. If ak > 0 (k = 1, 2, ..., n) thenP

cyclic

�

a1(a21+a1a2)(a2+a3)

a1+a3≤

nPk=1

ak.

Mihaly Bencze

PP29367. If x ∈�0, π2

�then

sin2 x√1 + cos2 x+ cos2 x

p1 + sin2 x ≤ 1 + sin3 x+ cos3 x.

Mihaly Bencze

PP29368. If x, y > 0 then

xy�xp2 (x2 + y2) + y2

��yp

2 (x2 + y2) + x2�≤

�2x3 + y3

� �2y3 + x3

�.

Mihaly Bencze


a4 +P b5c3

a4+P a8

bc3≥ 9

��

a3�

a

�2.

Mihaly Bencze

PP29370. If a, b, c > 0, a 6= b 6= c thenP a2(an−bn)

(a−b)bn−1 ≥ 3(n+1)�

a3�

a for all

n ∈ N∗.

Mihaly Bencze

PP29371. If a > 0, 0 < b ≤ xk ≤ c (k = 1, 2, ..., n) , thenP

cyclic

x1

x21+a2x2x3

≤ n2(b+c)2

8abcn�

k=1xk

.

Mihaly Bencze

PP29372. In all triangle ABC holdsPa2b2 cos2C ≤ 3

�s2 − r2 − 4Rr

�2 − 16s2Rr.

Mihaly Bencze



1).P

a2√b4 + c4 ≤ 2

√2��

s2 + r2 + 4Rr�2 − 20s2Rr

�

2).P

r2a

qr4b + r4c ≤

√2�2s4 − 5s2r (4R+ r)

�

3).P

sin4 A2

qsin8 B

2 + sin8 C2 ≤

√2

�2�s2+r2−8Rr

16R2

�2− 5r2(2R−r)

32R3

�

4).P

cos4 A2

qcos8 B

2 + cos8 C2 ≤

√2

�(s2+(4R+r)2)

2

128R4 − 5s2(4R+r)32R3

�

Mihaly Bencze

PP29374. If ak > 0 (k = 1, 2, ..., n) thenP

cyclic

r3a1a2 +

qa41+a42

2 ≤ 2nP

k=1

ak.

Mihaly Bencze


a2 ≥P ab+ 14

P (a2−b2)2

a2+b2.

Mihaly Bencze

PP29376. If ak > 0 (k = 1, 2, ..., n) then

n−12

nPk=1

a2k −P

1≤i<jnaiaj ≥ 1

4

P1≤i<jn

(a2i−a2j)2

a2i+a2j.

Mihaly Bencze


3P �

tg6A2 + tg6B2�tg2C2 ≥ 12r2(4R+r)2

s4+ 2

√3r2

s2.

Mihaly Bencze

PP29378. If a, b, c > 0, thenP

a2 + 2P a2b2

a2+b2≥ 2

Pab.

Mihaly Bencze

PP29379. If a, b, c, x, y > 0 thenP a

b+xc +y�

ab�

a2≥ 2

�1

x+1 +q

yx+1

�.

Mihaly Bencze

PP29380. If x, y, z > 0 and xyz = 1 thenP 1

(x+1)2+(y+1)2−2≤ 1

2 .

Mihaly Bencze



a2√b4 + c4 ≤ 2

√2P

a2b2 −√2abc

Pa.

Mihaly Bencze

PP29382. Determine all a, b ∈ R for which√a+

√b+

p2 (a+ b) ≥ 3 +

4(3−√6)

3

�ab+ 2 (a+ b)2 − 3

�.

Mihaly Bencze

PP29383. If x, y, z ∈ R then1).

P sin2 x3+sin2 y+sin2 z

+ cos2 x cos2 y cos2 z ≤ 1

2).P cos2 x

3+cos2 y+cos2 z+ sin2 x sin2 y sin2 z ≤ 1

Mihaly Bencze


1). (2s)32�√

6s−P√

a�≤ 4

�√3−

√2� �

s2 − 3r2 − 12Rr�

2). s32

�√3s−P

√s− a

�≤ 4

�√3−

√2� �

s2 − 3r2 − 12Rr�

3). (4R+ r)32

�p3 (4R+ r)−P√

ra

�≤ 4

�√3−

√2� �

(4R+ r)2 − 3s2�

4).�2R−r2R

� 32

�q3(2R−r)

2R −P sin A2

�≤ 4

�√3−

√2� � (4R+r)2−3s2

16R2

�

5).�4R+r2R

� 32

�q3(4R+r)

2R −P cos A2

�≤ 4

�√3−

√2� � (4R+r)2−3s2

16R2

�

Mihaly Bencze

PP29385. If a, b, c > 0 then

(P

a)32

�p3P

a−P√

a�≤ 4

�√3−

√2� �P

a2 −P

ab�.

Mihaly Bencze

PP29386. If x ∈�0, π2

�then

1). sinx+ cosx+√2−

√3 ≥ 4(3−

√6)

3

�sin2 x cos2 x− 1

�

2).√2 (sinx+ cosx)− 2 ≥ 4(3−

√6)

3

�sin2 2x− 1

�

Mihaly Bencze


PP29387. If a, b, c, x, y > 0 and a+ b+ c = 1 thenPcyclic

an+2

(b+x)(c+y) ≥1

3n−1(3x+1)(3y+1).

Mihaly Bencze


tgA2 tg

B2 ≥

√3 + 4

�√3−

√2� �3r(4R+r)

s2− 1

�.

Mihaly Bencze


1).P √

3+7 sinAsinA ≥ 147

2√6− s

2√6R

2).P √

3+7 sin2 A

sin2 A≥ 49

2√6

�1− s2−(2R+r)2

2R2

�

Mihaly Bencze


1).P √

3+7 cosAcosA ≥ 174

2√6− R+r

2√6R

2).P √

3+7 cos2 Acos2 A

≥ 492√6

�s2−r(4R+r)

2R2

�

Mihaly Bencze


1).P

�

3+7 sin2 A2

sin2 A2

≥ 494√6

�4R+rR

�

2).P

�

3+7 cos2 A2

cos2 A2

≥ 494√6

�2R−rR

�

Mihaly Bencze


1).Pp

ra (7r + 3ra) ≥ 49r(2R−r)√6

2).Pp

ha (7r + 3ha) ≥49r(s2+r2−2Rr)

4√6R

Mihaly Bencze

PP29393. If a, b, c > 0 thenP a(b+λc)2

b2+bc+c2≥ (λ+1)2

�

ab�

a for all λ ≥ 0.

Mihaly Bencze


PP29394. If a, b, c > 0 then

1). (a+b)3

a2+ab+b2+ 2(a+b)

3 ≥ 2a(a+2b)2a+b + 2b(b+2a)

2b+a

2).�

a(a+b)2

a2+ab+b2+ 2b

3

��b(a+b)2

a2+ab+b2+ 2a

3

�≥ 4ab

Mihaly Bencze

PP29395. If ak > 0 (k = 1, 2, ..., n) , then�nP

k=1

aλk

��nP

k=1

1aλk

�≥ n 3

s3n

�nP

k=1

ak

��nP

k=1

1ak

�− 2n3 for all λ ≥ 1

Mihaly Bencze

PP29396. If a, b, c > 0 and abc = 1 thenP a2(1+a2+a4)

b(1−b+b2)≥ 9

P 1a+b2+c

.

Mihaly Bencze


1).P 1√

a2+ab+b2≥ 1√

s2+r2+4Rr

�2 +

(s2+r2+4Rr)2+8s2Rr

2s2(s2+r2+2Rr)

�

2).P 1√

r2a+rarb+r2b

≥ 1s

�2 + s2+r(4R+r)

2R(4R+r)

�

Mihaly Bencze


1).P 1√

h2a+hahb+h2

b

≥ 2s

qR2r

�1 +

2Rr(5s2+r2+4Rr)(s2+r2+4Rr)(s2+r2+2Rr)

�

2).P 1√

(s−a)2+(s−a)(s−b)+(s−b)2≥ 1√

r(4R+r)

�2 +

r(s2+(4R+r)2)2s2R

�

Mihaly Bencze

PP29399. If a, b > 0 then√2ab+ a2

�2√

a2+ab+b2+ 1√

3a

�+√2ab+ b2

�2√

a2+ab+b2+ 1√

3b

�≥

≥ 4 + 22a+b

�92 + 2ab

a+b

�+ 2

2b+a

�92 + 2ab

a+b

�.

Mihaly Bencze

PP29400. If ak > 0 (k = 1, 2, ..., n) andnP

k=1

a2k ≤ 2n thennP

k=1

qa3k + 1 ≤ 2n.

Mihaly Bencze


PP29401. If ak > 0 (k = 1, 2, ..., n) andnP

k=1

qa3k + 1 ≥ 3n then

nPk=1

a2k ≥ 4n.

Mihaly Bencze


1). 4− r(4R+r)s2

≥Pq

1 + 8r3

h3a

2). 5− 4r(4R+r)s2

≥Pq1 + 8r3

r3a

Mihaly Bencze

PP29403. In all triangle ABC holdsPp

3√3 + 64 sin3A ≤

q√33

�4(s2−r(4R+r))

3R2 + 3

�.

Mihaly Bencze

PP29404. In all acute triangle ABC holdsP√

1 + 64 cos3A ≤ 11− 4(s2−(2R+r)2)R2 .

Mihaly Bencze


1).Pq

1 + 512 sin6 A2 ≤ 35 +

4(r2−s2)R2

2).Pq

1 + 51227 cos6 A

2 ≤ 3 +4((4R+r)2−s2)

9R2

Mihaly Bencze


�a21 + 2

� �a22 + 2

�= 4q�

a33 + 1� �

a34 + 1�

�a22 + 2

� �a23 + 2

�= 4q�

a34 + 1� �

a35 + 1�

−−−−−−−−−−−−−−−−−−�a2n + 2

� �a21 + 2

�= 4q�

a32 + 1� �

a33 + 1�

.

Mihaly Bencze


PP29407. If ak > 0 (k = 1, 2, ..., n) , then 2nλ2 +nP

k=1

a2k ≥ 2nP

k=1

qλa3k + λ4

for all λ ≥ 0.

Mihaly Bencze

PP29408. If ak > 0 (k = 1, 2, ..., n) andnP

k=1

1√a3k+λ3

≤ 1 then

nPk=1

a2k ≥ 2n√λ�n−

√λ�for λ > 0.

Mihaly Bencze

PP29409. If ak > 0 (k = 1, 2, ..., n) then

2nP

k=1

a3k+1

a2k+2

≤nP

k=1

qa3k + 1 ≤ 1

2

nPk=1

a2k + n.

Mihaly Bencze

PP29410. If x ∈�0, π2

�, then

2 sin2 x cos2 x+ (sinx+ cosx) (1− sinx cosx) +

2�p

1 + sin5 x+√1 + cos5 x

�≤ 6.

Mihaly Bencze

PP29411. If ak > 0 (k = 1, 2, ..., n) , thennP

k=1

ra3k + 2

qa5k + 1− 7

4 ≤ n2 +

nPk=1

a2k.

Mihaly Bencze

PP29412. If x, y, z > 0, thenP

cyclic

x4λ−2√2(y4+z4)+yz

≥ 19λ−1

�Px2

�2λ−2for all

λ ≥ 1.

Mihaly Bencze

PP29413. If xk > 0 (k = 1, 2, ..., n) thenP

cyclic

xn−11

�

(n−1)(x2n−22 +x2n−2

3 +...+x2n−2n )+x2x3...xn

≥ 1.

Mihaly Bencze


PP29414. If 0 < λ ≤ ak (k = 1, 2, ..., n) then

λnP

k=1

ak ≤P

a1a2 ≤ 1λ

Pa1a2a3.

Mihaly Bencze

PP29415. If 0 < λ ≤ ak (k = 1, 2, ..., n) , then

λn

�nP

k=1

ak

�n

≤ Qcyclic

(λ+ a1 + a2 + ...+ an−1)nQ

k=1

ak.

Mihaly Bencze

PP29416. If n,m, k ∈ N ∗ then��

1 + 1n

�n+�1 + 1

m

�m+�1 + 1

k

�k�3≤

��1 + 1

n

�n �1 + 1

m

�m+�1 + 1

k

�k��1 + 1

m

�m �1 + 1

k

�k+�1 + 1

n

�n� ·

·��

1 + 1k

�k �1 + 1

n

�n+�1 + 1

m

�m�.

Mihaly Bencze

PP29417. If ak > 0 (k = 1, 2, ..., n) , then�

nPk=1

ak

��nP

k=1

1ak

�≥ n2 + 2

n(n−1)

P

1≤i<j≤≤n

|ai−aj |√aiaj

!2

.

Mihaly Bencze

PP29418. If x, y, z > 0 and x 6= y 6= z thenP 1

(x−y)2+ 1

4

P 1xy ≥ 9

�

xy .

Mihaly Bencze

PP29419. If x, y, z > 0 thenPq

xy+z ≤

r2 (P

xy)�P 1

(x+y)2

�.

Mihaly Bencze

PP29420. If xk > 0 (k = 1, 2, ..., n), then P

1≤i<j≤nxixj

! P

1≤i<j≤n

1(xi+xj)

2

!≥

�n(n−1)

2

�2.

Mihaly Bencze

PP29421. If a, b, c > 0, then (a+ b+ c)15 ≥ 14348907a4b4c4�a3 + b3 + c3

�.

Mihaly Bencze



P 1r3a

≤ s8

4782969r11

2).P 1

h3a≤ 16s8

4782969r8R4

Mihaly Bencze

PP29423. In all triangle ABC holds r8 ≥ 4782969s6�s2 − 12Rr

�.

Mihaly Bencze

PP29424. If ak > 0 (k = 1, 2, ..., n) andnP

k=1

ak = n then

�nP

k=1

ank

��nQ

k=1

ak

�n+1

≤ n.

Mihaly Bencze

PP29425. If xk > 0 (k = 1, 2, ..., n) then

Pcyclic

x1x2

+2n+1

n�

k=1

xk

(x1+x2)(x2+x3)...(xn+x1)≥ n+ 2.

Mihaly Bencze

PP29426. If xk > 0 (k = 1, 2, ..., n) thenP x1

x2+x3+ 2n+1

�

(x1+x2)�

(x1+x2+x3)≥ n+ 2.

Mihaly Bencze

PP29427. If ak > 0 (k = 1, 2, ..., n) then

Pcyclic

r(a21+a22)(a1+a2)

2

a1a2+ 4a1a2 ≥ 2

√3

nPk=1

ak.

Mihaly Bencze

PP29428. If ak > 0 (k = 1, 2, ..., n) and λ > 0 then

12

nPk=1

ak +2n

(a1+λ)(a2+λ)...(an+λ) ≥ n− nλ2 + 1.

Mihaly Bencze


PP29429. In all triangle ABC holdsPs

�

ctgA2

�

ctgB2+�

ctgC2

≥q

32

PqtgA

2 tgB2 .

Mihaly Bencze

PP29430. If xk > 0 (k = 1, 2, ..., n) then

P x1x2

+2n

n�

k=1xk

(x1+x2)(x2+x3)...(xn+x1)≥ n+ 1.

Mihaly Bencze

PP29431. If a, b > 0 then a2+b2

a2+ab+b2+ 10

3 ≥ 4a(a+2b)(a+b)(2a+b) +

4b(b+2a)(a+b)(2b+a) .

Mihaly Bencze


1).P 1

�

ra(√rb+

√rc)

≥q

3√r

2s

P 1√ra

2).P 1

�

ha(√hb+

√hc)

≥q

3√R

2s√2

P 1√ha

Mihaly Bencze

PP29433. If ak > 0 (k = 1, 2, ..., n) thenP

cyclic

��a31+a322a1a2

�2+ a1a2

�≥ 2

nPk=1

a2k.

Mihaly Bencze

PP29434. If ak > 0 (k = 1, 2, ..., n) thenP

cyclic

r�a31+a322a1a2

�2+ a1a2 ≥

√2

nPk=1

ak.

Mihaly Bencze

PP29435. If ak > 0 (k = 1, 2, ..., n) thenP

cyclic

(a31+a32)2

a41+a21a22+a42

≥ 4(�

a1a2)2

2n�

k=1a2k+�

a1a2

.

Mihaly Bencze

PP29436. If a, b, c > 0, thenP ((a3+b3)c)

2

a4+a2b2+b4≥ 36a2b2c2

2�

a2+�

ab.

Mihaly Bencze


PP29437. If ak > 0 (k = 1, 2, ..., n) thenP

cyclic

pa21 − a1a2 + a22 ≥

nPk=1

ak.

Mihaly Bencze

PP29438. If ak > 0 (k = 1, 2, ..., n) thenP

cyclic

pa21 − a1a2 + a22 ≤

P a21a2.

Mihaly Bencze


k=1

3

r4(k2+k+1)

(2k3+3k2+3k+1)2≤ n

n+1 .

Mihaly Bencze

PP29440. If a, b, c > 0 and�

1ab

�3+�1bc

�3+�

1ca

�3= 4 then

P a2−ab+b2

(a3+b3)2≤ 1.

Mihaly Bencze

PP29441. If a, b, c > 0 thenP a4(b+c)√

(a2+ab+b2)(a2+ac+c2)≥ 1

3

Pa2 (b+ c) .

Mihaly Bencze


b ≥P√a2 − ab+ b2.

Mihaly Bencze

PP29443. If a, b > 0 then 2q

2a4+b4

+ 12

�1a2

+ 1b2

�+ 2

�a2

b + b2

a

�≥ 7.

Mihaly Bencze

PP29444. If ak > 0 (k = 1, 2, ..., n) then

Pcyclic

�a21a2

+a22a1

�pa21 + a1a2 + a22 ≥ 2

Ppa41 + a21a

22 + a42.

Mihaly Bencze

PP29445. If a, b > 0 then�2a3

a3+b3+ b3

2a3

��2b3

a3+b3+ a3

2b3

�≥ (2a+b)3(a+2b)3

36ab(2a2+b2)(a2+2b2).

Mihaly Bencze


PP29446. If a, b > 0 then 9�

2a3

a3+b3+ b3

2a3− 5

6

��2b3

a3+b3+ a3

2b3− 5

6

�≥

≥�a(a+2b)2a2+b2

+ 2a2+b2

a(a+2b)

��b(b+2a)2b2+a2

+ 2b2+a2

b(b+2a)

�.

Mihaly Bencze


P a3

b3+c3≥ 2s2

s2+r2+4Rr

2).P r3a

r3b+r3c

≥ (4R+r)2

2s2

3).P (sin A

2 )6

(sin B2 )

6+(sin C

2 )6 ≥ 2(2R−r)2

s2+r2−8Rr

4).P (cos A

2 )6

(cos B2 )

6+(cos C

2 )6 ≥ 4(4R+r)2

s2+(4R+r)2

Mihaly Bencze


�2a√3+ b2√

a2+ab+b2

��2b√3+ a2√

a2+ab+b2

�≥ a2 + ab+ b2.

Mihaly Bencze

PP29449. In all triangle ABC holds 4R+rs +

√3�

1

3n2 +1

P �ctgA

2

�n� 1k ≥ 2

√3

for all n, k ∈ N, n ≥ 2, k ≥ 6.

Mihaly Bencze

PP29450. If ak > 0 (k = 1, 2, ..., n) , thenPcyclic

a21�

(a21+a1a2+a22)(a21+a1a3+a23)≥ n

3 .

Mihaly Bencze


k=1

�3k2+6k+53k2+6k+2

+ 4k(k+2)(2k+1)(2k+3)

�≥ 2m.

Mihaly Bencze

PP29452. If a, b > 0 then a2+ab+b2

a+√ab+b

+ 8ab√ab

(a+b)(√a+

√b)

2 ≥ 2√ab.

Mihaly Bencze


PP29453. If x ∈�0, π2

�then

5(1+sinx)(1+cos x)−1 + 8 sin 2x

(2+sinx)(2+cosx)(sinx+cosx) ≥ 2.

Mihaly Bencze

PP29454. If x ∈ R, then 1−sin2 x cos2 x1+sin2 x cos2 x

+ sin2 2x

(1+sin2 x)(1+cos2 x)≥ 1.

Mihaly Bencze

PP29455. If a, b, c > 0 then 1 +P (a−b)(a−c)

b+c ≥ 2P a

√bc

b+c .

Mihaly Bencze

PP29456. If a, b > 0 then a2+2b2

b(2a+b) +b2+2a2

a(2b+a) +8ab

(a+b)2≥ 4.

Mihaly Bencze


b+c ≤ 12√abc

Pa2√a+ 1

2

P√ab.

Mihaly Bencze


b+c ≥Pq

(a2+bc)(b2+ca)(b+c)(c+a) .

Mihaly Bencze

PP29459. If xk > 0 (k = 1, 2, ..., n) andnQ

k=1

(xk + 1) = n then

Pcyclic

x41+3x3

1+5x21+7x1+4

x32+2x2

2+3x2+4≥ n n

√n.

Mihaly Bencze


b+c ≤ 2P

a+ 2P a2−bc

b+c .

Mihaly Bencze

PP29461. If x, y > 0 and xy + x+ y = 1 then

4x3+7x2+5x+24y2+3y+2

+ 4y3+7y2+5y+24x2+3x+2

≥ 2√2.

Mihaly Bencze


PP29462. If x, y, z > 0 and xyz + xy + yz + zx+ x+ y + z = 2 then5x3+6x2+5x+4

5y2+y+4+ 5y3+6y2+5y+4

5z2+z+4+ 5z3+6z2+5z+4

5x2+x+4≥ 3 3

√3.

Mihaly Bencze


b2+bc+c2≥

�

a�

ab .

Mihaly Bencze


7bc+ 3√

4(b6+c6)≥

�

a3�

ab .

Mihaly Bencze

PP29465. If a, b, c > 0 thenP ab2

7bc+ 3√

4(b6+c6)≥

�

ab3�

a .

Mihaly Bencze

PP29466. If ak > 0 (k = 1, 2, ..., n) thenP

cyclic

r10a1a2 +

3

q4�a61 + a62

�≤ 2

√3

nPk=1

ak.

Mihaly Bencze

PP29467. If ak > 0 (k = 1, 2, ..., n) , thenP

cyclic

r3

q4�a61 + a62

�− 2a1a2 ≤

√3P |a1 − a2| .

Mihaly Bencze

PP29468. If ak > 0 (k = 1, 2, ..., n) , thenP

cyclic

|a31−a32|7a1a2+

3�

4(a61+a62)≥ 1

3

Pcyclic

|a1 − a2| .

Mihaly Bencze

PP29469. If ak > 0 (k = 1, 2, ..., n) then

Pcyclic

a31−a32

7a1a2+3�

4(a61+a62)

!2

≥ 29

nPk=1

a2k − 29

Pa1a2.

Mihaly Bencze


PP29470. If ak > 0 (k = 1, 2, ..., n) thenP

cyclic

a21+a1a2+a22

7a2a3+3�

4(a62+a63)≥ n

3 .

Mihaly Bencze

PP29471. If ak > 0 (k = 1, 2, ..., n) then

3P

cyclic

��a31 − a32�� ≥ P

cyclic

|a1 − a2|�7a1a2 +

3

q4�a61 + a62

��.

Mihaly Bencze

PP29472. If x ∈ R thensin2 x

3√

4(64+cos12 x)+14 cos2 x+ cos2 x

3�

4(64+sin12 x)+14 sin2 x+ (sinx)

23+(cosx)

23

12 ≥ 7− 3√212 .

Mihaly Bencze

PP29473. If ak > 0 (k = 1, 2, ..., n) thenP

cyclic

a31+a32

a1a2+3�

4(a61+a62)≥ 2

3

nPk=1

ak.

Mihaly Bencze

PP29474. If a, b, c > 0 then√3 (P

a)112 ≥ 243 (

Pab)3 .

Mihaly Bencze


b4�

18ab−(a+b+c)√

3(a+b+c)c+2(a+b+c)2� ≥ 3

�

a2

(�

a)4.

Mihaly Bencze

PP29476. If x ∈�0, π2

�then

tg8x

2−√2+2 sin2 x cos2 x

+ 16sin8 x(4 sin2 x+2−cosx)

+ cos8 x16(4 cos2 x+2−sinx)

≥ 5−2 sin2 x cos2 x3 .

Mihaly Bencze

PP29477. If x, y, z > 0 such that xn + yn + zn = 3. Determine all n, k ∈ Nfor which (xy)n+k + (yz)n+k + (zx)n+k ≤ 3.

Mihaly Bencze

PP29478. If x ∈�0, π2

�then (sinx cosx)

52 + 2 4

√2�(sinx)

52 + (cosx)

52

�≤ 3.

Mihaly Bencze


PP29479. If x ∈�0, π2

�then

3�√

sinx+√cosx+ 4

√2�2

≥�2 + sin2 x cos2 x

�3.

Mihaly Bencze


�√23

PqctgA

2 ctgB2

�2

≥P�q

tgA2 ctg

B2 +

qtgA

2 ctgC2

�.

Mihaly Bencze

PP29481. If x ∈�0, π2

�then

23

�1

sinx + 1cosx + 1√

2

�2≥ (sinx+ cosx)

�1 +

√2

sinx cosx

�+ tgx+ ctgx.

Mihaly Bencze

PP29482. If x ∈�0, π2

�then

�sinx

√sinx

cos2 x√cosx

+ cosx√cosx

2 4√2+

4√8sin2 x

√sinx

�·

·�

cosx√cosx

sin2 x√sinx

+ sinx√sinx

2 4√2+

4√8cos2 x

√cosx

�≥ 9.

Mihaly Bencze

PP29483. If a, b > 0 then�2 4√a+

4√b

a(a+2b)

�2

+�2

4√b+ 4√a

b(b+2a)

�2

≥ a2+4ab+b2

3 .

Mihaly Bencze


1).P√

ra ≥q

9√3

2s

Pra

�√rb +

√rc�

2).P√

ha ≥q

9√Rr

2s√2

Pha

�√hb +

√hc�

Mihaly Bencze

PP29485. If xk ≥p√

2− 1 (k = 1, 2, ..., n) and λ0 =1+√√

2−1

1+4√√

2−1then

nPk=1

1+xk

1+√xk

≥ nλ0.

Mihaly Bencze



4

qtgA

2 tgB2 ≥ 9r(4R+r)

s3

q√3r (4R+ r).

Mihaly Bencze

PP29487. If ak > 0 (k = 1, 2, ..., n) andnP

k=1

a2k = n, then

nPk=1

1ak

≥r

nn−1

Pcyclic

a1+a2+...+an−1

an.

Mihaly Bencze

PP29488. If ai ≥ 1 (i = 1, 2, ..., n) , then

λ+ k

snQ

i=1ai

!nQ

i=1

λ+aiλ+ k

√ai

≥ λ+ 1

for all k ≥ 1, k ∈ N and for all λ > 0.

Mihaly Bencze

PP29489. If ak > 0 (k = 1, 2, ..., n) andnP

k=1

apk = n (p ≥ 1) then P

cyclic

ar1ar+s2

! P

cyclic

ar2ar+s1

!≥ n2 for all r, s > 0.

Mihaly Bencze

PP29490. If ak > 0 (k = 1, 2, ..., n) andnP

k=1

a3k = n thenP

cyclic

1a21(a22−a2a3+a23)

≥ n.

Mihaly Bencze

PP29491. If ai,λi > 0 (i = 1, 2, ..., n) andnP

i=1λi = 1 then

�aλ11 aλ2

2 ...aλnn

�(k+1)n+�aλ21 aλ3

2 ...aλ1n

�(k+1)n+ ...+

�aλn1 aλ1

2 ...aλn−1n

�(k+1)n≥

≥ anλ1+k1 anλ2+k

2 ...anλn+kn + anλ2+k

1 anλ3+k2 ...anλ1+k

n + ...

+anλn+k1 anλ1+k

2 ...anλn−1+kn .

Mihaly Bencze


PP29492. If a, b > 0 thenq

9 + 3(2a2+b2)(2b2+a2)a2b2

≥ 4 + ab +

ba .

Mihaly Bencze

PP29493. If aij > 0 (i = 1, 2, ...,m; j = 1, 2, ..., n) then�mPi=1

1ai1ai2 ...ain

�mPi=1

(ai1 + ai2 + ain)n ≥ m2nn.

Mihaly Bencze

PP29494. If a, b > 0 then 1 + ab +

ba ≥ (2a+b)(a+2b)

3ab .

Mihaly Bencze

PP29495. If ak > 0 (k = 1, 2, ..., n) thensn2(n−1)(n−2)

2 + n(n−1)2

�nP

k=1

a2k

��nP

k=1

1a2k

�≥ P

cyclic

a1+a2+...+an−1

an.

Mihaly Bencze

PP29496. If a, b > 0 then�1 +

√(2a2+b2)(2b2+a2)

ab

�2

≤�q

(2a+b)(a+2b)ab − 1

�2

.

Mihaly Bencze

PP29497. If ak > 0 (k = 1, 2, ..., n) then

�nP

k=1

ak

��nP

k=1

1ak

�≤ n

P a1a2.

Mihaly Bencze

PP29498. If xi > 0 (i = 1, 2, ..., n) and k ∈ {2, 3, ..., n} thenP x1x2+x3+...+xk

+2n+1

�

(x1+x2+...+xk−1)�

(x1+x2+...+xk)≥ n+ 2.

Mihaly Bencze

PP29499. If xi > 0 (i = 1, 2, ..., k) when k = n (n− 2) then

(n− 2)

1 +

vuut1 +

s�kP

i=1a2i

��kP

i=1

1a2i

� ≤

s�kP

i=1ai

��kP

i=1

1ai

�.

Mihaly Bencze



1).P� √

a+√b

a+√ab+b

�2≥ 4(5s2+r2+4Rr)

9s(s2+r2+4Rr)

2).P� √

s−a+√s−b

c+√

(s−a)(s−b)

�2

≥ 2(s2+r2+4Rr)9sRr

3).P� √

ra+√rb

ra+√rarb+rb

�2≥ 2((4R+r)2+s2)

9s2R

4).P� √

ha+√hb

ha+√hahb+hb

�2≥

2�

(s2+r2+4Rr)2+8s2Rr

�

9s2r(s2+r2+2Rr)

5).P�

sin A2+sin B

2

sin2 A2+sin A

2sin B

2+sin2 B

2

�2

≥ 16R(16R2−24Rr+5r2+s2)9((2R−r)(s2+r2−8Rr)−2Rr2)

6).P�

cos A2+cos B

2

cos2 A2+cos A

2cos B

2+cos2 B

2

�2

≥ 16R(5(4R+r)2+s2)9((4R+r)3+s2(2R+r))

Mihaly Bencze


1).P

(b+ c)�a+

√ab+b√

a+√b

�2≤ 9

8

�5s2 + r2 + 4Rr

�

2).P

a

�c+√

(s−a)(s−b)√s−a+

√s−b

�2

≤ 98

�s2 + r2 + 4Rr

�

3).P

(rb + rc)�ra+

√rarb+rb√

ra+√rb

�2≤ 9

8

�(4R+ r)2 + s2

�

4).P

(hb + hc)�ha+

√hahb+hb√

ha+√hb

�2≤

9�

(s2+r2+4Rr)2+8s2Rr

�

32R2

5).P �

sin2 B2 + sin2 C

2

�� sin2 A2+sin A

2sin B

2+sin2 B

2

sin A2+sin B

2

�2

≤ 9(16R2−24Rr+5r2+s2)128R2

6).P �

cos2 B2 + cos2 C

2

�� cos2 A2+cos A

2cos B

2+cos2 B

2

cos A2+cos B

2

�2

≤ 9(5(4R+r)2+s2)128R2

Mihaly Bencze


1).P�

a+√ab+b√

a+√b

�2≤ 9s

2

2).P�

c+√


√s−b

�2

≤ 9s4

3).P�

ra+√rarb+rb√

ra+√rb

�2≤ 9(4R+r)

4

4).P�

ha+√hahb+hb√

ha+√hb

�2≤ 9(s2+r2+4Rr)

8R

5).P�

sin2 A2+sin A

2sin B

2+sin2 B

2

sin A2+sin B

2

�2

≤ 9(2R−r)8R


6).P�

cos2 A2+cos A

2cos B

2+cos2 B

2

cos A2+cos B

2

�2

≤ 9(4R+r)8R

Mihaly Bencze

PP29503. If a, b, c > 0 and a 6= b 6= c thenP�

|a−b|+√

|ab+b−ca−b2|+|b−c|√|a−b|+

√|b−c|

�2

≤ 92 (max {a, b, c}−min {a, b, c}) .

Mihaly Bencze


k=0

�nk

� ��nk

�+q�

nk

�+ 1

�2

≤ 92

�2n +

�2nn

��.

Mihaly Bencze


k=1

�13k2−3k−3+

√36k4−12k3−17k2+3k+2

(√4k2−1+

√9k2−3k−2)

√36k4−12k3−17k2+3k+2

�2

≤ 9(3n2+3n+1)8(2n+1)(3n+1) .

Mihaly Bencze


k=1

�√k!((k+1)2+

√k2+k+1)√

k+√k2+k+1

�2

≤ 98 ((n+ 2)!− 1) .

Mihaly Bencze

PP29507. Prove that6481

nPk=1

�k2+k+

√k2+k+1√

k2+k+1

�2 �k2−k+

√k2−k+1√

k2−k+1

�2≤ n+

n(n+1)(2n+1)(3n2+3n+4)30 .

Mihaly Bencze

PP29508. Denote Fk and Lk the kth Fibonacci respective Lucas numbers.Prove that

1). 9 (Fn+2 + Ln+2 − 4) ≥ 8nP

k=1

�Fk+

√FkLk+Lk√

Fk+√Lk

�2

2). 9 (FnFn+1 + LnLn+1 − 9) ≥ 8nP

k=1

�F 2k+FkLk+L2

k

Fk+Lk

�2

Mihaly Bencze



k=1

�2k+1+

√k(k+1)√

k+√k+1

�2

≤ 9n (n+ 2) .

Mihaly Bencze


k=1

� √k+1√

k(k+√k+1)

�2

≥ 8n9(n+1) .

Mihaly Bencze

PP29511. Solve in�0, π2

�the following system:

8 (1 + sinx1 cosx1)2 = 9 (1 + sin 2x2)

8 (1 + sinx2 cosx2)2 = 9 (1 + sin 2x3)

−−−−−−−−−−−−−−−−8 (1 + sinxn cosxn)

2 = 9 (1 + sin 2x1)

.

Mihaly Bencze

PP29512. If ak > 0 (k = 1, 2, ..., n) , then 9nP

k=1

ak ≥ 4P

cyclic

�a1+

√a1a2+a2√

a1+√a2

�2.

Mihaly Bencze

PP29513. If ak > 0 (k = 1, 2, ..., n) , then

9

�nP

k=1

ak

� nP

k=1

ak +P

cyclic

√a1a2

!≥ 2

2

nPk=1

ak +P

cyclic

√a1a2

!2

.

Mihaly Bencze


8�a1 +

√a1a2 + a2

�2= 9 (a2 + a3)

�√a3 +

√a4�2

8�a2 +

√a2a3 + a3

�2= 9 (a3 + a4)

�√a4 +

√a5�2

−−−−−−−−−−−−−−−−−−−−−−8�an +

√ana1 + a1

�2= 9 (a1 + a2)

�√a2 +

√a3�2

.

Mihaly Bencze

PP29515. If a, b > 0 then (a+2b)(b+2a)

(a+b)4≤ 9

16ab .

Mihaly Bencze


PP29516. If a, b > 0 then 13√a2b

+ 13√ab2

≥ 4a+b .

Mihaly Bencze

PP29517. If ak > 0 (k = 1, 2, ..., n) , thenP

cyclic

1(a1+2a2)(a2+2a1)

≥ n3

9

�

n�

k=1

ak

�2 .

Mihaly Bencze


(a+b)(a+c) ≤9

4�

a .

Mihaly Bencze

PP29519. If ak > 0 (k = 1, 2, ..., n) , thenP an−1

1(a1+a2)(a1+a3)...(a1+an)

≤ n2

2n−1n�

k=1ak

.

Mihaly Bencze

PP29520. If a, b, c > 0 and λ > 0 thenP 1

(λa2+bc)n≥ 3n+1

(λ+1)n(�

a2)nfor all

n ∈ N.

Mihaly Bencze

PP29521. If a, b > 0 then2

a2(2a+b)2+ 2

b2(2b+a)2+ 1

(a2+2b2)2+ 1

(b2+2a2)2≥ (2a+b)2

9(2a6+b6)+ (2b+a)2

9(2b6+a6).

Mihaly Bencze

PP29522. If a, b, c > 0 and abc ≤ 1 then�2a + a

bc

�2+�2b +

bca

�2+�2c +

cab

�2 ≥ 27.

Mihaly Bencze

PP29523. If a, b, c > 0 ad λ > 0 thenP 1

(λa2+bc)2≥ 27

(λ+1)2(�

a2)2.

Mihaly Bencze

PP29524. In all triangle ABC holdsra

√s

rb√ra+4s

+ rrb√s

rc√rb+4s

+ src√r

rra√ra+4r

≥q

35

�4p

ras + 4

prbs + 4

prcs

�and his

permutations.

Mihaly Bencze



Rrha

hb

√2(Rha+8s)

+ hb

√r

hc

√s(hb+4sr)

+ hc

Rha

√r(hc+4r)

≥s

35

�4

qRha

2s + 4

qhb

sr + 4

qhc

r

�

and his permutations.

Mihaly Bencze

PP29526. If a, b, c > 0 then�P a4c4

b6√a4+4b4

��P a8

b4c2√a4+4c4

�≥ 3

5abc

p(P

a2b) (P

a2c).

Mihaly Bencze


aR

2b√

s(a+16s)+ br

c√

R(b+4R)+ 4cs

a√

r(c+4r)≥s

35

�4p

a4s +

4

qbR + 4

pcr

�and his

permutations.

Mihaly Bencze

PP29528. If a, b > 0 thena4(1+a4b8)b4√a4+4

+b4(1+b4a8)a4

√b4+4

+ 2a2b2

√1+4a4b4

≥ 2

q3((a+b)ab+1)

5 .

Mihaly Bencze

PP29529. If ak > 0 (k = 1, 2, ..., n) thenPcyclic

a21(a2+a3+...+an)

2√

a41+4≥ n2

√2m+1

(n−1)2�

(m+1)n�

k=1a2k−2m

n�

k=1ak+2n(m+1)

� for all

n,m ∈ N,n ≥ 2.

Mihaly Bencze

PP29530. Determine all a, b > 0 for which

ab +

b

4(√a+

√b)

2 +4(

√a+

√b)

2

a ≥ ab+ 4 (a+ b)�√

a+√b�2

.

Mihaly Bencze

PP29531. If a, b, c > 0 andP b

3a+1 ≤ 8, thenP ab

80a3+16a2+1≤ 1.

Mihaly Bencze



1).P�

rbra

�2≥ 81r2((4R+r)2−2s2)

s4

2).P�

hb

ha

�2≥ 81r4

��s2+r2+4Rr

4s2r2

�2− R

s2r3

�

Mihaly Bencze

PP29533. If x ∈�0, π2

�then tg4x+ cos4 x+ 1

sin4 x≥

�1− sin2 x cos2 x

�2.

Mihaly Bencze

PP29534. If a, b, c > 0 and abc=1 thenP 3a4+2b2−4a+11

(ab+a+1)2≥ 4.

Mihaly Bencze

PP29535. If a, b, c > 0 and abc = 1 thenP 1

(3a4+2b2−4a+11)n≥ 3

12n for all

n ∈ N.

Mihaly Bencze

PP29536. If a, b > 0 then1

3√3a4+2b2−4a+11+ 3

qa2b2

3a2b6+11a2b2−4a2b3+2+ 3

qa4b4

11a4b4+2a6b4−4a3b3+3≤ 3

q94 .

Mihaly Bencze

PP29537. If ak > 0 (k = 1, 2, ..., n) , thenQ

cyclic

3a41+2a22−4a1+11a1a2+a1+1 ≥ 4n.

Mihaly Bencze

PP29538. If x ∈�0, π2

�then

(1 + sinx+ cosx)2 + sin8 x+ cos8 x ≥ 53 + 2

3 sin2 2x.

Mihaly Bencze

PP29539. Determine all x, y > 0 for which

3

�x4 + y4 + 16 (x+ y)4 +

�√x+

√y +

p2 (x+ y)

�2�

≥

≥ 4�5x2 + 8xy + 5y2

� �2x2 + 5xy + 2y2

�.

Mihaly Bencze


PP29540. If xk > 0 (k = 1, 2, ..., n) andnP

k=1

xk = n then

nPk=1

x4k +

�nP

k=1

√xk

�2

≥ n+1n

�nP

k=1

x2k

� Pcyclic

x1x2

!.

Mihaly Bencze


1). 27r3P 1

r4a+�P 1√

ra

�2≥ 36

s2

�s2 − 2r (4R+ r)

�(4R+ r)

2). 27r3P 1

h4a+�P 1√

ha

�2≥ 9(s2−r2−4Rr)(s2+r2+4Rr)

2s4r

Mihaly Bencze

PP29542. If ak > 0 (k = 1, 2, ..., n) andnQ

k=1

ak = 1 then

Pcyclic

am1

am+12

�

r(am1 +r−1)≥ n

r for all m ∈ N∗, r > 1.

Mihaly Bencze

PP29543. If a, b > 0 then a4

b4√

5(a4+4)+ a5b9√

5(b4+4)+ 1

a7b2√5+4a4b4

≥ 35 .

Mihaly Bencze

PP29544. If x ∈ R then�2 + sin2 x

�p2 + sin2 x+

�2 + cos2 x

�√2 + cos2 x ≥ 3

√2− 1.

Mihaly Bencze

PP29545. Determine all a, b > 0 for whicha√a+ b+ b

√2a+ 3b+ 2 (a+ b)

√3a+ 2b ≥ 3

√2.

Mihaly Bencze


1).P 1

ra

q1ra

+ 1rb

≥√2

r√3r

2).P 1

ha

q1ha

+ 1hb

≥√2

r√3r

Mihaly Bencze


PP29547. If x, y > 0 then53xy (x+ y) + x4y3

x3y6+x3y3+1+ x3y4

x6y3+x3y3+1+ 1

xy(x3+y3+1)≥ 6.

Mihaly Bencze

PP29548. If ak > 0 (k = 1, 2, ..., n) andnP

k=1

ak = n thenP

cyclic

a1√a1 + a2 + ...+ an−1 ≥ n

√n− 1.

Mihaly Bencze

PP29549. Determine all a, b > 0 for which12ab+ 60 (a+ b)2 + 5ab (a+ b) ≥ a3 + b3 + 125 (a+ b)3 + 102.

Mihaly Bencze

PP29550. If xk > 0 (k = 1, 2, ..., n) and λ ≥ 1 andnQ

k=1

xk = 1 then

λnP

k=1

xk +P

cyclic

x1xn2+...+xn

n+1 ≥ λ+ 1.

Mihaly Bencze

PP29551. If ak > 0 (k = 1, 2, ..., n) then

Pcyclic

a1a2...an−1

n−1�

(λ+an−11 )(λ+an−1

2 )...(λ+an−1n−1)

≤n

�

cyclic

a1a2...an−1

λn+�

cyclic

a1a2...an−1for all λ > 0.

Mihaly Bencze

PP29552. Determine all x ∈ R for which17 sin2 x cos2 x ≥ sin6 x+ cos6 x+ 16.

Mihaly Bencze

PP29553. If ak > 0 (k = 1, 2, ..., n) and ai 6= aj(i = 1, 2, ..., n; j = 1, 2, ..., n, i 6= j) thenP

1≤i<j≤n

ai+aj(ai−aj)

2 + (n− 1)nP

k=1

1ak

≥ 9(n−1)n2

4n�

k=1ak

.

Mihaly Bencze


PP29554. If ak > 0 (k = 1, 2, ..., n) and ai 6= aj

(i = 1, 2, ..., n; j = 1, 2, ..., n, i 6= j) thenP

1≤i<j≤n

�1

(ai−aj)2 + 1

aiaj

�≥ 9(n−1)n2

8n�

k=1a2k

.

Mihaly Bencze

PP29555. If a, b, c > 0 and a 6= b 6= c thenP a+2b+3c

(a+2b−3c)2+ 9

P 1a + 3

P abc ≥ 3(37+4

√3+4

√6+24

√2)

2�

a .

Mihaly Bencze

PP29556. If ak > 0 (k = 1, 2, ..., n) and ai 6= aj(i = 1, 2, ..., n; j = 1, 2, ..., n, i 6= j) thenP1≤i<j≤n

�1

(ai−aj)2 + 1

aiaj

�≥ 9n2

4

�

n�

k=1ak

�2 .

Mihaly Bencze

PP29557. If a, b, c > 0 and a 6= b 6= c thenP 1(a+2b−3c)2

+ 2P 1

ab ≥�37 + 4

√3 + 4

√6 + 24

√2�P 1

(a+2b+3c)2.

Mihaly Bencze

PP29558. If a, b, c > 0 and a 6= b 6= c thenP a+b+2c

(a+b−2c)2+ 4

P 1a + 2

�

a2

abc ≥ 9(33+8√2)

4�

a .

Mihaly Bencze

PP29559. If ak > 0 (k = 1, 2, ..., n) and m ∈ N∗ thenPcyclic

a1+a2

(am1 3√a2+am2

3√a1)

3 ≥ n2

4n�

k=1amk

Mihaly Bencze

PP29560. If a, b, c > 0 thenP 1

(a+b−2c)2+ 2

P 1ab ≥

�33 + 8

√2�P 1

(a+b+2c)2

for all a 6= b 6= c.

Mihaly Bencze


PP29561. If xk ∈�0, π2

�(k = 1, 2, ..., n) then

Pcyclic

�1+sinx1 cosx2sinx1+cosx2

�2≤ 9n

8 .

Mihaly Bencze

PP29562. If a, b, c > 0 thenP a+b

(an 3√b+bn 3√a)

3 ≥ 94�

a3nfor all n ∈ N.

Mihaly Bencze


1).Q�

a+√ab+b√

a+√b

�2≤ 729s(s2+r2+2Rr)

256

2).Q�

c+√


√s−b

�2

≤ 729sRr128

3).Q�

ra+√rarb+rb√

ra+√rb

�2≤ 729s2R

128

4).Q�

ha+√hahb+hb√

ha+√hb

�2≤ 729s2r(s2+r2+2Rr)

512R2

5).Q�

sin2 A2+sin A

2sin B

2+sin2 B

2

sin A2+sin B

2

�2

≤ 729((2R−r)(s2+r2−8Rr)−2Rr2)16384R3

6).Q�

cos2 A2+cos A

2cos B

2+cos2 B

2

cos A2+cos B

2

�2

≤ 729((4R+r)3+s2(2R+r))16384R3

Mihaly Bencze


1).P

b�a+

√ac+c√

a+√c

�2≤ 9

4

p(s2 − r2 − 4Rr) (3s2 − r2 − 4Rr)

2).P

(s− b)

�b+√

(s−a)(s−c)√s−a+

√s−c

�2

≤ 94√2

p(s2 − 2r2 − 8Rr) (s2 − r2 − 4Rr)

3).P

rb

�ra+

√rarc+rc√

ra+√rc

�2≤ 9

4√2

r�(4R+ r)2 − 2s2

��(4R+ r)2 − s2

�

4).P

hb

�ha+

√hahc+hc√

ha+√hc

�2≤

≤ 9

4√2

vuut �

s2 + r2 + 4Rr

2R

�2

− 4s2r

R

! �s2 + r2 + 4Rr

2R

�2

− 2s2r

R

!

Mihaly Bencze



1).P 1

b+c

� √a+

√b

a+√ab+b

�2≥ 16

9(s2+r2+2Rr)

2).P 1

a

� √s−a+

√s−b

c+√

(s−a)(s−b)

�2

≥ 49Rr

3).P 1

ra+rb

� √ra+

√rb

ra+√rarb+rb

�2≥ 4(4R+r)

9s2R

4).P 1

ha+hb

� √ha+

√hb

ha+√hahb+hb

�2≥ 8R(s2+r2+4Rr)

9s2r(s2+r2+2Rr)

5).P 1

sin2 B2+sin2 C

2

�sin A

2+sin B

2

sin2 A2+sin A

2sin B

2+sin2 B

2

�2

≥ 256R2(2R−r)9((2R−r)(s2+r2−8Rr)−2Rr2)

6).P 1

cos2 B2+cos2 C

2

�cos A

2+cos B

2

sin2 A2+sin A

2sin B

2+sin2 B

2

�2

≥ 256R2(4R+r)

9((4R+r)3+s2(2R+r))

Mihaly Bencze

PP29566. If x ∈�0, π2

�then 4

3 + 1+ 3√2+3√sin2 x+

3√cos2 x

4−√2−(sinx+cosx)

≥ 163 sin2 x cos2 x.

Mihaly Bencze

PP29567. Determine all a, b > 0 for which3√a+

3√b+ 3

√2(a+b)

4−�√

a+√b+√

2(a+b)� + 8

3

�5a2 + 8ab+ 5b2

�≥ 12.

Mihaly Bencze


1).1+ 3√3r

� 13√ra

4−√3r

� 1√ra

+ 12 ≥ 48r(4R+r)s2

2).1+ 3√3r

� 13√ha

4−√3r

� 1√ha

≥ 12r(4R+r)s2

Mihaly Bencze

PP29569. Prove that1R

−1

�

n�

k=1

xk2+k−1

�

exdx

1+ex = nn+1 .

Mihaly Bencze


1).P a+4

√ab+b

(a−√ab+b)

2 ≤ 6(5s2+r2+4Rr)s(s2+r2+2Rr)


2).P c+4

√(s−a)(s−b)

�

c−√

(s−a)(s−b)�2 ≤ 3(s2+r2+4Rr)

sRr

3).P ra+4

√rarb+rb

(ra−√rarb+rb)

2 ≤ 3((4R+r)2+s2)s2R

4).P ha+4

√hahb+hb

(ha−√hahb+hb)

2 ≤3�

(s2+r2+4Rr)2+8s2Rr

�

s2r(s2+r2+2Rr)

5).P sin2 A

2+4 sin A

2sin B

2+sin2 B

2

(sin2 A2−sin A

2sin B

2+sin2 B

2 )2 ≤ 24R(16R2−24Rr+5r2+s2)

(2R−r)(s2+r2−8Rr)−2Rr2

6).P cos2 A

2+4 cos A

2cos B

2+cos2 B

2

(cos2 A2−cos A

2cos B

2+cos2 B

2 )2 ≤ 24R(5(4R+r)2+s2)

(4R+r)3+s2(2R+r)

Mihaly Bencze


1).P cos3 A

2

cos B2+cos C

2−cos A

2

≥ 4R+r2R

2).P cos5 A

2

cos B2+cos C

2−cos A

2

≥ (4R+r)2−s2

8R2

3).P m3

a

mb+mc−ma≥ 3

2

�s2 − r2 − 4Rr

�

4).P m5

a

mb+mc−ma≥ 9

8

�2�s2 − r2 − 4Rr

�2 −�s2 + r2 + 4Rr

�2+ 16s2Rr

�

Mihaly Bencze


1).P a2+ab+b2

a3+b3≤ 3(5s2+r2+4Rr)

2s(s2+r2+2Rr)

2).P (s−a)2+(s−a)(s−b)+(s−b)2

(s−a)3+(s−b)3≤ 3(s2+r2+4Rr)

4sRr

3).P r2a+rarb+r2

b

r3a+r3b

≤ 4((4R+r)2+s2)4s2R

4).P h2

a+hahb+h2b

h3a+h3

b

≤3�

(s2+r2+4Rr)2+8s2Rr

�

4s2r(s2+r2+2Rr)

5).P sin4 A

2+sin2 A

2sin2 B

2+sin4 B

2

sin6 A2+sin6 B

2

≤ 6R(16R2−24Rr+5r2+s2)(2R−r)(s2+r2−8Rr)−2Rr2

6).P cos4 A

2+cos2 A

2cos2 B

2+cos4 B

2

cos6 A2+cos6 B

2

≤ 6R(5(4R+r)2+s2)(4R+r)3+s2(2R+r)

Mihaly Bencze

PP29573. If ak > 0 (k = 1, 2, ..., n) thenP

1≤i<j≤n

�a3iaj

+a3jai

�≥ (n− 1)

nPk=1

a2k.

Mihaly Bencze


PP29574. If ak > 0 (k = 1, 2, ..., n) then

1).P

cyclic

(5a1+4a2)(5a1+a2)a1+a2

≥ 27nP

k=1

ak

2).P

cyclic

(5a1+4a2)(5a1+a2)8a1+a2

≥ 6nP

k=1

ak

Mihaly Bencze


1).P sinA

(1−sinA)2≥

√6�10− s2+r2+4Rr

2sr

�

2).P sin2 A

cos4 A≥

√6

�10 + 4R

r −�s2+r2+Rr

2sr

�2�

Mihaly Bencze


1).P cosA

(1−cosA)2≥

√6�10− s2+r2−4R2

s2−(2R+r)2

�

2).P cos2 A

sin4 A≥

√6

�10 + 8R(R+r)

s2−(2R+r)2−�

s2+r2−4R2

s2−(2R+r)2

�2�

Mihaly Bencze

PP29577. If ak > 0 (k = 1, 2, ..., n) andnP

k=1

ak = 1 then

P a51(a2+1)(a3+1) ≥

1n2(n+1)2

.

Mihaly Bencze


P rbrc(r+rb)(r+rc)r5a

≥ 1144r5

2).P hbhc

(r+hb)(r+hc)h5a≥ 1

144r5

Mihaly Bencze


P Arb

≥ π3r

2).P A

hb≥ π

3r

Mihaly Bencze


PP29580. If xk ∈�0, π2

�(k = 1, 2, ..., n) then

Pcyclic

(1−sinx1 cosx2)2

1+4 sinx1 cosx2≥ n

12 .

Mihaly Bencze


k=1

((k + 1)!)k2 ≤

�6((n−1)(n+2)!+2)n(n+1)(2n+1)

�n(n+1)(2n+1)6

.

Mihaly Bencze


k=1

k! ≤ 12((n−1)(n+2)!+2)(n+1)(n+3)(2n+1) .

Mihaly Bencze


P 2a2−3ab+2b2

a4+b4≤ 1

4Rr

2).P 2(s−a)2−3(s−a)(s−b)+2(s−b)2

(s−a)4+(s−b)4≤ 1

2r2

3).P 2r2a−3rarb+2r2

b

r4a+r4b

≤ 4R+r4s2r

4).P 2h2

a−3hahb+2h2b

h4a+h4

b

≤ s2+r2+4Rr8s2r2

5).P 2 sin4 A

2−3 sin2 A

2sin2 B

2+sin4 B

2

sin8 A2+sin8 B

2

≤ 4R(2R−r)r2

6).P 2 cos4 A

2−3 cos2 A

2cos2 B

2+cos4 B

2

cos8 A2+cos8 B

2

≤ 4R(4R+r)s2

Mihaly Bencze


6 +P �

tg3A2 ctgB2 + tg3B2 ctg

A2

�≥ 8(4s2r2−2(s2−r2−4Rr)(s2−r2−4Rr−4R2))

(s2−(2R+r)2)2 .

Mihaly Bencze

PP29585. In all triangle ABC holdsP (2a2−3ab+2b2)(a+b)

a4+b4≤ s2+r2+4Rr

4sRr .

Mihaly Bencze

PP29586. Prove that (n+1)(n+2)3 ≤ (n−1)(n+2)!+2

(n+1)!−1 .

Mihaly Bencze



k=1

k! ≤ 2((n+1)!−1)n+1 .

Mihaly Bencze


k=1

1(k+1)! ≤

2((n+1)!−1)(n+1)(n+1)! .

Mihaly Bencze


k=1

k! ≤ 3(n+1)(n+1)!n2+3n+5

.

Mihaly Bencze


k=1

14k4+1

≤ n2n2+2n+1

.

Mihaly Bencze

PP29591. In all triangle ABC holdsP (2a2−3ab+2b2)(a+b)2

a4+b4≤ s2+r2+10Rr

4Rr .

Mihaly Bencze


(a+b)(2a2−3ab+2b2)≥ (s2+r2+4Rr)

2+8s2Rr

s(s2+r2+2Rr).

Mihaly Bencze


(2a2−3ab+2b2)(b+c)(c+a)≥ 2(s2+r2−2Rr)

s2+r2+2Rr.

Mihaly Bencze

PP29594. If ak > 0 (k = 1, 2, ..., n) , thenP

1≤i<j≤n

a4i+a4jaiaj

≥ (n− 1)nP

k=1

a2k.

Mihaly Bencze

PP29595. If ak > 0 (k = 1, 2, ..., n) , thenP

1≤i<j≤naiaj

�2a2i − 3aiaj + 2a2j

�≤ n−1

2

nPk=1

a2k.

Mihaly Bencze


PP29596. In all scalene triangle ABC holds

1).P

w2a

qm2

a−h2a

w2a−h2

a= s2 + r2 + 4Rr

2).P w2

ahb

ha

qm2

a−h2a

w2a−h2

a= s2 + r2 + 4Rr

3).P

w2aha

qm2

a−h2a

w2a−h2

a=

(s2+r2+4Rr)2−16s2Rr

2R

4).P w2

araha

qm2

a−h2a

w2a−h2

a= 2R (4R+ r)

5).P w2

ar2a

ha

qm2

a−h2a

w2a−h2

a= 2R

�(4R− r)2 − 2s2

�

Mihaly Bencze

PP29597. If a, b ∈ Z such that a2 + b2 is divisible by a+ b2, then2a3b2 + ab2 + 3b4 is divisible by a+ b2.

Mihaly Bencze


2P

ab sin A2 sin B

2 ≥�s+ s(R−2r)2

4R(R+2r)

�2− (4R+r)r2+(2R−3r)s2

2R .

Mihaly Bencze

PP29599. In all acute triangle ABC holdsP ln(9−cosA)−ln(cosA)

ln(9−cosA)−ln(3−cosA) ≤3(s2+r2−4R2)2(s2−(2R+r)2)

.

Mihaly Bencze


1).P ln(8+cos2 A

2 )−2 ln(sin A2 )

ln(8+cos2 A2 )−ln(2+cos2 A

2 )≤ 3(s2+r2−8Rr)

2r2

2).P ln(8+sin2 A

2 )−2 ln(cos A2 )

ln(8+sin2 A2 )−ln(2+sin2 A

2 )≤ 3(s2+(4R+r)2)

2s2

Mihaly Bencze

PP29601. If x, y ∈ [0, 1] then�3− x2

� �3− y2

�sin (x+ y) ≥ 3 (x+ y) (3− xy) cosx cos y.

Mihaly Bencze

PP29602. In all triangle ABC holdsP ln(9−sinA)−ln(sinA)

ln(9−sinA)−ln(3−sinA) ≤3(s2+r2+4Rr)

4sr .

Mihaly Bencze


PP29603. In all acute triangle ABC holds 9− π2

3 ≥ 3(R+r)R + 2

PA sinA.

Mihaly Bencze

PP29604. If x ∈�0, π2

�then sinx

tg(sinx) +cosx

tg(cosx) ≤53 .

Mihaly Bencze

PP29605. In all acute triangle ABC holdsP �2 + sin2A

�tg (cosA) ≥ 3(R+r)

R .

Mihaly Bencze


P �3− sin4 A

2

�tg

�sin2 A

2

�≥ 3(2R−r)

2R

2).P �

3− cos4 A2

�tg

�cos2 A

2

�≥ 3(4R+r)

2R

Mihaly Bencze

PP29607. If xk ∈ [0, 1] (k = 1, 2, ..., n) thennP

k=1

tgxk ≥3n2

n�

k=1xk

3n2−�

n�

k=1xk

�2 .

Mihaly Bencze

PP29608. If x ∈�0, π2

�then�

3− sin4 x�tg

�sin2 x

�+�3− cos4 x

�tg

�cos2 x

�≥ 3.

Mihaly Bencze


�2(a+b)2

ab

�2a2+b2

+�2(a+b)2

ab

�a2+2b2

≤�3a2+2ab+b2

a(a+2b)

�(2a+b)2

+�a2+2ab+3b2

b(b+2a)

�(a+2b)2

.

Mihaly Bencze

PP29610. If ak ∈ [0, 1] (k = 1, 2, ..., n) thennP

k=1

aλk�3− a2k

�tgak ≥ 3

nλ

�nP

k=1

ak

�λ+1

for all λ ∈ (−∞,−1] ∪ [0,+∞) .

Mihaly Bencze



Aλ−1�3π2 −A2

�tgA

π ≥ πλ+1

3λ−2 for allλ ∈ (−∞,−1] ∪ [1,+∞) .

Mihaly Bencze


2 + cos2A�tg (sinA) ≥ 3s

R .

Mihaly Bencze


k=1

�9−

�kn

�4�tg k

n ≥ 3(n+1)(7n+1)4n .

Mihaly Bencze


k=1

kctg kn ≤ 16n2−3n−1

18 .

Mihaly Bencze

PP29615. If 0 ≤ a ≤ b ≤ 1 then�cos acos b

� 23 ≥ 3−a2

3−b2.

Mihaly Bencze

PP29616. If a, b > 0 then 9�4a3 + b (a+ b)3

��4b3 + a (a+ b)3

�≥

≥ (2a+ b)2 (a+ 2b)2 (a+ b)4 .

Mihaly Bencze

PP29617. If a, b, c > 0, λ > 0 such thatP

a = 9λ+1 thenp

λP

a2 + (λ2 + λ+ 1)P

ab ≤P√a+ λb.

Mihaly Bencze

PP29618. If xk ∈ R (k = 1, 2, ..., n) , xi 6= xj(i 6= j) andnP

k=1

xk = 1 then

Pcyclic

ex1−ex2x1−x2

≥ ne1n .

Mihaly Bencze

PP29619. Let ABCD be a tetrahedron inscribed in a sphere with radius R.The insphere of the tetrahedron have the incenter I and inradius r. DenoteRA, RB, RC , RD the radii of the circumsphere of tetrahedronsIBCD, IACD, IABD, IABC.


Prove that V ol[RARBRCRD]V ol[ABCD] = R

3r .

Mihaly Bencze

PP29620. If a, b, c > 0 and λ > 0 such thatP

a = 3λ+1 thenP 1

1+2(a+λb)(b+λc) ≥2

1+(a+λb)(b+λc)(c+λa) .

Mihaly Bencze

PP29621. If x ∈�0, π2

�then 1+ 4

1+sin 2x+4 cos2 x+ 4

1+sin 2x+4 sin2 x≥ 40

9(1+sin 2x) .

Mihaly Bencze

PP29622. Let A1A2...An be a convex polygon, and M a random point inspace. Denote x = A1MA2∡ = A2MA3∡ = ... = AnMA1∡. DenoteR1, R2, ..., Rn the circumradii of triangles A1MA2, A2MA3, ..., AnMA1.

Determine x such that

�nP

k=1

Rk

�2

≥nP

k=1

MA2k.

Mihaly Bencze

PP29623. If a, b > 0 and (a+ b)√a = 2, then (2a+ b)3 + 5a2b ≥ 32.

Mihaly Bencze

PP29624. If a, b, c > 0 and a3 + ab (a+ b)− 2a− b = 0 then2a2

�2 + a2

�≥ (2 + ab)

�2 + a2 − b2

�.

Mihaly Bencze

PP29625. In all triangle ABC holdsP (a+b)2

c ≤ 6√3R2.

Mihaly Bencze


�r3a − r3

� �r3b − r3

� �r3c − r3

�≥ 18252r9

2).�h3a − r3

� �h3b − r3

� �h3c − r3

�≥ 18252r9

Mihaly Bencze


3π2 −A2�tgA

π ≥ 3π2.

Mihaly Bencze


PP29628. If a, b > 0 and n ≥ 3 then�(2a+ b)n−1 + n− 1

��(a+ 2b)n−1 + n− 1

�≥ 9n2ab

n√2a2+5ab+2b2

, for all n ∈ N∗.

Mihaly Bencze

PP29629. If x, y, z > 0 then

27(ex−1)(ey−1)(ez−1)xyz

�ex+y+z

3 −1x+y+z

�3

≥ 64

�

ex+y2 −1

��

ey+z2 −1

��

ez+x2 −1

�

(x+y)(y+z)(z+x)

2

.

Mihaly Bencze

PP29630. If x ∈�0, π2

�then 16 + 7

sin4 x cos4 x≤ 4

sin2 x cos2 x.

Mihaly Bencze

PP29631. If a, b, c > 0 then 6P

(a− bc)2 ≥ (P |a− b|)2 .

Mihaly Bencze

PP29632. Find all ak, bk ∈ R (k = 1, 2, ..., n) such that

3x5+4x4+5x3+6x2+7x+8(x2−x+1)n

=nP

k=1

akx+bk(x2−x+1)k

.

Mihaly Bencze

PP29633. If x > 0 then ch6x+ sh6x+ 1ch2x

+ 1sh2x

+ ch3xshx + sh3x

chx ≥ 32 .

Mihaly Bencze

PP29634. If x, y, z, t > 0 then

2x+y+z+t + 2x+y + 2x+z + 2x+t + 2y+z + 2y+t + 2z+t + 1 ≥≥ 2x + 2y + 2z + 2t + xy+yz+zx

√512xyz.

Mihaly Bencze

PP29635. If ak ∈ (−λ,λ) ,λ > 0 (k = 1, 2, ..., n) , thenP |a1+a2|+|an+a1|λ−a21

≥ 8P |a+a2|

4λ−(a1+a2)2 .

Mihaly Bencze


P 1cos2 A

≥ 4P 1

4−(sinA+sinB)2

2).P 1

sin2 A≥ 4

P 14−(cosA+cosB)2

Mihaly Bencze


PP29637. If ai ∈ (−λ,λ) , (i = 1, 2, ..., n) , λ > 0 and k ∈ {1, 2, ..., n} , thennP

i=1

1λ−a2i

≥ k2P 1

λk2−(a1+...+ak)2 .

Mihaly Bencze

PP29638. In all convex quadrilateral ABCD holds

1).P 1

cos2 A≥ 9

P 19−(sinA+sinB+sinC)2

2).P 1

sin2 A≥ 9

P 19−(cosA+cosB+cosC)2

Mihaly Bencze

PP29639. In all convex polygon A1A2...An holds

1).nP

i=1

1cos2 Ai

≥ k2P 1

k2−(sinA1+...+sinAk)2

2).nP

i=1

1sin2 Ai

≥ k2P 1

k2−(cosA1+...+cosAk)2

Mihaly Bencze

PP29640. If xk ∈ (0, 1) (k = 1, 2, ..., n) and λ ∈ (−∞, 0] ∪ [1,+∞) and

k ∈ {1, 2, ..., n} , thennP

i=1

11−xλ

i

≥ kλP 1

kλ−(x1+...+xk)λ .

Mihaly Bencze

PP29641. If x, y, ak > 0 (k = 1, 2, ..., n) then

n

snQ

k=1

(xak + y) ≥ y +nx

n�

k=1ak

�

cyclic

a1a2...an−1.

Mihaly Bencze

PP29642. If x > 0 then144 (x cosx− sinx)2 (x sinx+ cosx− 1)2 ≤ x6

�4x2 − sin2 2x

�.

Mihaly Bencze

PP29643. If x > 0 then 3 (x sinx+ cosx− 1)2 + 3 (sinx− x cosx)2 ≤ x4.

Mihaly Bencze


PP29644. If ai > 0 (i = 1, 2, ..., n) and k, p ∈ N∗ thennQ

i=1

�a2(k+p)i − a2ki + 2p+ 1

�≥

�nP

i=1

2p+1

qa2pi

�2p+1

.

Mihaly Bencze

PP29645. Let ABCDE be a convex pentagon. Prove that:

√AB2 +BC2 + CD2 +

√BC2 + CD2 +DA2 +

√CD2 +DA2 +AB2+

+√DA2 +AB2 +BC2 +

√BC2 + CD2 +DE2 +

√CD2 +DE2 + EB2+

+√DE2 + EB2 +BC2 +

√EB2 +BC2 + CD2 +

√CD2 +DE2 + EA2+

+√DE2 + EA2 +AC2 +

√EA2 +AC2 + CD2 +

√AC2 + CD2 +DE2+

+√DE2 + EA2 + EB2 +

√EA2 + EB2 +BD2 +

√EB2 +BD2 +DE2+

+√BD2 +DE2 + EA2 +

√EA2 +AB2 +BC2 +

√AB2 +BC2 +DE2+

+√BC2 +DE2 + EA2 +

√DE2 + EA2 +AB2 ≥

≥ 2√3�√

AC ·BD +√BD · CE +

√CE ·DA+

√DA · EB +

√EB ·AC

�.

Mihaly Bencze

PP29646. If ak > 0 (k = 1, 2, ..., n) then

P a1a2+a3+...+an

≥

�

n�

k=1ak

�2

2�

1≤i<j≤n

aiaj≥ n

n−1 .

Mihaly Bencze

PP29647. Let ABCD be a cyclic quadrilateral. Prove thatsinA sinB + sinB sinC + sinC sinD + sinD sinA ≤≤ 4

abcd (−a+ b+ c+ d) (a− b+ c+ d) (a+ b− c+ d) (a+ b+ c− d) .

Mihaly Bencze

PP29648. Let ABCDE be a convex pentagon inscribed in circle with radiusR. Prove that3 (AB +BC + CD +DE + EA) +DA+ EB +AC +BD + CE ≥≥ 2

q2√2

R ((AB ·BC · CD ·DA)38 + (BC · CD ·DE · ED)

38 +

+(CD ·DE · EA ·AC)38 + (DE · EA ·AB ·BD)

38 + (EA ·AB ·BC · CE)

38 ).

Mihaly Bencze


PP29649. Determine all cyclic quadrilaterals ABCD in which

cos A2 + cos B

2 + cos C2 + cos D

2 ≤ 44

qArea2[ABCD]

4abcd .

Mihaly Bencze

PP29650. In the cyclic quadrilateral ABCD holds

�a2 − b2 − c2 + d2 + F

� �b2 − c2 − d2 + a2 + F

� �c2 − d2 − a2 + b2 + F

�·

·�d2 − a2 − b2 + c2 + F

�≤ 64 (ad+ bc)4 where F = Area [ABCD] .

Mihaly Bencze

PP29651. If ak > 0 (k = 1, 2, ..., n) , thenP

cyclic

a21−a1a2+a22a22+a2a3+a23

≥ n3 .

Mihaly Bencze

PP29652. If ak > 0 (k = 1, 2, ..., n) , thenP (a21−a1a2+a22)

2

(a22+a2a3+a23)(a23+a3a4+a24)≥ n

9 .

Mihaly Bencze


k=1

3k2+3k+1k(k3+2k2+2k+1)

≤ 3nn+1 .

Mihaly Bencze

PP29654. If a, b, c > 0 then 9 ≥ (2�

a2+�

ab)2

2�

a4−�

a2b2.

Mihaly Bencze

PP29655. If ak > 0 (k = 1, 2, ..., n) thenP a41+a21a

22+a42

(a22+a2a3+a23)(a23+a3a4+a24)≥ n

3 .

Mihaly Bencze


1).P m2

a−mamb+m2b

m2b+mbmc+m2

c< 3

2).P 2+cosA+cosB−2 cos A

2cos B

2

2+cosB+cosC−2 cos B2cos C

2

< 3

Mihaly Bencze


PP29657. If ak > 0 (k = 1, 2, ..., n) , thenP�

a21−a1a2+a22a22+a2a3+a23

�λ≥ n

3λfor all

λ ∈ (−∞, 0] ∪ [1,+∞) .

Mihaly Bencze


1).Q

(ma +mb) ≥ 89 (P

ma) (P

mamb)2). 9

Qcos

�π−A4

�Qcos A−B

4 ≥�P

cos A2

� �Pcos A

2 cos B2

�

Mihaly Bencze


1).P 1

cosA(5+4 cosA) ≥67 − 18

49 ln

�2(s2−(2R+r)2)

R2

�

2).P cos2 A

1+5 cosA ≥ 314 + 9

98 ln

�2(s2−(2R+r)2)

R2

�

Mihaly Bencze


P 1ra(5+2ra)

≥ 37 − 9

49 ln�s2r

�

2).P r2a

2+5ra≥ 3

7 + 949 ln

�s2r

�

Mihaly Bencze


1).P 1

sinA(5√3+4 sinA)

≥ 221 − 2

49 ln�

4sr3√3R2

�

2).P sin2 A

5 sinA+√3≥ 2

√3

7 + 6√3

49 ln�

4sr3√3R2

�

Mihaly Bencze

PP29662. If ak > 0 (k = 1, 2, ..., n) , thennP

k=1

1xk(5+2xk)

≥ n7 − 9

49 ln

�nQ

k=1

ak

�.

Mihaly Bencze


PP29663. Prove that

1).nP

k=1

(k+1)2

k(7k+5) ≥n7 + 9

49 ln (n+ 1)

2).nP

k=1

k2

(k+1)(7k+2) ≥n7 − 9

49 ln (n+ 1)

Mihaly Bencze

PP29664. If ak > 0 (k = 1, 2, ..., n) , thenP a22

a1(2a1+5a2)≥ n

7 .

Mihaly Bencze

PP29665. Prove that

1).nP

k=1

k2k

(k+1)k(5kk+2(k+1)k)≥ n

7 − 949 ln

�(n+1)n

n!

�

2).nP

k=1

(k+1)2k

kk(2kk+5(k+1)k)≥ n

7 + 949 ln

�(n+1)n

n!

�

Mihaly Bencze

PP29666. If λ > 0 thennP

k=1

1kλ(5+2kλ)

≥ n7 − 9λ

49 ln (n!) .

Mihaly Bencze

PP29667. If ak > 0 (k = 1, 2, ..., n) , then

4P

cyclic

a23(a1+a2)(2a1+2a2+10a3)

≥ n7 − 9

49 ln�

12nQ a1+a2

a3

�.

Mihaly Bencze

PP29668. If xk > 0 (k = 1, 2, ..., n) , thenP 1

ch2xk(5+2ch2xk)≥ n

7 − 9n49 ln

�1 + 1

n

nPk=1

sh2xk

�.

Mihaly Bencze

PP29669. If xk ≥ 1 (k = 1, 2, ..., n) , then

nQk=1

�7xk

5+2xk

� 495

�nQ

k=1

xxk

k

�9

≥

7n�

k=1xk

5n+2n�

k=1

xk

495 �

1n

nPk=1

xk

�9n�

k=1xk

. If

0 < xk < 1 (k = 1, 2, ..., n) , then holds the reverse inequality.

Mihaly Bencze



1). 15

Pln 7

2+5 sinA − 949

P ln(sinA)sinA ≥ 144

49

�s2+r2+4Rr

2sr − 3�

2). 15

Pln 7

2+5 sin2 A− 18

49

P ln(sinA)

sin2 A≥ 144

49

��s2+r2+Rr

2sr

�2− 4R

r − 3

�

Mihaly Bencze

PP29671. If x, y, z ≥ 1 then�Q 7x5+2x

� 15(Q

xx)949

�7�

x15+2

�

x

�3·��

x3

� 949

�

x≥Q

�7(x+y)5+x+y

� 25 �x+y

2

� 949

(x+y).

If 0 < x, y, z ≤ 1 then holds the reverse inequality.

Mihaly Bencze


1). 15

Pln 7

2+5 cosA − 949

P ln(cosA)cosA ≥ 144

49

�s2+r2−4R2

s2−(2R+r)2− 3

�

2). 15

Pln 7

2+5 cos2 A− 18

49

P ln(cosA)cos2 A

≥ 14449

��s2+r2−4R2

s2−(2R+r)2

�2− 8R(R+r)

s2−(2R+r)2− 3

�

Mihaly Bencze


k=1

k45k

(2k+5)49≥ e360n(n−1)

(7n·n!)49 .

Mihaly Bencze


k=1

k90k2

(2k2+5)49≥ e120n(n−1)(2n+5)

(7n(n!)2)49 .

Mihaly Bencze


k=1

(1+ 1k )

45(1+ 1k )

(7k+2)49≥ (n+1)720

(7n·(n+1)!)49.

Mihaly Bencze


k=1

k2+k+17k2+7k+1

�k2+k+1k2+k

� 45(k2+k+1)k2+k ≥ 7−ne

720nn+1 .

Mihaly Bencze


PP29677. If x ≥ 1 then 495 ln 7x

2x+5 + 9x lnx ≥ 144 (x− 1). What happens ifx ∈ (0, 1)?

Mihaly Bencze

PP29678. In all triangle ABC holdsQ ( a+b

c )45(a+b)

c

(2a+2b+5c)49≥ e

360

�

s2+r2

Rr−8

�

(686s(s2+r2+2Rr))49.

Mihaly Bencze


�

ra+rbrc

�

45(ra+rb)rc

(2ra+2rb+5rc)49 ≥ e

720

�

2(2R−r)r −3

�

(4s2R)49.

Mihaly Bencze

PP29680. If xk ≥ 1 (k = 1, 2, ..., n) , then

49P x1−1

x2−1 ln7x1

5+2x1+ 45

P�x1−1x2−1

�x1 lnx1 ≥ 720

�nP

k=1

xk − n

�.

Mihaly Bencze

PP29681. Prove thatn+1Pk=2

1

(49 ln 7k2k+5

+45k ln k)(49 ln 7k+72k+7

+45(k+1) ln(k+1))≤ n

158400(n+1) .

Mihaly Bencze


P2sinA (sinB + sinC)1−sinA ≤ 4

PsinA

2).P

2cosA (cosB + cosC)1−cosA ≤ 4P

cosA if ABC is acute

3).P

2sin2 A

�sin2B + sin2C

�cos2 A ≤ 4P

sin2A

4).P

2cos2 A

�cos2B + cos2C

�sin2 A ≤ 4P

cos2A

Mihaly Bencze

PP29683. In all quadrilateral ABCD holdsP(sinA)sinB (sinA+ sinB)sinC (sinA+ sinB + sinC)sinD ≥ 1.

Mihaly Bencze

PP29684. If x ∈ R then (sinx cosx)2 sin2 x + (sinx cosx)2 cos

2 x ≥ 1.

Mihaly Bencze



1).P

(sinA+ sinB)sinC ≥ 2

2).P

(cosA+ cosB)cosC ≥ 2 if ABC is acute

Mihaly Bencze

PP29686. If ak ∈ (0, 1) (k = 1, 2, ..., n) thenPcyclic

aa21 (a1 + a2)a3 ... (a1 + a2 + ...+ an−1)

an ≥ 1.

Mihaly Bencze


1).P

(sinA)sinB (sinA+ sinB)sinC ≥ 1

2).P

(cosA)cosB (cosA+ cosB)cosC ≥ 1 if ABC is acute

Mihaly Bencze


1).P

(sinA+ sinB)sinC ≥ 2

2).P

(cosA+ cosB)cosC ≥ 2 if ABC is acute

Mihaly Bencze

PP29689. If ak ∈ (0, 1) (k = 1, 2, ..., n) , thenaa21 (a1 + a2)

a3 ... (a1 + a2 + ...+ an−1)an ≥ a1

a1+a2+...+an.

Mihaly Bencze

PP29690. If ak ∈�0, 1

n−1

�(k = 1, 2, ..., n) , then

Pcyclic

aa2+a3+...+an1 ≥ 1.

Mihaly Bencze

PP29691. If ak ∈ (0, 1) (k = 1, 2, ..., n) thenP(a1 + a2 + ...+ an−1)

an ≥ n− 1.

Mihaly Bencze

PP29692. If x > 1 thennP

k=1

xk−1−1xk+1

≤ n(n−1)(x−1)4x .

Mihaly Bencze


PP29693. If a, b > 0 then bn+1−an+1

(n+1)(b−a) +√akbk

�bn−k+1−an−k+1

(n−k+1)(b−a)

�≤ an + bn, for

all n, k ∈ N.

Mihaly Bencze


(xn−1−1)dx(xn+1)(x−1) ≥

n−12 ln b

a .

Mihaly Bencze

PP29695. If xk > 1 (k = 1, 2, ..., n) and n ≥ 2 then

nPk=1

xk(xmk+1)

xm−1k

−1≥

2

�

n�

k=1xk

�2

(m−1)

�

n�

k=1

xk−n

� for all m ≥ 2.

Mihaly Bencze

PP29696. If λ, ai > 0 (i = 1, 2, ..., n) and k ≥ 2 thenP

cyclic

(λ+a1)k

λ2+λka2+k(k−1)

2a22

≥ nλk−2.

Mihaly Bencze


1).P ab

a2−ab+b2≥ s2+9r2+18Rr

2(s2−3r2−6Rr)

2).P rarb

r2a−rarb+r2b

≤ 6s2R(4R+r)2−2s2

− 1

Mihaly Bencze

PP29698. Determine all λk, µk > 0 (k = 1, 2, 3, 4) such that for all a, b, c > 0holds

�Paλ1bµ1

� �Paλ2bµ2

�≥

�Paλ3bµ3

� �Paλ4bµ4

�.

Mihaly Bencze


1).P (1+cosn A)(1−cosA) cos2 A

1−cosn−1 A≥ 2(s2+r2−4R2)

(n−1)(s2−(2R+r)2)

2).P (1+cos2n A) cos4 A sin2 A

1−cos2n−2 A≥ 2

n−1

��s2+r2−4R2

s2−(2R+r)2

�2− 8R(R+r)

s2−(2R+r)2

�for all

n ≥ 2

Mihaly Bencze



1).P (1+sinn A)(1−sinA) sin2 A

1−sinn−1 A≥ s2+r2+4Rr

(n−1)sr

2).P (1+sin2n A) sin4 A cos2 A

1−sin2n−2 A≥ 2

n−1

��s2+r2+Rr

2sr

�2− 4R

r

�for all n ≥ 2

Mihaly Bencze

PP29701. In all acute triangle ABC holdsP tg2A−sin2 A

B2−sin2 B≥ 9.

Mihaly Bencze

PP29702. If xk ∈�0, π2

�(k = 1, 2, ..., n) then

nPk=1

tgxk ≥ 12

nPk=1

sin 2xk +nP

k=1

x3k.

Mihaly Bencze


A3 ≤ 2srs2−(2R+r)2

− srR2 .

Mihaly Bencze

PP29704. If x ∈�0, π2

�then tgx ≥ x3 + sin 2x

2 .

Mihaly Bencze


A�tg2A+ 2 sin2A

�≥ π3

3 .

Mihaly Bencze

PP29706. In all acute triangle ABC holdsP tg2A+2 sin2 A

A ≥ 3π.

Mihaly Bencze

PP29707. In all acute triangle ABC holdsP tg2A+2 sin2 A

B2 ≥ 9.

Mihaly Bencze

PP29708. In all acute triangle ABC holdss2−r(4R+r)

R2 +4s2r2−2(s2−r2−4Rr)(s2−r2−4Rr−4R2)

(s2−(2R+r)2)2 ≥ π2.

Mihaly Bencze


PP29709. If xk ∈�π4 ,

π2

�(k = 1, 2, ..., n) , then

nPk=1

tg2xk + 2nP

k=1

sin2 xk ≥ 3nP

k=1

x2k + 2nP

k=1

xk.

Mihaly Bencze

PP29710. If a, b, c, d > 0 and x, y, z, t ∈ N ∗ then

��

a −b 0 00 b −c 00 0 c −d1 1 1 1 + d

��≥ abcd+ x+y+z+t

((ax)x(by)y(cz)z(dt)t)1

x+y+z+t.

Mihaly Bencze


��

ha −hb 0 00 hb −hc 00 0 hc −hd1 1 1 1 + hd

��+

��

ra −rb 0 00 rb −rc 00 0 rc −rd1 1 1 1 + rd

��=

=�1 + 1

r

�hahbhchd +

�1 + 2

r

�rarbrcrd.

Mihaly Bencze


��

cosx+ cos y − cosx+ cos y − sinx+ sin y sinx+ sin y− cosx+ cos y cosx+ cos y sinx+ sin y − sinx+ sin y− sinx+ cos y sinx+ sin y cosx+ cos y − cosx+ cos ysinx+ sin y − sinx+ sin y − cosx+ cos y cosx+ cos y

��=

= 16 cos 2x cos 2y.

Mihaly Bencze

PP29713. Solve in Z the equation x+y1+3z + y+z

1+3x + z+x1+3y = 3

2 .

Mihaly Bencze


PP29714. If a, b, c > 0 and Δ1 =

��

1 a a3

1 b b3

1 c c3

��;

Δ2 =

��

a2 b2 c2

b2 + c2 c2 + a2 a2 + b2

bc ca ab

��then Δ1+Δ2

(b−a)(a−c)(b−c) ≥2√3(P

a)2 , for all

a 6= b 6= c.



P a2+b2

a4+6a2b2+b4≤ 1

8Rr

2).P r2a+r2

b

r4a+6r2ar2b+r4

b

≤ 4R+r4s2r

Mihaly Bencze

PP29716. In all triangle ABC holds1). s2+r2+4Rr

27sRr ≥P 2b−a(2a+b)(a+2b) +

16s

2). 427r ≥

P 2rb−ra(2ra+rb)(ra+2rb)

+ 13(4R+r)

Mihaly Bencze

PP29717. If a, b, c > 0 thenP 14a+1 + 3

P 12a+2b+1 ≤ 2

P 13a+b+1 + 2

P 1a+3b+1 + 12

P (a−b)4

(2a+2b+1)5.

Mihaly Bencze

PP29718. If a, b, c > 0 then13

P 1a+b+2c +

P 15(b+c)+2a ≥P 1

5a+4b+3c +P 1

5a+3b+4c .

Mihaly Bencze

PP29719. If a, b, c > 0 then�P a

c

� �48 +

�P ab

�3� ≥ 25�P a

b

�2.

Mihaly Bencze


1 ≥ (s2−r2−4Rr)(s2−(2R+r)2)2s2r2

+ 2R2−s2+(2R+r)2

s2−r(4R+r).

Mihaly Bencze


PP29721. In all acute triangle ABC holds 3 ≥ 9+(�

ctgA)2

3+�

ctg2A.

Mihaly Bencze

PP29722. If a, b, c > 0 then 3310 ≤P 8a+7b

14a+13b +P 7a+8b

13a+14b ≤ 103 .

Mihaly Bencze

PP29723. If a, b, c > 0 thenP a2k−1+b2k−1

(1+c)2≥ (

�

a)2k−1

9k−1(1+(�

a)2).

Mihaly Bencze

PP29724. Determine all triangle ABC in which (P

cosActgB)2 ≥P cos2A.

Mihaly Bencze

PP29725. If xk ≥ 1 (k = 1, 2, ..., n) , then 3nP

k=1

x2k(3xk−1)

x2k+xk+1

≥

�

3n�

k=1

xk−n

�2

n+n�

k=1xk

.

Mihaly Bencze

PP29726. If xk ≥ 1 (k = 1, 2, ..., n) , then 3nP

k=1

x2k

x2k+xk+1

≥

�

3n�

k=1xk−n

�2

n�

k=1

x2k+2

n�

k=1

xk−n.

Mihaly Bencze


k=1

k(3k−1)(k2+k+1)

≥ n3(n+1) .

Mihaly Bencze

PP29728. Prove that

nQk=1

(3k − 1)�k2 + k + 1

�≤ 3n (n!)2 (n+ 1)!.

Mihaly Bencze

PP29729. If x, y ∈�0, π2

�then x3y3 (sinx sin y + cos (x+ y)) ≤ sin3 x sin3 y.

Mihaly Bencze


PP29730. In all acute triangle ABC holdsPtg2A+ 2

Psin2A ≥ 3

�A2 +B2 + C2

�.

Mihaly Bencze

PP29731. If xk ∈�0, π2

�(k = 1, 2, ..., n) , then sin

�nP

k=1

xk

�+

nQk=1

sinxk ≥ 0.

Mihaly Bencze

PP29732. If x, y, z ∈�0, π2

�, then

sinx sin y sin z + sin (x+ y + z) ≤P sinx sin3 y sin3 zy3z3

.

Mihaly Bencze

PP29733. If x, y, z ∈�0, π2

�, then

cosx cos y cos z − cos (x+ y + z) ≤P sinx sin y sin3 zz3

.

Mihaly Bencze

PP29734. If x ∈�0, π2

�, then 1

3 sin2 x cos2 x≥ x2 +

�π2 − x

�2.

Mihaly Bencze

PP29735. In all acute triangle ABC holdsP �tg2A+ 2 sin2A

� �tg2B + 2 sin2B

�≥ 9

PA2B2.

Mihaly Bencze

PP29736. In triangle ABC holds A ≤ B ≤ C ≤ π2 . Prove that

P Atg2A+2 sin2 A

≤ (π−B)2

4πAC .

Mihaly Bencze


k=1

(k − ln k) ≤ 1n!e

n(n−1)2 .

Mihaly Bencze

PP29738. If xi > 0 (i = 1, 2, ..., n) and k ∈ {1, 2, ..., n} , thenP

cyclic

�ex1+x2+...+xk−1 + (x2 + ...+ xk) ln (x1 + ...+ xk)

�≥ k2

n

�nP

i=1xi

�2

.

Mihaly Bencze



k=1

(k−ln k)(k+1−ln(k+1))e2k−1 ≤ n

n+1 .

Mihaly Bencze

PP29740. In all triangle ABC holdsP eA

A−lnA ≥ eπ.

Mihaly Bencze

PP29741. If ak > 0 (k = 1, 2, ..., n) , then

nQk=1

�eak + (1 + ak) ln (1 + ak)

�≥ 1 + n

snQ

k=1

ak

!2n

.

Mihaly Bencze

PP29742. If ak > 0 (k = 1, 2, ..., n) , thennQ

k=1

�eak(1+ak) +

�1 + ak + a2k

�ln�1 + ak + a2k

��≥

≥ 1 + n

snQ

k=1

ak + n

snQ

k=1

a2k

!2n

.

Mihaly Bencze

PP29743. Let A1A2...An be a convex polygon. Prove that

nPk=1

eAk + e ln

�nQ

k=1

Ak

�≥ (2 (n− 2)π − n) e.

Mihaly Bencze

PP29744. If ak > 0 (k = 1, 2, ..., n) , thennP

k=1

eak + ln

�nQ

k=1

(1 + ak)

�≥ n+ 2

nPk=1

ak.

Mihaly Bencze

PP29745. If xi > 0 (i = 1, 2, ..., n) and k ∈ N∗ thenP x1

x2+P

x1ek(x2−1) ≥ k+1

kk

k+1

nPi=1

xi.

Mihaly Bencze


PP29746. If ak > 0 (k = 1, 2, ..., n) and An = 1n

nPk=1

ak and Gn = n

snQ

k=1

ak

then eGn−1 + lnGn + 2An ≤ eAn−1 + lnAn + 2Gn.

Mihaly Bencze

PP29747. If xi > 0 (i = 1, 2, ..., n) and k ∈ N∗ thenP

x1ekx2 +

P x11+x2

≥ k+1

kk

k+1

nPi=1

xi.

Mihaly Bencze

PP29748. Prove thatbRa

(ex−1+x lnx)(ex−1+lnx+1)dxx4+1

≥ 12 ln

1+b4

1+a4for all

1 ≤ a ≤ b.

Mihaly Bencze


k=1

1

(ek+ln(k+1))(ek+1+ln(k+2))≤ n

3(2n+3) .

Mihaly Bencze

PP29750. Let A1A2...An be a convex polygon. Prove thatnP

k=1

eπAk

−1+ n lnπ ≥ n(n+2)

n−2 + ln

�nQ

k=1

Ak

�.

Mihaly Bencze

PP29751. If x ∈ R then esin2 x + ecos

2 x + ln��1 + sin2 x

� �1 + cos2 x

��≥ 4.

Mihaly Bencze

PP29752. If x ∈�0, π2

�then

esinx−1 + ecosx−1 + ln�(sinx)sinx (cosx)cosx

�≥ 1.

Mihaly Bencze


eA+BC + 3 lnπ ≥ 15 + ln (ABC) .

Mihaly Bencze



1).P

e1

cosA−1 − ln

�s2−(2R+r)2

4R2

�+ 3 ≥ 2(s2+r2−4R2)

s2−(2R+r)2

2).P

etg2A − 2 ln

�s2−(2R+r)2

4R2

�+ 3 ≥ 2

�s2+r2−4R2

s2−(2R+r)2

�2− 16R(R+r)

s2−(2R+r)2

Mihaly Bencze

PP29755. If ak > 0 (k = 1, 2, ..., n) thennQ

k=1

(eak + ln (1 + ak) + 1) ≥ 2n

1 + n

snQ

k=1

ak

!n

.

Mihaly Bencze


(ex−1+x lnx)dxx6+1

≥ 13arctg

b3−a3

1+a3b3for all 0 < a ≤ b.

Mihaly Bencze


Pe

1sinA

−1 − ln�

sr2R2

�+ 3 ≥ s2+r2+4Rr

sr

2).P

ectg2A − 2 ln

�sr2R2

�+ 3 ≥ 2

�s2+r2+Rr

2sr

�2− 8R

r

Mihaly Bencze

PP29758. In all acute triangle ABC holdsPp(esinA−1 + sinA ln (sinA)) (ecosA−1 + cosA ln (cosA)) ≥ sr

R2 .

Mihaly Bencze

PP29759. In all triangle ABC holdsPr�

e− cos2 A2 + 2 sin2 A

2 ln�sin A

2

��e− sin2 A

2 + 2 cos2 A2 ln

�cos A

2

��≥

≥ s2−r(4R+r)8R2 .

Mihaly Bencze


1).P

esin3 A

2−1 + 3

Psin3 A

2 ln�sin A

2

�≥ (2R−r)((4R+r)2−3s2)+6Rr2

32R3

2).P

ecos3 A

2−1 + 3

Pcos3 A

2 ln�cos A

2

�≥ (4R+r)3−3s2(2R+r)

32R3

Mihaly Bencze


PP29761. If x ∈ R thene− sin2 x + e− cos2 x + ln

�(sinx)2 sin

2 x (cosx)2 cos2 x�≥ 1− 2 sin2 x cos2 x.

Mihaly Bencze


1).P

e− cos2 A2 + 2

Psin2 A

2 ln�sin A

2

�≥ 8R2+r2−s2

8R2

2).P

e− sin2 A2 + 2

Pcos2 A

2 ln�cos A

2

�≥ (4R+r)2−s2

8R2

Mihaly Bencze

PP29763. Let ABCD be a convex quadrilateral andM1 ∈ (AB) , N1 ∈ (BC) ,M3 ∈ (CD) , N3 ∈ (DA) . The quadrilateralsM1M2M3M4 and N1N2N3N4 are the intersections of triangles AM3B andCM1D respective BN3C and AN1D. Determine all quadrilaterals ABCD forwhich Area [M1M2M3M4] +Area [N1N2N3N4] ≤ 1

2Area [ABCD] .

Mihaly Bencze


k=1

1kp ≥ (

√4n+5−2)

2p

n2p−1 for all n ≥ 2 and p ∈ N.

Mihaly Bencze


k=1

1√k≥ 2

m

�√m2n+m2 + 1−m

�for all n ≥ 2 and

m ≥ 2; n,m ∈ N.

Mihaly Bencze

PP29766. If x, y, z > 0 then (P√

x)4 ≥ 16

Pxy.

Mihaly Bencze

PP29767. Determine all ak > 0 (k = 1, 2, ..., n) for which

√n+ 1 ≤

s

an +

ran−1 +

q...+

pa2 +

√a1 ≤

√2n− 1 for all n ≥ 2.

Mihaly Bencze


PP29768. If ak =

s

k +

rk − 1 +

q...+

p2 +

√1 then

n−13(2n+1) ≤

nPk=2

1a2ka2k+1

≤ n2(n+1) .

Mihaly Bencze

PP29769. If a, b, c > 0 then determine all xk, yk > 0 (k = 1, 2, 3) such thatP

cyclic

(ax1bx2cx3)1

x1+x2+x3 ≥ Pcyclic

(ay1by2cy3)1

y1+y2+y3 .

Mihaly Bencze

PP29770. If 0 < ak < m (k = 1, 2, ..., n) , then�nP

k=1

1m−ak

��nP

k=1

1ak

�≥

�2nm

�2.

Mihaly Bencze

PP29771. If 0 < λ < ak (k = 1, 2, ..., n) , thenP

cyclic

a21a2−λ ≥ 4nλ.

Mihaly Bencze

PP29772. If a, b, c, d > 0 and a 6= b 6= c 6= d then��

1 1 1 1a b c da2 b2 c2 d21a2

1b2

1c2

1d2

��≥ 4

4√

(abcd)5|(a− b) (a− c) (a− d) (b− c) (b− d) (c− d)| .

Mihaly Bencze


1).P a2b

a3+b≤ s

2).P r2ab

r3a+rb≤ 4R+r

2

Mihaly Bencze

PP29774. If ak > 0 (k = 1, 2, ..., n) thenP

cyclic

a21a2a31+a2

≤ 12

nPk=1

ak.

Mihaly Bencze



1).P ab2

(a3+b)(b3+c)≤ 5s2+r2+4Rr

64sRr

2).P rar2b

(r3a+rb)(r3b+rc)≤ (4R+r)2+s2

16s2r

Mihaly Bencze

PP29776. If x, y, z ∈ C such that min {|x| , |y| , |z|} ≥ e then

(|x+ y|+ |y + z|+ |z + x|)|x|+|y|+|z|+|x+y+z| ≥≥ (|x|+ |y|+ |z|+ |x+ y + z|)|x+y|+|y+z|+|z+x|.If max {|x| , |y| , |z|} ≤ e then holds the reverse inequality.

Mihaly Bencze


k=1

k−1k ≤ n1− 1

n .

Mihaly Bencze

PP29778. If x ≥ e then (1 + lnx)(1+lnx)x√xln x ≤

�x√xlnx

�(1+lnx)1+ln x

. If

x ∈ (0, e) then holds the reverse inequality.

Mihaly Bencze

PP29779. If n ≥ 3 then

�nP

k=2

logk (k + 1)

�n

< n

n�

k=2logk(k+1)

.

Mihaly Bencze

PP29780. If n ≥ 2 then�

4n

n+1

�(n+1)(2n)!≥

�(2n)!

(n!)2

�4n(n!)2

.

Mihaly Bencze

PP29781. If n ≥ 6 then�n2

�nn+1·n!3n < (n!)

n2n

6n <�n3

�nn+1·n!2n .

Mihaly Bencze

PP29782. If n ≥ 3 then�2n3

�3n2+3n+1 ≤�3n2 + 3n+ 1

�2n3

.

Mihaly Bencze


PP29783. If e ≤ a1 ≤ a2 ≤ ... ≤ an then�a1+a2+...+ak

k

� (a1+a2+...+an)(ak+1+ak+2+...+an)n(n−k) >

>�a1+a2+...+an

n

� (a1+a2+...+ak)(ak+1+ak+2+...+an)k(n−k) >

>�ak+1+ak+2+...+an

n−k

� (a1+a2+...+ak)(a1+a2+...+an)

kn.

If 0 < a1 ≤ a2 ≤ ... ≤ an ≤ e then holds the reverse inequality.

Mihaly Bencze


2a+b +P b

a(a+2b) ≤s2+r2+4Rr

4sRr .

Mihaly Bencze


a�

rb2a+c +

rc2a+b

�≤ 4R+ r.

Mihaly Bencze


a

�1

2b+a + 12c+a

�≤ 1

2Rr .

Mihaly Bencze

PP29787. In all triangle ABC holds12 + 5

Psin4 A

2 + 5P

cos4 A2 ≥ 8

Psin6 A

2 + 8P

cos6 A2 .

Mihaly Bencze

PP29788. In all acute triangle ABC holds6 + 5

P 1sin2 A cos2 A

+ 2P 1

sinA + 2P 1

cosA ≤ 8P 1

sin3 A+ 8

P 1cos3 A

.

Mihaly Bencze


a

�1

(2a+b) sin2 C2

+ 1(2a+2) sin2 B

2

�≤ s2+r2−8Rr

r2.

Mihaly Bencze

PP29790. In all acute triangle ABC holds27 + 2

PsinA+ 2

PcosA ≥ 8

Psin3A+ 8

Pcos3A.

Mihaly Bencze



6 + 2P 1

sin2 A2cos2 A

2

+ 5P 1

sin4 A2

+ 5P 1

cos4 A2

≤ 8P 1

sin6 A2

+ 8P 1

cos6 A2

.

Mihaly Bencze


10√3�s2−r(4R+r)

R2

�+ 12s

R + 9√3 ≥ 16s(s2−3r2−6Rr)

R3 .

Mihaly Bencze


27 + 4rR ≥ 10(s2−(2R+r)2)

R2 +16((R+r)((4R+r)2−3s2)+3R(s2−(2R+r)2))

R3 .

Mihaly Bencze


11 +10(8R2+r2−s2)

R2 ≥ 4rR +

16((2R−r)((4R+r)2−3s2)+6Rr2)R3 .

Mihaly Bencze


225 + 36rR +

30((4R+r)2−s2)R2 ≥ 16((4R+r)3−3s2(2R+r))

R3 .

Mihaly Bencze

PP29796. If xk ∈ [0, 1] (k = 1, 2, ..., n), then

94

nPk=1

11+2xk

≥ 2n

�nP

k=1

11+xk

�2

+

�

n�

k=1xk

�2

n�

k=1x2k+2

n�

k=1xk+n

.

Mihaly Bencze

PP29797. If xk ∈ [0, 1] (k = 1, 2, ..., n), then 94

nPk=1

1+2xk

2+x2k

≥

�

n+2n�

k=1xk

�2

n�

k=1x2k+2

n�

k=1xk+n

.

Mihaly Bencze


PP29798. If xk ∈ (0, 1) (k = 1, 2, ..., n), then

n+ 274

nPk=1

x2k

(1−xk)2(1+2xk)

≥ 1n

�nP

k=1

1+2x2k

1−x2k

�2

.

Mihaly Bencze

PP29799. Determine all x, y ∈ R for which�5x2 + 2y + 1

� �5y2 + 2x+ 1

�≥ 64x3y3.

Mihaly Bencze

PP29800. If xk ∈ [0, 1] (k = 1, 2, ..., n) then 32

nPk=1

(2+x2k)

32

√1+2xk

≥

�

2n+n�

k=1

x2k

�2

n+n�

k=1xk

.

Mihaly Bencze

PP29801. If xk ∈ [0, 1] (k = 1, 2, ..., n) then

32

nPk=1

x2k

�

(1+2xk)(2+x2k)

≥

�

n�

k=1

xk

�2

n+n�

k=1xk

.

Mihaly Bencze


3√2 ≥

q�1 + 2 sin2 x

� �2 + sin4 x

�+p(1 + 2 cos2 x) (2 + cos4 x).

Mihaly Bencze

PP29803. Prove that 3n(n+2)2(n+1) ≥

nPk=1

√(k2+k+2)(2k4+4k3+2k2+1)

(k(k+1))32

.

Mihaly Bencze


k=1

1√k(k+2)(2k2+1)

≥ 2n3(n+1) .

Mihaly Bencze


(1 + 2 sinA)�2 + sin2A

�≤ 3(s+3R)

2R .

Mihaly Bencze



Pp(1 + 2 cosA) (2 + cos2A) ≤ 3(4R+r)

2R .

Mihaly Bencze

PP29807. If xk ∈ [0, 1] (k = 1, 2, ..., n) then

nQk=1

(1 + 2xk)�2 + x2k

�≤

�32

�1 + 1

n

nPk=1

xk

��2n

.

Mihaly Bencze


1).5(s2−4Rr−r2)

4s2r3≥P 1

r3a

2).5(3s2−r2−4Rr)

16s2r3≥P 1

h3a

Mihaly Bencze

PP29809. If 0 ≤ a ≤ b ≤ 1 thenbRa

(x2+2)dx(x+1)2

≤ 98 ln

1+2b1+2a .

Mihaly Bencze


k=1

2k2+1k(k+1)3

≤ 9n8(n+2) .

Mihaly Bencze


1). s2−3r2

4Rr ≥ 1 + 8�s2−r2−4Rrs2+r2+4Rr

�2

2). 1 + s2−12Rrr2

≥ 4�s2−2r2−8Rr

r(4R+r)

�2

3). 1− 12Rr + (4R+r)3

s2r≥ 4

��4R+r

s

�2 − 2�2

4). 4 +(2R−r)((4R+r)2−3s2)

2Rr2≥ 16

�8R2+r2−s2

s2+r2−8Rr

�

5). (4R+r)3−3s2r2s2R

≥ 2 + 16�(4R+r)2−s2

s2+(4R+r)2

�2

Mihaly Bencze



1).P (a+b)2(a+c)2

b+c ≥ 16sRr(5s2+r2+4Rr)s2+r2+2Rr

2).P (ra+rb)

2(ra+rc)2

rb+rc≥ 2r

�6s2 + (4R+ r)2

�

Mihaly Bencze

PP29813. If a, b, c > 0 thenP(a+ b)2 (a+ c)2 ≥ 12abc (a+ b+ c) + 1

9

�P ��a2 + ab+ ac− bc��2 .

Mihaly Bencze

PP29814. If a, b, c > 0 then

12abc (a+ b+ c) +P �

a2 + ab+ ac− bc�2 ≤P

�a+ b+c

2

�4.

Mihaly Bencze

PP29815. In all triangle ABC holdsP (a+b)2(a+c)2

bc = 16s2 +P (b2+bc+ba−ca)

2

ca .

Mihaly Bencze

PP29816. In all scalene triangle ABC holdsP�

cos A2

sin B−C2

�2λ

≥ 2λ

3λ−1 for all

λ ≥ 1.

Mihaly Bencze

PP29817. In all scalene triangles ABC holdsP a2+c2

(a−b)2≥ 3.

Mihaly Bencze

PP29818. If a, b, c > 0, a 6= b 6= c thenP a2+c2+(a+b)2

(a−b)2≥ 5.

Mihaly Bencze

PP29819. If a, b, c ∈ R, a 6= b 6= c thennP

k=1

��a

a−b + k�2

+�

bb−c + k

�2+�

cc−a + k

�2�

≥ n(2n2+6n+7)3 .

Mihaly Bencze


PP29820. If a, b, c > 0 then�√a2 + b2 + c2 + b

√3�(a+ c) ≤ 4

√3

3

�a2 + b2 + c2

�, and his permutations.

Mihaly Bencze


�1√ha

+ 1√hc

��1√hb

+ 1√hd

�≤ 1

r

2).�

1√ra

+ 1√rc

��1√rb

+ 1√rd

�≤ 2

r

Mihaly Bencze

PP29822. If a, b, c > 0 then�a2 + b2 + 1

� �b2 + c2 + 1

� �c2 + a2 + 1

�≥ abc

�2 + 3

√abc

�3.

Mihaly Bencze

PP29823. If ak > 0 (k = 1, 2, ..., n) , then

Qcyclic

�a21 + a22 + 1

�≥

nQk=1

ak

2 + n

snQ

k=1

ak

!n

.

Mihaly Bencze


1). 1√ra

+q

rrarc

+ 1√rb

≤ 2√r

2). 1√ha

+q

rhahc

+ 1√hb

≤ 2√r

Mihaly Bencze

PP29825. If a, b, c > 0 thenP√

a+ 2b+ 3c ≥2√2�√

a(c+a)+√

b(a+b)+√

c(b+c)�

√a+b+c

.

Mihaly Bencze

PP29826. If λ, ak > 0 (k = 1, 2, ..., n) thenP

cyclic

a31+λa2+λa3

≥ 3nλ+1

3

q�λ2

�2.

Mihaly Bencze


PP29827. If xk > 0 (k = 1, 2, ..., n) , thenP x1

x2+P

x1ex2−1 ≥ 2

nPk=1

xk.

Mihaly Bencze

PP29828. If xk > 0 (k = 1, 2, ..., n), thenP x2

1x2

+P

x21ex2−1 ≥ 2

n

�nP

k=1

xk

�2

.

Mihaly Bencze

PP29829. If x ∈ R then etg2x + ectg

2x ≥ 3.

Mihaly Bencze

PP29830. Let f : [0, 1] → (0,+∞) be a continuous function, thennP

k=2

1k−1

1R0

kpf (x) dx ≤ n−1

n

1R0

f (x) dx+ lnn.

Mihaly Bencze

PP29831. If a, b, c, d > 0 and k, p ∈ N ∗ then�a2 + c2

� �b2 + d2

�≥�

ak+p√akbp + c

k+p√ckdp

�·�b

k+p√apbk + d

k+p√cpdk

�≥ (ac+ bd)2 .

Mihaly Bencze

PP29832. If xk > 0 (k = 1, 2, ..., n) and a, b > 0 then

min

�nP

k=1

x3k

ax2k+b

;nP

k=1

x3k

bx2k+a

�≥ 1

a+b

nPk=1

xk.

Mihaly Bencze

PP29833. If ai > 0 (i = 1, 2, ..., n) and p, k ∈ N thennP

i=1apn+ki ≥

�nP

i=1aki

��nQ

i=1ai

�p

.

Mihaly Bencze


1). 3√3

2 ≤Pq

rar+ra

≤ 36√s2r√

r+3√s2r

2). 3√3

2 ≤Pq

ha

r+ha≤ 3 6√R√

3√R+3√2s2r2

Mihaly Bencze


PP29835. If a, b, c > 0 thenP 1

a2(b2(b2+c2)+c2(a+b)2)≤ 1

2a2b2c2.

Mihaly Bencze


1).P ra

√ra

ra√ra+2r

√rb

≤ 4R+r5r

2).P ha

√ha

ha

√ha+2r

√hb

≤ 3(s2+r2+4Rr)10R

Mihaly Bencze

PP29837. If a, b, c, d, e > 0 thenP a2

(b+c+d+e)(a−b)(a−c)(a−d)(a−e) ≤1

(a+b+c+d+e)2, for all a 6= b 6= c 6= d 6= e.

Mihaly Bencze

PP29838. If ak > 0 (k = 1, 2, ..., n) , thenQ �2 (1 + a1a2) + a21 − a22

�≤ 2n (1 +

Pa1a2)

n .

Mihaly Bencze


1). 3√3R√

s+3R≤P 1√

1+sinA≤ 3

6√2R2√

3√2R2+ 3√sr

2). 3√6R√

8R−r≤P 1

�

1+sin2 A2

≤ 36√16R2√

3√16R2+

3√r2

3). 3√6R√

10R+r≤P 1

�

1+cos2 A2

≤ 36√16R2√

3√16R2+

3√s2

Mihaly Bencze

PP29840. Prove thatPa8 +

Pa6b2 + 4

Pa4b4 + 2abc

Pa3b2 ≥ 4

Pa6c2 + 4a2b2c2

Pab.

Mihaly Bencze

PP29841. If 0 < ak ≤ 1 (k = 1, 2, ..., n) thenPcyclic

1√2(1+a1a2)+a21−a22

≥ n√n

�

�

�

�2

�

n+�

cyclic

a1a2

�

.

Mihaly Bencze


PP29842. Determine all a, b, c > 0 for which 2abc+ 1 ≥ ab+ bc+ ca.

Mihaly Bencze

PP29843. If a, b, c > 0 then 3 + 5P

a2 ≥ 2P

a+ 4P

ab.

Mihaly Bencze


a4 + 2P

a2b2 ≥ 2abcP

a+ 2P

ab3.

Mihaly Bencze

PP29845. If a, b, c > 0 thena2 + b2 + c2 ≥ ab+ bc+ ca+ (a− b) (b− c) + (b− c) (c− a) + (c− a) (a− b) .

Mihaly Bencze


a2 ≥ 12

Pab+ 1

2

P ��a2 − bc�� .

Mihaly Bencze

PP29847. If k ≥ 1 and a, b, c > 0 then

3k−1P �

a2 − bc�2k ≥

��Pa2�2 − (

Pab)2

�k.

Mihaly Bencze


atg2A2 ≥ 2m where m is a positiveroot of the equation x3 + sx2 − 2sRr = 0.

Mihaly Bencze

PP29849. If a, b, c, d, e > 0 then√a2 + b2 − ab+

pb2 + c2 −

√2bc+

+pc2 + d2 −

√3ad+

qd2 + e2 − 1

2

�√6 +

√2�de ≥

pa2 + e2 +

√3ae.

Mihaly Bencze

PP29850. In triangle ABC holds 0 < A ≤ B ≤ C ≤ π2 . Prove that

sinCC ≤ s

πR ≤ sinAA .

Mihaly Bencze


PP29851. If ak ∈�0, 5π12

�(k = 1, 2, ..., n) , then 1 ≤

n�

k=1tgak

n�

k=1

ak

≤ 12(2+√3)

5π .

Mihaly Bencze

PP29852. If 0 < m ≤ xk ≤ M (k = 1, 2, ..., n) and 0 < a < b then

a+m2

b+m2 ≤na+

n�

k=1a2k

nb+n�

k=1b2k

≤ a+M2

b+M2 .

Mihaly Bencze

PP29853. Complex numbers a, b, c are pairwise distinct and satisfy|a| = |b| = |c| = 1 and |a+ b+ c| ≤ 1. Prove that1). |a− b|+ |b− c|+ |c− a| ≥ |a+ b|+ |b+ c|+ |c+ a|2).

��a2 + bc��+

��b2 + ca��+

��c2 + ab�� ≥ |a+ b|+ |b+ c|+ |c+ a|

Mihaly Bencze

PP29854. Let ABC be a triangle in which 0 < m ≤ t1

sinu ≤ M where

t ∈ {a, b, c} , u ∈ {sinA, sinB, sinC} . Prove that m ≤ (4sRr)Rr ≤ M.

Mihaly Bencze

PP29855. If 0 < m ≤ ak ≤ M ≤ 1 (k = 1, 2, ..., n) , then

2m1+m2 ≤

n�

k=1ak

n+n�

k=1a2k

≤ 2M1+M2 .

Mihaly Bencze

PP29856. If ak ∈�0, π2

�(k = 1, 2, ..., n) then 2

π ≤n�

k=1sin ak

n�

k=1ak

≤ 1.

Mihaly Bencze

PP29857. If 0 < a1 ≤ a2 ≤ ... ≤ an ≤ π2 then sin an

an≤

n�

k=1sin ak

n�

k=1ak

≤ sin a1a1

.

Mihaly Bencze


PP29858. If x > 0 then nn+1 ≤

nPk=1

�

k�

i=1ixi−1

��

k+1�

i=1ixi−1

�

�

k−1�

i=0(n−i)xi

��

k�

i=0(n−i)xi

� ≤ n(n+1)(n+2)3 .

Mihaly Bencze

PP29859. Prove that nn+1 ≤

nPk=1

�

k�

i=1ixi−1

��

k+1�

i=1ixi−1

�

�

k�

i=1i2xi−1

��

k+1�

i=1i2xi−1

� ≤ n for all x ≥ 0.

Mihaly Bencze

PP29860. Let ABC be a triangle in which 0 < A ≤ B ≤ C ≤ π2 . Prove that

tgA ≤ sR+r ≤ tgC.

Mihaly Bencze

PP29861. In all acute triangle ABC holds 2 ≤P sinA ≤ π.

Mihaly Bencze

PP29862. Let A1A2...An be a tangential polygon with inradius R andsemiperimeter s. Prove that1). s

R ≥ Pcyclic

ctgA1+A2+...+Ak

2k ≥ nctg (n−2)πn

2). sR ≥ k

(n−1k−1)

P1≤i1<...<ik≤n

ctgAi1

+Ai2+...+Aik

2k ≥ nctg (n−2)πn where

k ∈ {1, 2, ..., n}

Mihaly Bencze

PP29863. Let ABCD be a circumscribed quadrilateral around a circle ofradius R, and denote s his semiperimeter. Prove1). s

R ≥P ctg 2π−A6 ≥ 4

2). sR ≥ 2

3

�ctgA+B

4 + ctgA+C4 + ctgA+D

4 + ctgB+C4 + ctgB+D

4 + ctgC+D4

�≥ 4

Mihaly Bencze

PP29864. Let ABCD be a circumscribed quadrilateral around a circle ofradius R, and denote s his perimeter. Prove that 4s

R + 3P

ctg 2π−A6 ≥

≥P

ctgA2 +4

�ctgA+B

4 + ctgA+C4 + ctgA+D

4 + ctgB+C4 + ctgB+D

4 + ctgC+D4

�.

Mihaly Bencze



2R−r

�x5 +

Psin10 A

2

�+ 1

4R+r

�x5 +

Pcos10 A

2

�≥ 3x(x+1)

2R for all x > 0.

Mihaly Bencze

PP29866. Let ABCD be a cyclic quadrilateral. Prove thatPtg5A2 ≥P tgA

2 .

Mihaly Bencze

PP29867. In all tetrahedron ABCD holds1). r

P 1h5a≥ 1

hahbhchd

2). r2

P 1r5a

≥ 1rarbrcrd

Mihaly Bencze

PP29868. In all triangle ABC holds1). x5 +

P 1h5a≥ x(xr+1)R

2s2r3

2). x5 +P 1

r5a≥ x(xr+1)

s2r2for all x > 0

Mihaly Bencze

PP29869. If xk > 0 (k = 1, 2, ..., n) , thennP

k=1

x4k

3x3k+7

≥

�

n�

k=1xk

�4

3

�

n�

k=1xk

�3

+7n3

.

Mihaly Bencze

PP29870. If xk > 0 (k = 1, 2, ..., n) , then 10nP

k=1

x4k

3x3k+7

≥nP

k=1

xk.

Mihaly Bencze

PP29871. If a, b, c > 0 and λ ≥ 0 thenaλ

√a2 + 2bc+ bλ

√b2 + 2ca+ cλ

√c2 + 2ab ≤ (a+ b+ c)

√a2λ + b2λ + c2λ.

Mihaly Bencze

PP29872. If a, b, c > 0 and x, y ≥ 0 thenPcyclic

bxcy 3pa3 + bc (2a+ 3b+ 3c) ≤ (

Pa) 3p(P

a3x) (P

a3y).

Mihaly Bencze


PP29873. Prove that1). sin π

4n+2 ≥ 12n 2). tg π

4n ≤ 1n

3). tg π4n − sin π

4n < 12n for all n ∈ N∗

Ionel Tudor

PP29874. Prove that sin 4π15 − sin π

15 =√15−

√3

4 . Solve in N∗ the equation

sin 4πn − sin π

n =√n−

√3

4 .

Ionel Tudor

PP29875. Solve in N∗ the equation tg4 πn − 4tg3 πn − 14tg2 πn − 4tg πn + 1 = 0

Ionel Tudor

PP29876. 1). Prove that 2m ≥ m for all m ∈ N.2). Solve in N the equation 2n − 3n = 2015.

Ionel Tudor

PP29877. If d > 1 dividet 2016, then the equation 22016 + 32016 = xd havenot solution in N.

Ionel Tudor

PP29878. Let be N = 52017. Determine the last 5 digits of N. Determinethe number of digits of N.

Ionel Tudor

PP29879. Let be N = 20177...7 which have the last 2017 digits equal with7. Determine bn in corp Z2017.

Ionel Tudor

PP29880. Determine all n,m ∈ N for which 4n+ 1 is prime and√m+

pm+

√m+ n is natural.

Ionel Tudor

PP29881. Solve in Z the equation�x3 − 1

�y2 +

�y3 − 1

�x2 = x

�y4 − y2 + 1

�+ y

�x4 − x2 + 1

�.

Mihaly Bencze



4�x51 + 1

�= 5 5

√4x2

4�x52 + 1

�= 5 5

√4x3

−−−−−−−−−4�x5n + 1

�= 5 5

√4x1

.

Mihaly Bencze and Ionel Tudor

PP29883. Let ABCD be a tetrahedron, and denote RA, RB, RC , RD thecircumradii of triangles BCD, CDA, ABD, ABC. Prove thatP

R2nA ≥ 1

33n·4n−1 (AB +AC +AD +BC +BD + CD)2n for all n ∈ N∗.


PP29884. Let ABCD be a tetrahedron and denote RA, RB, RC , RD

respective SA, SB, SC , SD the circumradii respective the semiperimeters oftriangles BCD, CDA, DAB, ABC. Prove that24316

P (RARBRC)2

SBSC≥ 3

√AB2 ·AC2 ·AD2 ·BC2 ·BD2 · CD2.

Mihaly Bencze

PP29885. Let ABCD be a tetrahedron and denote RA, RB, RC , RD

respective SA, SB, SC , SD the circumradii respective the semiperimeters of

triangles BCD, CDA, DAB, ABC. Prove thatP RA

SB≥ 1

6√3(P

SA)�P 1

SA

�.

Mihaly Bencze

PP29886. Determine a polynomial P ∈ Z [x] of minimal degree for whichx = 7

√8 + 7

√16 is a roote.

Ionel Tudor

PP29887. Solve in R the equation 4x5 − 5 5√4x+ 4 = 0.

Ionel Tudor

PP29888. Let ABCD be a tetrahedron, and denote RA, RB, RC , RD thecircumradii of triangles BCD, CDA, ABD, ABC. Prove that729 (RARBRCRD)

3 ≥ (AB ·AC ·AD ·BC ·BD · CD)2 .

Mihaly Bencze


PP29889. Determine all n ∈ N∗ for which

1).nQ

k=1

(1 + Fk)1+Fk ≤ eFn+2−1

√eFnFn+1

2).nQ

k=1

(1 + Lk)1+Lk ≤ eLn+2−3

√eLnLn+1−2

Mihaly Bencze

PP29890. Determine all x, y > 1 for which(1 + lnx)1+ln y + (1 + ln y)1+lnx ≤ x

pyln y + y

√xlnx.

Mihaly Bencze

PP29891. If x, y > 1 then

arctg�(1+lnx)1+ln x

√xln x

�+ arctg

�(1+ln y)1+ln y√

yln y

�≤ arctg x+y

xy−1 .

Mihaly Bencze

PP29892. Solve in Z the equation 2x

3y+5z + 2y

3z+5x + 2z

3x+5y = 34 .

Mihaly Bencze

PP29893. If in triangle ABC we have A ≤ B ≤ C thenP (ln eπ

A )ln eπ

A

( πA)

12 ln π

A

≤�3(π−B)

2√AC

�2.

Mihaly Bencze

PP29894. Let A1A2...An be a convex polygon such that

α ≤ Ak ≤ β (k = 1, 2, ..., n) . Prove thatnP

k=1

�

ln eπAk

�ln eπAk

�

πAk

� 12 ln π

Ak

≤ n2(α+β)2

4(n−2)αβ .

Mihaly Bencze

PP29895. If xk > 0 (k = 1, 2, ..., n) , thennQ

k=1

2x1+...+2xk−1+xk+2xk+1+...+2xn

2x1+...+2xk−1+3xk+2xk+1+...+2xn≥ 1

3 .

Mihaly Bencze


PP29896. If x, y, z > 0 then

Px2

�1 + x+ x2

�≥Pxy +

Pxy (x+ y − z) +

Pxy

�x2 + y2 − z2

�.

Mihaly Bencze


1). 5 (P√

a)�P√

ab�≥ (P√

a)3+ 18

√abc

2). 5�P√

ma

� �P√mamb

�≥

�P√ma

�3+ 18

√mambmc

3). 5

�Pqcos A

2

��Pqcos A

2 cos B2

�≥

�Pqcos A

2

�3

+ 9p

sR

Mihaly Bencze

PP29898. If a, b, c > 0 then(5P

a)�P

a2 + 3P

ab�≥ 4 (

Pa)3 + 9

Q(a+ b) .

Mihaly Bencze

PP29899. If a, b, c > 0 thenP��

ab

�4+�bc

�3+ c

a

�≥ 3 +

P a(a2c2+b4)b2c3

.

Mihaly Bencze

PP29900. Determinr all n, k ∈ N for whichP (ab)n

ck≥P ab

c for alla, b, c > 0.

Mihaly Bencze

PP29901. If x, y, z, a, b, c > 0 such that ax+ by = z, by + cz = x,

cz + ax = y then 16abc (P

ab)�P 1

a+b+2

�≤ 1.

Mihaly Bencze

PP29902. If xk > 0 (k = 1, 2, ..., n) and r ∈ N thennP

k=1

xn+rk ≥

�nQ

k=1

xk

��nP

k=1

xrk

�.

Mihaly Bencze


PP29903. If ak ∈�0, π2

�(k = 1, 2, ..., n) , then

nPk=1

�kP

i=1sin ai

��kP

i=1cos ai

�≤ n(n+1)(2n+1)

12 sin

12n�

k=1

�

n(n+1)2

− k(k−1)2

�

ak

n(n+1)(2n+1)

.

Mihaly Bencze

PP29904. Let ABC be a triangle and denote the angle α,β, γ the angle of

triangle formed by ma,mb,mc. Prove thatP√

mamb cos γ ≤√22

Pma.

Mihaly Bencze

PP29905. In all acute triangle ABC holdsPapb (cosB cosC) ≤ 1

2

�s2 + r (4R+ r)

�.

Mihaly Bencze


cos A2 ≥P

qr2R + 2 sin2 A

2 .

Mihaly Bencze

PP29907. If ak ∈ R, ak + k > 0 (k = 1, 2, ..., n) andnP

k=1

kak ≤ n(n+1)(3n2−5n−4)24 , n ∈ N,n ≥ 3, then

nPk=1

kak+k ≥ 2.

Mihaly Bencze

PP29908. If ak > 0 (k = 1, 2, ..., n) and m, r ∈ N thenP

cyclic

am+r+11am2 ar3

≥nP

k=1

ak.

Mihaly Bencze

PP29909. If xk > 0 (k = 1, 2, ..., n) , thennP

k=1

xk

1+√xk

≤√n

n�

k=1xk

√n+

�

n�

k=1xk

.

Mihaly Bencze

PP29910. If a, b, c > 0 then 2√2P a

c ≥√2Pp

ab +

Pq�cb

�2+�ba

�2.

Mihaly Bencze


PP29911. If ak > 0 (k = 1, 2, ..., n) , thenP a1(4a31+3a1a22+2a32)

a2(a1+2a2)(2a21+a22)≥ n.

Mihaly Bencze

PP29912. In all acute triangle ABC holdsP sinA cos3 B

(sin2 A−3 sin2 A sin2 B+2 sin2 B) sinB+

+2P cos2 A cosB

sinA(sinA cosB+sinC) ≥ 1.

Mihaly Bencze


k=1

√k4+4k3+10k2+12k+5

(3k2+6k+5)(k2+2k+2)≤ n(3n+5)

6(n+1)(n+2) .

Mihaly Bencze

PP29914. If n ≥ 6 then��

n3

�n+ n! +

�n2

�n� ��n3

�n+

�n2

�n� ≥≥ 12

�n2

6

�n ��n3

�2n+ (n!)2 +

�n2

�2n�.

Mihaly Bencze

PP29915. Prove that

1).nP

k=1

Fk+2(Fk+Fk+2)2

Fk≥ 6FnFn+1 + 2F 2

n+1 + F 2n+2 − 3

2).nP

k=1

Lk+2(Lk+Lk+2)2

Lk≥ 6LnLn+1 + 2L2

n+1 + L2n+2 − 6

Mihaly Bencze

PP29916. Prove that

1).nP

k=1

(√Fk+

√Fk+2)

2(√Fk+

√Fk+1+

√Fk+2)

2

√FkFk+2

≥ 12 (2Fn+1 + 4Fn+2 − 9)

2).nP

k=1

(√Lk+

√Lk+2)

2(√Lk+

√Lk+1+

√Lk+2)

2

√LkLk+2

≥ 12 (2Ln+1 + 4Ln+2 − 9)

Mihaly Bencze

PP29917. Prove that 2nQ

k=1

�√k+

√k+1+

√k+2√

k+2−√k

�4≥ 81nn! ((n+ 1)!)2 (n+ 2)!.

Mihaly Bencze



k=1

qk+1

k3(3k2+6k+5)≥ n√

3(n+1).

Mihaly Bencze

PP29919. If 0 < a1 ≤ a2 ≤ ... ≤ an then√32

nPk=1

ak ≥P

cyclic

�

a1a3(a21+a22+a23)a1+a3

.

Mihaly Bencze

PP29920. If 0 < a1 ≤ a2 ≤ ... ≤ an then 1√3

nPk=1

ak ≥ Pcyclic

�

a1a3(a21+a22+a23)a1+a2+a3

.

Mihaly Bencze

PP29921. If a, b, c > 0 then

�q3abc�

ab +q

13

Pa��q

3abc�

ab +6√abc+

q13

Pa�≥

≥ 2√3 4

qabc

�

a�

ab

q3abc�

ab +3√abc+ 1

3

Pa.

Mihaly Bencze

PP29922. If 0 < a1 ≤ a2 ≤ ... ≤ an then

136

Pcyclic

(√a1+

√a3)

2(√a1+

√a2+

√a3)

2

√a1a3

≥nP

k=1

ak.

Mihaly Bencze

PP29923. If a, b, c > 0 then 2 (P

a)2 (P

ab)2 + 9Q

(a+ b) ≥≥ 8 (

Pa) (P

ab) + 3 (P

ab)3 + 3abc (P

a)3 .

Mihaly Bencze


�q2aba+b +

4√ab+

qa+b2

��q2aba+b +

qa+b2

�≥

≥ 2√3 4√abq

2aba+b +

√ab+ a+b

2 .

Mihaly Bencze


PP29925. If ak ≥ 1 (k = 1, 2, ..., n) , thennP

k=1

11+a2

k

≥ Pcyclic

11+a1a2

+

�

�

√(a1a2−1)|a1−a2|

�2

2�

(a21+1)(a22+1)(a1a2+1).

Mihaly Bencze

PP29926. If x, y, z, a, b, c, d > 0 thenp(a2 + b2)x2 − 2axy + 2bxz + y2 + z2+

+p(d2 + 1) y2 + (c2 + 1) z2 − 2 (c+ d) yz ≥

≥p

(a2 + b2)x2 + d2y2 + c2z2 + 2bdxy − 2acxz.

Mihaly Bencze


P ra√r2+r2a

≤ 3�

1+( rs )

43

2).P ha√

r2+h2a

≤ 3�

1+�

Rr

2s2

� 43

Mihaly Bencze

PP29928. If ak ≥√2 (k = 1, 2, ..., n) , then

nPk=1

1√1+a2

k

≥ n�

�

�

�1+ n

�

n�

k=1a2k

. If

0 < ak <√2 (k = 1, 2, ..., n) then holds the reverse inequality.

Mihaly Bencze


P 1√1+sin2 A

≤ 3�

1+�

sr

2R2

� 32

2).P 1√

1+cos2 A≤ 3

�

1+�

s2−(2R+r)2

4R2

�

23

3).P 1

�

1+sin4 A2

≤ 3�

1+( r4R)

43

4).P 1

�

1+cos4 A2

≤ 3�

1+( s4R)

43

Mihaly Bencze


PP29930. If x, y, z ∈�0, π2

�then

Pcosx ≤P

qcosx cos ycos(x−y) .

Mihaly Bencze

PP29931. If a, b, c > 0 thenP (a+b)(a−b)2

(a2+b2)�

a+b+√

2(a2+b2)� ≤ 1 + 8(a+b)(b+c)(c+a)√

2(a2+b2)(b2+c2)(c2+a2).

Mihaly Bencze

PP29932. If ak ∈ [0, 1] (k = 1, 2, ..., n) , thennP

k=1

1√1+a2

k

≤ Pcyclic

1√1+a1a2

.

Mihaly Bencze

PP29933. If n,m, k ∈ N ∗ then 4e2P �

1 + 1n

�n �e−

�1 + 1

n

�n� ≤

≤ 1 + 16e3

�1 + 1

n

�2 �1 + 1

m

�2 �1 + 1

k

�2.

Mihaly Bencze

PP29934. If x ∈�0, π2

�then 1√

1+sin2 x+ 1√

1+cos2 x≤ 2√

1+sinx cosx.

Mihaly Bencze

PP29935. If a, b, c > 0 thenP √

ab(√a−

√b)

2

a+b ≤ 18 + 16abc

(a+b)(b+c)(c+a) .

Mihaly Bencze

PP29936. If x, y, z > 0 (x 6= y 6= z) then 8P (x−y)((x+y)(lnx−ln y)−2(x−y))

(x+y)2(lnx−ln y)2≤

≤ 1+128(x−y)(y−z)(z−x)(x+y)(y+z)(z+x)(lnx−ln y)(ln y−ln z)(ln z−lnx) .

Mihaly Bencze

PP29937. Denote Fk, Lk, Pk the kth Fibonacci, Lucas and Pell numbers.Prove that

nPk=1

Fk+1Fk+2

F 2k

+nP

k=1

Lk+1Lk+2

L2k

+ 2P Pk+1Pk+2

P 2k

≤ n4 + 4

nPk=1

FkLkPk

Fk+2Lk+2Pk+2.

Mihaly Bencze


PP29938. In all triangle ABC holds1). 4r

P ra−rr2a

≤ 1 +�4rs

�2

2). 4rP ha−r

h2a

≤ 1 + 8Rr2

s2

Mihaly Bencze


Psin2A ≤ 1

4 +�rR

�2

2).P

cos2A ≤ 14 +

�sR

�2

Mihaly Bencze

PP29940. If a, b, c > 0 thenP (a+2b)(b+2c)

(a+3b+2c)2≤ 1

4 + 4(a+2b)(b+2c)(c+2a)(a+3b+2c)(b+3c+2a)(c+3a+2b) .

Mihaly Bencze

PP29941. In all triangle ABC holdsPsin2 2A ≤ 1 + 8

�sin2A sin2B sin2C + cos2A cos2B cos2C

�.

Mihaly Bencze


1).P (ab)2

(a+b)3≥ 1

16s3

�(s2+r2+4Rr)

2+8s2Rr

s2+r2+2Rr

�2

2).P (rarb)

2

(ra+rb)3 ≥ 1

2(4R+r)

�s2+r(4R+r)

4R

�2

Mihaly Bencze


P ab(a+b)2

≤ s2+r2+34Rr4(s2+r2+2Rr)

2).P rarb

(ra+rb)2 ≤ R+4r

4R

3).P sin2 A

2sin2 B

2

(sin2 A2+sin2 B

2 )2 ≤ 1

4 + 8Rr2

(2R−r)(s2+r2−8Rr)−2Rr2

4).P cos2 A

2cos2 B

2

(cos2 A2+cos2 B

2 )2 ≤ 1

4 + 8Rs2

(4R+r)3+s2(2R+r)

Mihaly Bencze

PP29944. In all triangle ABC holdsP (−a+b+c)(a−b+c)

c2≤ R+4r

R .

Mihaly Bencze


PP29945. If a, b, c > 0 thenPq

aba+b ≤

r2 (P

a)�14 + 4abc

(a+b)(b+c)(c+a)

�.

Mihaly Bencze

PP29946. If a, b, c > 0 such that a3 + b3 + c3 ≥ nn√

(a+b)(b+c)(c+a)then

a+ b+ c ≥ 3√n+ 1 for all n ∈ N∗.

Mihaly Bencze

PP29947. If a, b, c ∈ [0,λ] where λ > 0 such that a+ b+ c = 2λ thena2 + b2 + c2 ≤ 2λ2.

Mihaly Bencze

PP29948. If a, b, c,λ > 0 such that (a+ b) (b+ c) (c+ a) ≥ λ6

a3+b3+c3then

a+ b+ c ≥ 2λ.

Mihaly Bencze

PP29949. If a, b, c > 0 such that (a+ b) (b+ c) (c+ a) ≥ nn√a3+b3+c3

then

a+ b+ c ≥ 3√n+ 1 for all n ∈ N∗.

Mihaly Bencze

PP29950. If ak > 0 (k = 1, 2, ..., n) such that λa1 ≥ a2 + a3 + ...+ an where

λ ∈ [1, 3) then

�nP

k=1

ak

��nP

k=1

1ak

�≥ (3− λ)n2 + 2 (λ− 3)n+ 4.

Mihaly Bencze

PP29951. If a, b, c > 0 and x, y, z, t > 0 such that x2P

a2 + 3y2 = tabc thenP 1a3+b3+z3

≤ t6xyz .

Mihaly Bencze


Q �a4 + 3

�≥ 4

�1 + 2

�s2 − r2 − 4Rr

��2.

Mihaly Bencze



a2 + 3 sin2 A2

�+Q �

a2 + 3 cos2 A2

�≥

≥ r2

4R2

�1 +

P asin A

2

�2

+ s2

4R2

�1 +

P acos A

2

�2

.

Mihaly Bencze

PP29954. In all triangle ABC holdsQ �a2 + 6b2 + 6c2

�≥ 256 (

Pambmc +mambmc)

2 .

Mihaly Bencze

PP29955. If ak, bk > 0 (k = 1, 2, ..., n) then

nQk=1

�a2k + nb2k

�≥ n

nQ

k=1

bk +P

cyclic

a1b2b3...bn

!2

.

Mihaly Bencze

PP29956. If ak ∈ {z ∈ C| |z| = 1} (k = 1, 2, ..., n) , then�

2n�

k=1

ak

�2

n+�

cyclic

a1a2+P

cyclic

�a1−a21−a1a2

�2≥ n.

Mihaly Bencze


a2 + 3r2a�≥ 4 (

Parbrc + rarbrc)

2 .

Mihaly Bencze

PP29958. If 0 ≤ ak ≤�nk

�(k = 0, 1, ..., n) , then

nQk=1

ak +nQ

k=1

��nk

�− ak

�≤ (n!)n+1

�

n�

k=1k!

�2 .

Mihaly Bencze

PP29959. If x, y ∈ R then�

eix+eiy

1+ei(x+y)

�2+�

eix−eiy

1−ei(x+y)

�2≥ 1 + sin (x− y) .

Mihaly Bencze

PP29960. If xk ∈ R (k = 1, 2, ..., n) thennQ

k=1

sin2 xk +nQ

k=1

cos2 xk ≤ 1.

Mihaly Bencze


PP29961. If 0 < ak ≤ bk ≤ ck (k = 1, 2, ..., n) thennQ

k=1

(bk − ak) (ck − ak) (ck − bk) ≥

≥

nY

k=1

bk −nY

k=1

ak

! nY

k=1

ck −nY

k=1

ak

! nY

k=1

ck −nY

k=1

bk

!.

Mihaly Bencze

PP29962. If x ∈�0, π2

�then

sin2 x(1+2 sinx)1+cosx + cos2 x(1+2 cosx)

1+sinx + 4 (1− sinx) cos2 x+ 4 (1− cosx) sin2 x ≤ 12.

Mihaly Bencze


sin4 x(1+2 sin2 x)1+cos2 x

+cos4 x(1+2 cos2 x)

1+sin2 x≤ 8 + 12 sin2 x cos2 x.

Mihaly Bencze

PP29964. If ln�√

2− 1�≤ xk ≤ ln

�√2 + 1

�(k = 1, 2, ..., n) , thenP

cyclic

��sh2x1 − sh2x2�� ≤ 2

Pcyclic

1(chx1chx2)

2 .

Mihaly Bencze

PP29965. If x ∈�0, π2

�then

(sinx+ cosx)6 (1− sinx cosx)4 ≥ (sinx cosx)5 .

Mihaly Bencze

PP29966. In all triangle ABC holdsP ��sin A−B

2

�� ≤P 1(1+cosA)(1+cosB) cos C

2

.

Mihaly Bencze


P |ra − rb| (r + ra) (r + rb) ≤ 2s2�s2 − 2r (4R+ r)

�

2).P

|ha − hb| (r + ha) (r + hb) ≤4s2r2(s2−r2−4Rr)

R2

Mihaly Bencze


PP29968. If ak > e (k = 1, 2, ..., n) , then

1).nQ

k=1

(ln ak)a2k ≥

ln

n�

k=1a2k

n�

k=1ak

n�

k=1a2k

2).Q

cyclic

(ln a1)a1a2 ≥

ln

�

cyclic

a1a2

n�

k=1ak

�

cyclic

a1a2

Mihaly Bencze

PP29969. In all triangle ABC holdsP ��sin A−B

2

�� ≤P 1(1+sinA)(1+sinB) sin C

2

.

Mihaly Bencze


1).Q �

ln esinA

� 1sin2 A ≥

�ln

e� 1

sin2 A� 1

sinA

�� 1sin2 A

2).Q �

ln esinA

� 1sinB sinC ≥

�ln

e� 1

sinA sinB� 1

sinA

�� 1sinA sinB

Mihaly Bencze

PP29971. In all triangle ABC holds�ln e2

�

sinA3 sinA sinB sinC

��

sinA≤Q

�ln e2

sinB sinC

�sinA.

Mihaly Bencze

PP29972. If ak > e (k = 1, 2, ..., n) and λk > 0 (k = 1, 2, ..., n) , then

nQk=1

(ln ak)λkak ≥

ln

n�

k=1λkak

n�

k=1

λk

n�

k=1λkak

. If ak ∈ (1, e) (k = 1, 2, ..., n) , then


Mihaly Bencze

PP29973. If a, b, c ∈�1e , 1

�, then

�ln a+b+c

3abc

�a+b+c ≥�ln 1

bc

�a �ln 1

ca

�b �ln 1

ab

�c.

Mihaly Bencze


PP29974. If ak ∈�1e , 1

�(k = 1, 2, ..., n) , then

ln

n�

k=1ak

nn�

k=1

ak

n�

k=1ak

≥ Qcyclic

�ln 1

a2a3...an

�a1.

Mihaly Bencze


�ln e

�

cosA�

cos2 A

��

cosA≤Q

�ln e

cosA

�cosA.

Mihaly Bencze


1).�ln s2

r(4R+r)

� er ≤Q

�ln era

r

� 1rb

2).�ln s2

s2−2r(4R+r)

� er ≤Q

�ln era

r

� 1ra

Mihaly Bencze

PP29977. In all triangle ABC holds�ln e

�

sinA�

sinA sinB

��

sinA≤

≤�ln e

sinA

�sinB �ln e

sinB

�sinC �ln e

sinC

�sinA.

Mihaly Bencze

PP29978. In all acute triangle ABC holds�ln e

�

cosA�

cosA cosB

��

cosA≤

≤�ln e

cosA

�cosB �ln e

cosB

�cosC �ln e

cosC

�cosA.

Mihaly Bencze

PP29979. In all triangle ABC holds�ln e

�

sinA�

sin2 A

��

sinA≤

≤�ln e

sinA

�sinA �ln e

sinB

�sinB �ln e

sinC

�sinC.

Mihaly Bencze


PP29980. If λk > 0, ak ∈�1e , 1

�(k = 1, 2, ..., n), then

ln

n�

k=1λk

n�

k=1λkak

n�

k=1λk

≥

≥nQ

k=1

�ln 1

ak

�λk

. If 0 < ak ≤ 1e (k = 1, 2, ..., n) , then holds the reverse

inequality.

Mihaly Bencze

PP29981. If ak ∈�1e , 1

�(k = 1, 2, ..., n) , then

ln

n�

k=1ak

�

cyclic

a1a2

n�

k=1

ak

≥

≥�ln 1

a1

�a2 �ln 1

a2

�a3...

�ln 1

an

�a1.

If ak ∈�0, 1e

�(k = 1, 2, ..., n) then holds the reverse inequality.

Mihaly Bencze

PP29982. Solve in Q the following equation: 2a

b+c +2b

c+a + 2c

a+b = 3

Mihaly Bencze

PP29983. If xk > 0 (k = 1, 2, ..., n) andnQ

k=1

xk = 1 then determine all λ ∈ R

for whichnP

k=1

1n−1+xλ

k

≤ 1.

Mihaly Bencze


k=1

(−1)k+1 (nk)k ≥ 2n

n+1 .

Mihaly Bencze

PP29985. If xk > 0 (k = 1, 2, ..., n) andnP

k=1

xk (2xk + 1) = nn+1 then

nPk=1

(xk+1)2

25xk+1 ≥ n2

3(n+1) .

Mihaly Bencze



k=1

hk + 1

k+√k2+2

i= n(n+1)

2 when [·] denote the

integer part.

Mihaly Bencze


1).P a

b+c ≥ 5s2−6r2−24Rr3(s2−r2−4Rr)

2).P b+c−a

a ≥ 11s2−24r2−96Rr3(s2−2r2−8Rr)

3).P ra

rb+rc≥ 11(4R+r)2−24s2

6((4R+r)2−2s2)

4).P sin2 A

2

sin2 B2+sin2 C

2

≥ 44R2+5r2−6s2+4Rr3(8R2+r2−s2)

5).P cos2 A

2

cos2 B2+cos2 C

2

≥ 5(4R+r)2−6s2

3((4R+r)2−s2)

Mihaly Bencze

PP29988. Solve the equation x5 − 5x3 − 5x = a10+1a5

, when a ∈ R∗.

Mihaly Bencze


1).P sinA

sinB ≥ 1 + sr − sr

2R2

2).P�

sin A2

sin B2

�2

≥ 2− r2R −

�r4R

�2

3).P�

cos A2

cos B2

�2

≥ 3 + r2R −

�s4R

�2

Mihaly Bencze


s−arb

≤p

sr .

Mihaly Bencze

PP29991. Solve the equation x4 − 4x2 =�a4−1a2

�2when a ∈ R\ {−1, 0, 1} .

Mihaly Bencze


PP29992. Prove that

1).nP

k=1

k2(V kn )

2

n2k+2Lk≥ 1

Ln+2−3

2).nP

k=1

�kV k

n

nk+1Lk

�2≥ 1

LnLn+1−2 where Lk denote the kth Lucas number.

Mihaly Bencze


sin2 A(1+sinB sinC)≤ 3(s2+r2+4Rr−4sr)R2

2s2r2.

Mihaly Bencze

PP29994. If λ ∈ (−∞, 0] ∪ [1,+∞) thennP

k=1

�kV k

n

nk+1

�λ≥ 1

n2λ−1 .

Mihaly Bencze

PP29995. Prove that

1).nP

k=1

k2(V kn )

2

n2k+2Fk≥ 1

Fn+2−1

2).nP

k=1

�kV k

n

nk+1Fk

�2≥ 1

FnFn+1 where Fk denote the kth Fibonacci number.

Mihaly Bencze

PP29996. Let ABC be a triangle, determine all x, y ∈ R for which

cos (xA+ yB) cos (xB + yC) cos (xC + yA) ≥�2rR

�2.

Mihaly Bencze

PP29997. If ak > 1 (k = 1, 2, ...,m) then�

n�

k=1

(ak+1)

�m

−�

n�

k=1

ak

�m

�

n�

k=1ak

�m−2 n�

k=1(ak+1)

�

n�

k=1(ak+1)−

n�

k=1ak

� <

<mPj=1

�nQ

k=1

1+aik

1+(ak+1)i

�<

�

n�

k=1(ak+1)

�m

−�

n�

k=1ak

�m

�

n�

k=1ak

�m−1� n�

k=1(ak+1)−

n�

k=1ak

� .

Mihaly Bencze


PP29998. Let ABCD be a convex quadrilateral in which A∡ = C∡ = 90◦

and M ∈ BD. Prove that 1AM2 + 1

CM2 ≤ 1AB2 + 1

BC2 + 1CD2 + 1

DA2 .

Mihaly Bencze

PP29999. If xk > 0 (k = 1, 2, ..., n) andnP

k=1

xk = 1 thennP

k=1

xk

3x2k+2

≥ n2

2n2+3.

Determine all a, b > 0 for whichnP

k=1

xk

ax2k+b

≥ n2

bn2+a.

Mihaly Bencze

PP30000. Let ABC be a triangle, D ∈ (BC) , the bisectors of angles B andC meet the cevian AD in points M and N. Prove that�MDMA

�2+�NDNA

�2 ≥ sin2 Asin2 B+sin2 C

.

Mihaly Bencze


k=2

�

2

�

3√

...√k

k(k+1) < n−16(n+1) .

Mihaly Bencze

PP30002. If a, b ∈ R such thatab

�a2 − 3b2

� �3a2 − b2

�+ 4 = 2a

�a2 − 3b

�+ 2b

�3a2 − b2

�then exist x ∈ R

such that a =√2 sinx and b =

√2 cosx.

Mihaly Bencze

PP30003. If a, b > 0 then computeR (sinx−cosx)dx

2aex+b sinx .

Mihaly Bencze

PP30004. If x1 = 1 and xn+1 = λ+ 1xn

for all n ≥ 1 then determine all

λ ∈ R for whichnP

k=1

[xk] = 2n, when [·] denote the integer part.

Mihaly Bencze



P 1(4r−ha)(4r−hb)(4r−hc)

= 0

2).P 1

(2r−ra)(2r−rb)(2r−rc)= 0

Mihaly Bencze

PP30006. If Ak ∈ M3 (R) (k = 1, 2, ..., n) and AiAj = AjAi

(i, j ∈ {1, 2, ..., n}) then

Pcyclic

det�A2

1 +A1A2 +A22

�+P

cyclic

det�A2

1 −A1A2 +A22

�≥ 3

nPk=1

(detAk)2 .

Mihaly Bencze

PP30007. Let ABC be a triangle and M ∈ Int (ABC) . Prove thatP 1

sin2 AMB2

≥ 2(�

MA)2

s2−r(4R+r).

Mihaly Bencze


cos2�aA+bB+cC

a+b+c

�+ cos2

�bA+cB+aC

a+b+c

�+ cos2

�cA+aB+cC

a+b+c

�≤ 3− s2

3R2 .

Mihaly Bencze


1). 2P√

ab ≤P�a�ba

� rhc + b

�ab

� rhc

�≤ 2

Pa

2). 2P√

ab ≤P�a�ba

� rrc + b

�ab

� rrc

�≤ 2

Pa

Mihaly Bencze

PP30010. If 0 < a < b and 0 < c < d < 1 then 2 (d− c)√ab ≤

≤ acbd−adbc

ln b−ln a

�1

ac+d−1 + 1bc+d−1

�≤ (a+ b) (d− c) .

Mihaly Bencze


1). 2P√

ab ≤P�a�ba

�sinC+ b

�ab

�sinC�≤ 2

Pa

2). 2P√

ab ≤P�a�ba

�sin2 C2 + b

�ab

�sin2 C2

�≤ 2

Pa


3). 2P√

ab ≤P�a�ba

�cos2 C2 + b

�ab

�cos2 C2

�≤ 2

Pa

Mihaly Bencze

PP30012. In all acute triangle ABC holds 32sRr ≤≤Q

�a�ba

�cosC+ b

�ab

�cosC� ≤ 2s�s2 + r2 + 2Rr

�.

Mihaly Bencze

PP30013. If λ > 0 then in all triangle ABC holds1).

Pλ

rra ≤ λ+ 2 2).

Pλ

rha ≤ λ+ 2

Mihaly Bencze

PP30014. If λ > 0 then in all tetrahedron ABCD holds1).

Pλ

rha ≤ λ+ 3 2).

Pλ

r2ra ≤ λ+ 3

Mihaly Bencze

PP30015. If x, y, z > 0 and x+ y + z = 1 then8 + 27xyz ≥ 27 (xy + yz + zx) .

Mihaly Bencze

PP30016. In all triangle ABC holds64Q

(ma +mb)2 ≥ (

Pma)

3Q (3ma + 3mb −mc) .

Mihaly Bencze

PP30017. Prove that

�nP

k=1

�k+1k

� 1k+1

�= n where [·] denote the integer part.

Mihaly Bencze


P a2

b3+c3≥ 3s

2(s2−r2−4Rr)

2).P r2a

r3b+r3c

≥ 3(4R+r)

2((4R+r)2−2s2)

3).P sin4 A

2

sin6 B2+sin6 C

2

≥ 6R(2R−r)8R2+r2−s2

4).P cos4 A

2

cos6 B2+cos6 C

2

≥ 6R(4R+r)

(4R+r)2−s2

Mihaly Bencze


PP30019. If a, b, c > 0 thenP 1

(a+b)(b+c) ≤9

4(ab+bc+ca) .

Mihaly Bencze

PP30020. If ak ∈ (0, 1) (k = 1, 2, ..., n) then

nPk=1

logak

�1n

nPi=1

ai

�≥

n3n�

k=1ak

�

n�

k=1ak

�n .

Mihaly Bencze


1).P

logsinAs3R ≥ 27Rr

2s2

2).P

logsin2 A2

2R−r6R ≥ 27r2R

2(2R−r)3

3).P

logcos2 A2

4R+r6R ≥ 27s2R

2(4R+r)3

Mihaly Bencze

PP30022. Compute the integer part of the expressionnP

k=1

3

q1 + 1

k! +�1k!

�2.

Mihaly Bencze


1).P 1

r2a(2r2a+rbrc)

≥ 13s2r2

2).P 1

h2a(2h

2a+hbhc)

≥ R6s2r3

Mihaly Bencze

PP30024. Let ABC be a triangle. Determine all cevians ca, cb, cc for whichP(a+ b) ca ≤ 2

√3�s2 − r2 − 4Rr

�.

Mihaly Bencze

PP30025. If a ∈ (0, 1) ∪ (1,+∞) then solve in R the following system

(a− 1)2 x21 +�a2 − 1

�x2 + 2a = 2ax3+1

(a− 1)2 x22 +�a2 − 1

�x3 + 2a = 2ax4+1

−−−−−−−−−−−−−−−−−(a− 1)2 x2n +

�a2 − 1

�x1 + 2a = 2ax2+1

.

Gyorgy Szollosy and Mihaly Bencze


PP30026. If ak > 0 (k = 1, 2, ..., n) andnP

k=1

ak = n thenP a21

a1+a22≥ n

n−1 .

Mihaly Bencze

PP30027. Determine all A ∈ M3 (R) for which

A3 +A2 +A =

1 1 11 1 11 1 1

.

Mihaly Bencze

PP30028. In all triangle ABC holds1). 1√

2(s2−r2−4Rr)≤P a

bc+2(s2−r2−4Rr)≤ 1√

s2−r2−4Rr

2). 1√s2−2r2−8Rr

≤P s−a(s−b)(s−c)+s2−2r2−8Rr

≤q

2s2−2r2−8Rr

3). 1√(4R+r)2−2s2

≤P rarbrc+(4R+r)2−2s2

≤q

2(4R+r)2−2s2

4). 4√2R√

8R2+r2−s2≤P sin2 A

2

sin2 B2sin2 C

2+ 8R2+r2−s2

8R2

≤ 8R√8R2+r2−s2

5). 4√2R√

(4R+r)2−s2≤P cos2 A

2

cos2 B2cos2 C

2+

(4R+r)2−s2

8R2

≤ 8R√(4R+r)2−s2

Mihaly Bencze

PP30029. In all acute triangle ABC holdsP cosA

1+cosA cosB ≤ 52 − s2−(2R+r)2

4R2 .

Mihaly Bencze


r2a(r2+rbrc)

≤ 5s2−2r2

2s4r2.

Mihaly Bencze


h2a(r

2+hbhc)≤ (5s2−Rr)R

4s4r3.

Mihaly Bencze

PP30032. In all triangle ABC holds 1sr ≤P 1

ra(r+√rbrc)

≤√2

sr .

Mihaly Bencze


PP30033. In all triangle ABC holds 1sr

qR2r ≤P 1

ha(r+hbhc)≤ 1

sr

qRr .

Mihaly Bencze


1).Pq

a+bc ≤ 3

2

qs2+r2+2Rr

2Rr

2).Pq

ab+c−a ≤ 3

qR2r

3).Pq

ra+rbrc

≤ 3q

Rr

4).Pr

sin2 A2+sin2 B

2

sin2 C2

≤ 32r

q(2R−r)(s2+r2−8Rr)−2Rr2

2R

5).Pr

cos2 A2+cos2 B

2

cos2 C2

≤ 32s

q(4R+r)3+s2(2R+r)

2R

Mihaly Bencze


1).P a3

b(a2+bc+b2)≥ 2(s2−r2−4Rr)

s2+r2+4Rr

2).P r3a

rb(r2a+rbrc+r2b)

≥ (4R+r)2−2s2

s2

3).P sin6 A

2

sin2 B2 (sin

4 A2+sin2 B

2sin2 C

2+sin4 B

2 )≥ 2(8R2+r2−s2)

s2+r2−8Rr

4).P cos6 A

2

cos2 B2 (cos4

A2+cos2 B

2cos2 C

2+cos4 B

2 )≥ 2((4R+r)2−s2)

s2+(4R+r)2

Mihaly Bencze

PP30036. If x > 0, n, k ∈ N, n ≥ 2, k ≥ 2 thennP

i=1

�k

qx+ i−1

n

�≤

�nk−1 [nx]

� 1k where [·] denote the integer part.

Mihaly Bencze

PP30037. If ak ≥ 1 (k = 1, 2, ..., n) , then

1).P

cyclic

a51+4

a42−2a32+2a22−a2+1≥ 5n

2).P

cyclic

(a51+4)(a52+4)(a43−2a33+2a23−a3+1)(a44−2a34+2a24−a4+1)

≥ 25n

Mihaly Bencze



1).Pq

a+bc ≥

√2 +

qs2+r2+2Rr

2Rr

2).Pq

ab+c−a ≥ 1 +

q2Rr

3).Pq

ra+rbrc

≥√2 + 2

qRr

4).Pr

sin2 A2+sin2 B

2

sin2 C2

≥√2 + 1

r

q(2R−r)(s2+r2−8Rr)−2R2

2R

5).Pr

cos2 A2+cos2 B

2

cos2 C2

≥√2 + 1

s

q(4R+r)3+s2(2R+r)

2R

Mihaly Bencze


x21 + n− 1 ≤ x2 + x3 + ...+ xnx22 + n− 1 ≤ x3 + x4 + ...+ x1−−−−−−−−−−−−−−x2n + n− 1 ≤ x1 + x2 + ...+ xn−1

.

Mihaly Bencze

PP30040. Let be zk ∈ C (k = 1, 2, ..., n) such that |z1| = |z2| = ... = |zn| .Prove that z1, z2, ..., zn are the affixum of a regular n-gon, if and only ifz1 |z2 − z3| = z2 |z3 − z4| = ... = zn |z1 − z2| .

Mihaly Bencze

PP30041. In all triangle ABC holdsP (4+sin5 A) sin4 B

sin5 A(sin4 B−2 sin3 B+2 sin2 B−sinB+1)≥ 15.

Mihaly Bencze

PP30042. If ak ≥ 1 (k = 1, 2, ..., n) thenP

cyclic

(a41−2a31+2a21−a1+1)(a2+1)

a51+4≤ 1

5

�n+

nPk=1

ak

�.

Mihaly Bencze

PP30043. Solve in R the following equation�arccos x1

2 + arccos 2x2� �

arccos x22 + arccos 2x3

�...

�arccos xn

2 + arccos 2x1�=

=�2π3

�2.

Mihaly Bencze


PP30044. If ak ∈ [0, 1] (k = 1, 2, ..., n) , thennP

k=1

2+aka2k+ak+1

≥ n2 + 1

2

Pcyclic

2+a1a21+a1a2+a21a

22.

Mihaly Bencze

PP30045. ComputeR (3x4+6x3+x2+3x−1)dx

x8+2x6+3x4+6x2+12x+10.

Mihaly Bencze

PP30046. If a, b, c,λ > 0 thenP a4+(a2+b2)c2

b+λc ≥ (�

a)3

3(λ+1) .

Mihaly Bencze

PP30047. If ak > 0 (k = 1, 2, ..., n) thenP a1

an−11 +(n−1)a2a3...an

≤

�

n�

k=1ak

�2

n2n�

k=1ak

.

Mihaly Bencze

PP30048. If xk > 0 (k = 1, 2, ..., n) andnP

k=1

x2k = n then

nPk=1

x3k

xk+λ ≥nP

k=1

xk

xk+λ for all λ > 0.

Mihaly Bencze

PP30049. If x, y, z,λ > 0 thenP x3

√3x+λ

√x2+y2+z2

≥√

x2+y2+z2

3

P x√3x+λ

√x2+y2+z2

.

Mihaly Bencze

PP30050. If xk > 0 (k = 1, 2, ..., n) and λ > 0 andnP

k=1

x2k = n, thenP xk

xk+λ ≤ nλ+1 .

Mihaly Bencze

PP30051. The triangle ABC is equilateral if and only ifaλ+b(1−λ)

c = bλ+c(1−λ)a = cλ+a(−λ)

b when λ ∈ (0, 1).

Mihaly Bencze


PP30052. If 0 < a < 1 then solve the following system:

xax2

1 = axa3

xax3

2 = axa4

−−−−−xa

x1

n = axa2

.

Mihaly Bencze

PP30053. In triangle ABC we have a ≤ b ≤ c. Prove that for all λ ≥ 1holds aλ + hλa ≤ bλ + hλb ≤ cλ + hλc .

Mihaly Bencze

PP30054. If a > b ≥ 2, a, b ∈ N then [a, b] + 2nP

k=1

[a+ k, b+ k] +

+ [a+ n− 1, b+ n+ 1] ≥ 2n√a−b

(a+ b+ n+ 1) .

Mihaly Bencze

PP30055. Denote a (n) the smallest natural number which don’t divide n.

Compute∞Pn=1

11+d2(n)

.

Mihaly Bencze


sinA cos A2

+ 4 ≥ 2P 1

cos A2cos π−A

4

.

Mihaly Bencze

PP30057. In all convex polygon A1A2...An holdsnP

k=1

1

sinAk cosAk2

≥ n

sin�

(n−2)πn

�

cos�

(n−2)π2n

� .

Mihaly Bencze

PP30058. If x, y, z ∈ (0, 1) thenP 1

x(1−x) ≥ 8.

Mihaly Bencze


sin2 A2cos2 A

2

≥ maxn

72R8R−r ;

72R10R+r

o.

Mihaly Bencze


PP30060. If ak > 0 (k = 1, 2, ..., n) , thenP a1

a22(a2+a3)3 ≥ n3

(n−1)3(�

a1a2)2 .

Mihaly Bencze


1).P a

b+c ≥ 32 +

27(s2−7r2−10Rr)64s2

2).P b+c−a

a ≥ 3 +27(s2−13Rr+r2)

4s2

3).P ra

rb+rc≥ 3

2 +27((4R+r)3−s2(8R−r))

8(4R+r)3

Mihaly Bencze


1).P a

b ≥q

s2+r2

Rr − 5

2).P b+c−a

a−b+c ≥q

8Rr − 7

3).P ra

rb≥q

8Rr − 7

4).P�

sin A2

sin B2

�2

≥q

(2R−r)(s2+r2−8Rr)Rr2

− 9

5).P�

cos A2

cos B2

�2

≥q

(4R+r)(s2+(4R+r)2)Rs2

− 9

Mihaly Bencze

PP30063. In all triangle ABC in which A ≤ B ≤ C holds

1).�e sin2 A

2

�sin2 B2sin2 C

2 ≤�e sin2 B

2

�sin2 C2sin2 A

2 ≤�e sin2 C

2

�sin2 A2sin2 B

2

2).�e cos2 A

2

�cos2 B2cos2 C

2 ≥�e cos2 B

2

�cos2 C2cos2 A

2 ≥�e cos2 C

2

�cos2 A2cos2 B

2

Mihaly Bencze

PP30064. Let ABC be a triangle in which A ≤ B ≤ C ≤ π2 then

(e sinA)sinB sinC ≤ (e sinB)sinA sinC ≤ (e sinC)sinA sinB .

Mihaly Bencze

PP30065. Let ABC be a triangle in which A ≤ B ≤ C ≤ π2 then

(e cosA)cosB cosC ≥ (e cosB)cosA cosC ≥ (e cosC)cosA cosB .

Mihaly Bencze


PP30066. If xk ≥ e (k = 1, 2, ..., n) then

�nP

k=1

xk

��nP

k=1

1xk

�n2

≤

≤ n2

�

n�

k=1xk

��

n�

k=1

1xk

�

. If xk ∈ (0, e) (k = 1, 2, ..., n) then holds the reverseinequality.

Mihaly Bencze

PP30067. If 0 < a1 ≤ a2 ≤ ... ≤ an ≤ e thenaa2a3...an1 ≥ aa1a3a4...an2 ≥ ...a

a1a2...an−1n .

Mihaly Bencze

PP30068. If xk, yk ≥ e (k = 1, 2, ..., n) and p∈ N∗, p ≥ 2 then

�nP

k=1

(xk + yk)p

��

n�

k=1

xpk

� 1p+

�

n�

k=1

ypk

� 1p

≤

≤ �

nPk=1

xpk

� 1p

+

�nP

k=1

ypk

� 1p

!p

�

n�

k=1

(xk+yk)p

� 1p

. If xk, yk ∈ (0, e)

(k = 1, 2, ..., n) then holds the reverse inequality.

Mihaly Bencze

PP30069. If ak, bk ≥ e (k = 1, 2, ..., n) then

�nP

k=1

akbk

�2

�

n�

k=1a2k

��

n�

k=1b2k

�

≥

≥��

nPk=1

a2k

��nP

k=1

b2k

��

n�

k=1akbk

�2

. If ak, bk ∈ (0, e) (k = 1, 2, ..., n) then


Mihaly Bencze

PP30070. If a, b, c > 0 then9�a2 + b2 + c2

�2 ≥ 3P �

a2 + b2 − c2�2

+ 8 (P

ab)2 .

Mihaly Bencze

PP30071. If a, b, c ≥ e then (P

ab)�

a2 ≥�P

a2�� ab

. If a, b, c ∈ (0, 1) thenholds the reverse inequality.

Mihaly Bencze


PP30072. If xk ∈ (0, 1) (k = 1, 2, ..., n) then

1). ne−2n ≤

nPk=1

x√xk

k ≤ n

2). ne−4n ≤Px

√x1

1 x√x2

2 ≤ n

3). e−2n

nPk=1

xk ≤Px1x√x2

2 ≤nP

k=1

xk

Mihaly Bencze

PP30073. If a, b, c > 0 then 16P

a2b2 ≤P�a2 + b2 − c2

�2+ 5

�Pa2�2

.

Mihaly Bencze

PP30074. If ak ≥ 1 (k = 1, 2, ..., n) thennP

i,j=1

(aiaj)2

ai+aj−1 ≥�

nPk=1

ak

�2

.

Mihaly Bencze

PP30075. If ak ≥ 1 (k = 1, 2, ..., n) thennP

i,j,k=1

(aiajak)2

aiaj+ajak+akai−ai−aj−ak+1 ≥�

nPk=1

ak

�3

.

Mihaly Bencze

PP30076. If x1, y1 ∈ N∗ such that x1 <√7y1 and

xn+1 + xnyn = yn+1

�x2n + 1

�for all n ≥ 1, then prove that

nPk=1

xk

yk< n

√7.

Mihaly Bencze

PP30077. If a, b ∈ N ∗ and a < b√7 then a4 + 2a2 + 2 < ab

�a2 + 1

�√7.

Mihaly Bencze

PP30078. If a, b, c ∈ N ∗ and ab+ bc+ ca ≥ 1358786 thena3+b3+c3

3 ≥ 2019 + abc.

Mihaly Bencze

PP30079. If k ∈ N ∗ thenP

i,j=1

(ij)k+1

(i+j−1)k≥ (n+1)2k+2

22k+2nk−2 .

Mihaly Bencze



1).P

n

snQ

k=1

(a+ λk) ≤ 3 n

snQ

k=1

�2s3 + λk

�

2).P

n

snQ

k=1

(s− a+ λk) ≤ 3 n

snQ

k=1

�s3 + λk

�

3).P

n

snQ

k=1

(ra + λk) ≤ 3 n

snQ

k=1

�4R+r

3 + λk

�

4).P

n

snQ

k=1

�sin2 A

2 + λk

�≤ 3 n

snQ

k=1

�1− r

2R + λk

�

5).P

n

snQ

k=1

�cos2 A

2 + λk

�≤ 3 n

snQ

k=1

�2 + r

2R + λk

�for all λk > 0

(k = 1, 2, ..., n)

Mihaly Bencze

PP30081. If x, y, z,λk > 0 (k = 1, 2, ..., n) then

n

snQ

k=1

(x+ λk) + n

snQ

k=1

(y + λk)+

+ n

snQ

k=1

(z + λk) + 3 n

snQ

k=1

�x+y+z3 + λk

�≤

≤ 2 n

snQ

k=1

�x+y2 + λk

�+ 2 n

snQ

k=1

�y+z2 + λk

�+ 2 n

snQ

k=1

�z+x2 + λk

�.

Mihaly Bencze

PP30082. If λ > 0 and xk ∈�1λ ,λ

�(k = 1, 2, ..., n) then determine

max

��nQ

k=1

(1− xk)

�� .

Mihaly Bencze

PP30083. If xk ∈�12 , 2

�(k = 1, 2, ..., n), then

1).nP

k=1

(1− xk)2 ≤ 1

2

nPk=1

xk

2).

��nQ

k=1

(1− xk)

�� ≤s

12n

nQk=1

xk

Mihaly Bencze



1).�λ+ 1

r

�4 ≥ 125�

1hahbhchd

+ λP 1

hahbhc

�

2).�λ+ 2

r

�4 ≥ 125�

1rarbrcrd

+ λP 1

rarbrc

�for all λ > 0

Mihaly Bencze


1).�x+ y + 1

r

�4 ≥ 125�x+ys2r

+ xy(4R+r)s2r

�

2).�x+ y + 1

r

�4 ≥ 125

�R(x+y)2s2r2

+xy(s2+r2+4Rr)

4s2r2

�for all x, y > 0

Mihaly Bencze


PctgA

2 ≤P 1sin A

2sin B

2

2). 2P

tgA2 ≤P 1

cos A2cos B

2

Mihaly Bencze

PP30087. If ak ≥ 1 (k = 1, 2, ..., n) thenP

cyclic

a1a2 ≥ 2nP

k=1

qa2k − 1.

Mihaly Bencze


��

1 1 1ra rb rcr4a − r3a r4b − r3b r4c − r3c

��=

= (rb − ra) (rc − ra) (rc − rb)�(4R+ r)2 − s2 − 4R− r

�.

Mihaly Bencze

PP30089. In all acute triangle ABC holds1). 2

PctgA ≤P 1

sinA sinB2). 2

PtgA ≤P 1

cosA cosB

Mihaly Bencze


PP30090. In all triangle ABC holdsP�n√sin2A+

n√sin2B

�n+P�

n√cos2A+

n√cos2B

�n≤

≤ 3 +�P n

√sin2A

�n+�P n

√cos2A

�nfor all n ∈ N∗.

Mihaly Bencze


��

1 1 1a b ca4 − a3 b4 − b3 c4 − c3

��=

= (b− a) (c− a) (c− b)�3s2 − r2 − 12Rr − 2s

�.

Mihaly Bencze

PP30092. Let ABCD be a convex pentagon and denote M,N,K,L, T themidpoints of AC,AD,BD,BE,EC. Prove that3 (AB +BC + CD +DE + EA) >

> AC +BE + EC +BD +DA+ 2 (MK +NT + LN + LM) .

Mihaly Bencze

PP30093. In all triangle ABC holds3 +

p3 + 2

Pcos (A−B) ≥ 2

Pcos A−B

2 .

Mihaly Bencze


1).P �

sin A2 + sin B

2

�2+P �

cos A2 + cos B

2

�2 ≤ 3 +�P

sin A2

�2+

�Pcos A

2

�2

2).P �

sin2 A2 + sin2 B

2

�2+P �

cos2 A2 + cos2 B

2

�2 ≤≤ 12R2+4Rr+r2−s2

4R2 +�P

sin2 A2

�2+�P

cos2 A2

�2

Mihaly Bencze

PP30095. Prove that 2P

1≤i<j≤n

1ij > ln2 n− 4(n−2)

5(2n+1) for all n ≥ 2.

Mihaly Bencze


1≤i<j≤n

1ij < (1 + lnn)2 − 830n−89

252(2n+1) for all n ≥ 1.

Mihaly Bencze


PP30097. If ak > 0 (k = 1, 2, ..., n) thenP

cyclic

pa21 − a1a2 + a22 ≤

≤nP

k=1

ak +

snP

k=1

a2k −P

a1a2.

Mihaly Bencze

PP30098. In all acute triangle ABC holdsP(cosA+ cosB)2 +

P(sinA+ sinB)2 ≤

≤ 3 + (P

cosA)2 + (P

sinA)2 .

Mihaly Bencze


P√a2 − ab+ b2 ≤ 2s+

√s2 − 3r2 − 12Rr

2). 12

Pq3 (a− b)2 + c2 ≤ s+

√s2 − 3r2 − 12Rr

3).Pq

r2a − rarb + r2b ≤ 4R+ r +q(4R+ r)2 − 3s2

4).Pq

sin4 A2 − sin2 A

2 sin2 B2 + sin4 B

2 ≤ 2R−r2R +

√(4R+r)2−3s2

4R

5).Pq

cos4 A2 − cos2 A

2 cos2 B2 + cos4 B

2 ≤ 4R+r2R +

√(4R+r)2−3s2

4R

Mihaly Bencze


k=1

�2�1 + 1

2 + ...+ 1k

�+ 1

k+1 + 1k+2 + ...+ 1

2k

�<

<

�2nPk=1

1k

�2

− 1 < 2536

nPk=1

�2�1 + 1

2 + ...+ 1k

�+ 1

k+1 + 1k+2 + ...+ 1

2k

�.

Mihaly Bencze


k=1

�1 + 1

13

� �1 + 1

23

� �1 + 1

33

�...

�1 + 1

(k(k+1))3

�< n(3n+2)

n+1 .

Mihaly Bencze


k=1

�1 + 1

22+ 1

32+ ...+ 1

(k(k+1))2

�< n(2n+1)

n+1 .

Mihaly Bencze


PP30103. If Ak =kP

i=1

2i−12i then n

16(n+1) <nP

k=1

A2kA

2k+1 <

9n64(n+1) .

Mihaly Bencze

PP30104. Prove that

nPk=1

rk2+5k+3√

(k+2)(k+3)�

(k+2)√k+

√(k2−1)(k+3)

� ≤rnq

n(n+2)n+3 .

Mihaly Bencze

PP30105. If Ak =kQ

i=1

2i−12i then

nPk=1

A2kA

2k+1 <

n4(3n+4) .

Mihaly Bencze

PP30106. If x, y, z > 0 thenP x

1+√x+

√3�

x√3+√

�

x≤P

√2(x+y)√2+

√x+y

.

Mihaly Bencze


1≤i<j≤n

1√ij

≥ 4n2

(1+√n)

2 − ln en.

Mihaly Bencze


1).P a

1+√a≤ 2

√3s√

3+√2s

2).P s−a

1+√s−a

≤√3s√

3+√s

3).P ra

1+√ra

≤√3(4R+r)√3+

√4R+r

4).P sin2 A

2

1+sin A2

≤√6(2R−r)√

6R+√

R(2R−r)

5).P sin4 A

2

1+sin2 A2

≤√3(8R2+r2−s2)

2R�

4R√3+√

2(8R2+r2−s2)�

6).P cos2 A

2

1+cos A2

≤√3(4R+r)

2√3R+

√2R(4R+r)

7).P cos4 A

2

1+cos2 A2

≤√3((4R+r)2−s2)

2R�

4√3+

�

2R((4R+r)2−s2)�

Mihaly Bencze



wa ≤r

2(4R+r)R

P�aba+b

�2≤ 1

2

q�4 + r

R

�(3s2 − r2 − 4Rr).

Mihaly Bencze


b+cbc wa ≤

r(4R+r)((s2+r2+4Rr)2+8s2Rr)

sR(s2+r2+2Rr).

Mihaly Bencze


wa ≤ 2sqP ab

(a+b)2≤ s

√3.

Mihaly Bencze


P√b+ cwa ≤

r2s((s2+r2+4Rr)2+8s2Rr)

s2+r2+2Rr.

Mihaly Bencze

PP30113. In all triangle ABC holdsP 2(s−a)(s−b)+c(s−c)√

ab≤ 2s.

Mihaly Bencze

PP30114. In all triangle ABC holds Rr ≥ s2+r2−2Rr

6Rr ≥ s2+r2+10Rr12Rr ≥ 2. A

refinement of Euler’s inequality.

Mihaly Bencze

PP30115. In all tetrahedron ABCD holds1). 4 + r2

4

�1

rarb+ 1

rarc+ 1

rard+ 1

rbrc+ 1

rbrd+ 1

rcrd

�+ r4

8rarbrcrd≥

≥ 2 + r3

8 (ra + rb + rc + rd)

4). 4 + r2�

1hahb

+ 1hahc

+ 1hahd

+ 1hbhc

+ 1hbhd

+ 1hchd

�≥

≥ 2 + r3

hahbhchd(ha + hb + hc + hd)

Mihaly Bencze


a+ b− c ≤ (3+2λ2)√2s

2λ for allλ > 0.

Mihaly Bencze


PP30117. Let A1A2...An be a convex polygon. Prove thatP

cyclic

q(a1 + a2 + ...+ an−1)

2 − a2n ≤pn (n− 2)

nPk=1

ak.

Mihaly Bencze

PP30118. In all acute triangle ABC holds1).

PHA2 ≥ 4

�s2 + r2 − 4Rr

�

2).P HA

HB+HC ≥ (R+r)2

2(s2+r2−4R2)

Mihaly Bencze


Paw2

c ≤ 2s3

3

2).P

cwa ≤√2s√s2 − r2 − 4Rr

Mihaly Bencze

PP30120. In all triangle ABC holds 2P a(b+c)

b+c−a + 8s ≥P (a+b)(a+b+2c)c .

Mihaly Bencze


1).P�

a(b+c)

(b+c−a) sin A2

�2

≥ 32s2R2R−r

2).P�

a(b+c)

(b+c−a) cos A2

�2

≥ 32s2R4R+r

Mihaly Bencze


a(b+c)(b+c−a)ma

�2≥ 32s2

3(s2−r2−4Rr).

Mihaly Bencze


a(b+c)(b+c−a)ra

�2≥ 16s2

(4R+r)2−2s2.

Mihaly Bencze

PP30124. In triangle ABC the centroid belongs to the incircle. Prove that

12s2R+P

r3a =�

s2

4R

�3.

Mihaly Bencze


PP30125. In triangle ABC the centroid belongs to the incircle. Prove that1

a2b2+ 1

b2c2+ 1

c2a2= 3

32R2r2.

Mihaly Bencze

PP30126. Let Ta, Tb, Tc be the circumradii of triangles BGC, CGA, AGB.

Prove that 54s2r2TaTbTc ≤�s2 − r2 − 4Rr

�3.

Mihaly Bencze

PP30127. Denote N the Nagel’s point in triangle ABC. Prove that�Psin2 A

2

� �Pcos2 A

2

�= (2NO+3r)(4NO+9r)

4(NO+2r)2.

Mihaly Bencze

PP30128. In all triangle ABC holdss2−r(4R+r)

Rr ≤P cos A2

sin B2sin C

2

≤ 31−1λ

�Paλ

� 1λ for all λ ∈ (−∞, 0] ∪ [1,+∞) .

Mihaly Bencze


Pm2

ar2a ≥ (4R+ r) s2r

2).P 1

m2a≤ 4R+r

s2r

3).P m2

a

ra≥ s2−2r(4R+r)

r

4).P m2

a

rar2c≥ (4R+r)2−2s2

s2r

5).P

m4a ≥ s2

�s2 − 2r (4R+ r)

�

6).Q �

m2a +m2

b

�≥ 4s4Rr

7).P m2

a+m2b

(ra+rb)2 ≥ 1

2

�4R+r

s

�2

Mihaly Bencze


1).P m2

a

rb≥ 4R+ r

2).P m2

a

hb≥ 3 3

qR2r

3).P m6

a

r3b

≥ (4R+ r)3 − 12s2R

Mihaly Bencze



Pmamb ≤ 3

r3�s2r2 + 1

32

P(a2 − b2)2

�.

Mihaly Bencze


1).P ra

rb+rc= (4R+r)3+(8R+r)s2

4s2R

2). (4R+r)2

2s2≤P ra

rb+rc≤ R

r − 12

Mihaly Bencze

PP30133. In all triangle ABC holds 33√s2r ≤Pma ≤ 3

2

p2 (s2 − r2 − 4Rr).

Mihaly Bencze


sin A+B4 ≥ r

4R .

Mihaly Bencze

PP30135. In all triangle ABC holds 3 3

q�r4R

�2 ≤P 1AI·BI ≤ 1

r2

Psin A+B

4 .

Mihaly Bencze

PP30136. Let A1B1C1 denote the Morley triangle of triangle ABC. DenoteRm and r the circumradii of triangle A1B1C1 and the inradi of triangle

ABC. Prove that Rm ≤ 2√3r3 .

Mihaly Bencze


P a2

bmb≥ 2

√3

2).P b2+λc2

ama≥ 2 (λ+ 1)

√3 for all λ ≥ 0

3).P a2+xb2+yc2

ama≥ 2 (x+ y + 1)

√3 for all x, y ≥ 0

Mihaly Bencze


3p

a2b2m2a ≤ 8

3 3√3

�s2 − r2 − 4Rr

�.

Mihaly Bencze


PP30139. In all triangle ABC holdsQ�√3ama +

2s2

3

�≤ 8

�s2 − r2 − 4Rr

�3.

Mihaly Bencze

PP30140. In all triangle ABC holdsP (b+c) cos A

2wa

≥ 8s2

s2+r2+4Rr.

Mihaly Bencze

PP30141. Denote AD, BE, CF the Gergonne’s cevians in triangle ABC.

Prove that AD2 +BE2 + CF 2 =2(s2+r2)r

R + 2s2 + 7r2 − 4Rr.

Mihaly Bencze

PP30142. In all triangle ABC holds 2R(4R+r)2

2R−r ≤P�

rasin A

2

�2

≤ s2 +�9R2

�2.

Mihaly Bencze

PP30143. Denote AD, BE, CF the Gergonne’s cevians in triangle ABC.Prove that

AD ·ra+BE ·rb+CF ·rc ≤r�

(4R+ r)2 − s2��

2(s2+r2)rR + 2s2 + 7r2 − 4Rr

�.

Mihaly Bencze


P(ahb + bha) ≤ s2−3r2

2 − 6Rr +(s2+r2+4Rr)

2−24s2Rr

2R2 .

Mihaly Bencze

PP30145. In all triangle ABC holdsP ab+4sr

b2(4sRr+4(a+c)sr+ac2)≥ 3

8sRr .

Mihaly Bencze

PP30146. In all triangle ABC holdsP bma+2sr

b2((mb+mc)ac+2(a+c)sr)≥ 3

8sRr .

Mihaly Bencze



1).P AI2

b+c =(s2+r2+4Rr)

2−4Rr(3s2+r(4R+r))2s(s2+r2+2Rr)

2).P

AI ≤p3 (s2 + r2 − 8Rr)

Mihaly Bencze

PP30148. Prove that

2nP

k=1

(ln(k+1))p+1−(ln k)p+1

(2k+1) ln(1+ 1k )

< (ln (n+ 1))p+1 < 12

nPk=1

(2k+1)((ln(k+1))p+1−(ln k)p+1)k(k+1)

for all p ∈ N.

Mihaly Bencze

PP30149. Solve in Z the equation 1x+yz + 1

y+zx + 1z+xy = 19

35 .

Mihaly Bencze

PP30150. If x ≥ 1 then�1 + 1

x

�x ≥ e4(x2+1)arctgx

πx(2x+1) .

Mihaly Bencze


k=1

ln2 k+ln k ln(k+1)+ln2(k+1)2k+1 < ln3 (n+ 1) <

< 12

nPk=1

(2k+1)(ln2 k+ln k ln(k+1)+ln2(k+1))k(k+1) .

Mihaly Bencze

PP30152. If 0 < x ≤ y then ln y(x+1)x(y+1) <

4(y−x)(2x+1)(2y+1) .

Mihaly Bencze

PP30153. If 1 ≤ a ≤ b then 2b+12a+1 ≤ aa(b+1)b+1

bb(a+1)a+1 ≤q

b(b+1)a(a+1) .

Mihaly Bencze

PP30154. Prove that π4−902985840 <

∞Pk=1

�e1+

12+...+ 1

k−γ − k − 1

2

�4< π4

2985840

where γ = 0, 57... denote the Euler constant.

Mihaly Bencze


PP30155. If 0 < x ≤ y then (y−x)(2xy+x+y+1)2xy(x+1)(y+1) ≥ ln y(x+1)

z(y+1) .

Mihaly Bencze

PP30156. If Hk = 1 + 12 + ...+ 1

k then

� ∞Pn=1

Hn

n·2n

�2

= 512

� ∞Pn=0

H2n

(n+1)2

�.

Mihaly Bencze

PP30157. If ξ denote the Reiemann zeta function, then� ∞Pn=1

(ξ (2n)− 1)

�2

+ 7

� ∞Pn=1

(ξ (2n+ 1)− 1)

�2

= 1.

Mihaly Bencze

PP30158. Prove that�36n2 + 12n+ 1

�� nPk=0

�6n+36k

�B6k

�·�

nPk=0

�6n+56k+2

�B6k+2

�=

=�36n2 + 48n+ 15

�� nPk=0

�6n+16k−2

�B6k−2

�2

where Bk denote the kth Bernoulli

numbers.

Mihaly Bencze

PP30159. If γ = 0, 57... denote the Euler constant thenn(n+2)

2 + 124 ln

n+1e <

nPk=1

e1+12+...+ 1

k−γ < n(n+2)

2 + 124 lnne for all n ≥ 1.

Mihaly Bencze

PP30160. Prove that π2−63456 <

∞Pk=1

�e1+

12+...+ 1

k−γ − k − 1

2

�2< π2

3456 where

γ = 0, 57... denote the Euler constant.

Mihaly Bencze

PP30161. Prove that P

1≤i1≤n+1

1i1

!λ

+

P

1≤i1<i2≤n+1

1i1i2

!λ

+ ...+

P

1≤i1<i2<...<in≤n+1

1i1i2...in

!λ

≤

≤ n�n+ 1− 1

(n+1)!

�λfor all λ ∈ [0, 1] .

Mihaly Bencze



k=0

(nk)(2n−1

k )≥ 2

(n+1)√n+1

.

Mihaly Bencze

PP30163. If a > 1 thennP

k=0

�

a2k+1

�

(nk)2

2k≥ (2nn )

2

1a−1

− 2n+1

a2n+1−1

.

Mihaly Bencze

PP30164. If a1 = 3, a2 = 5 and 2an+1 =12

�a2n + 1

�for all n ≥ 1 then

compute limn→∞

n

�12 − (an+1 − 1)

nPk=1

11+ak

�.

Mihaly Bencze


n

�2−

nPk=2

1Fk−1Fk+1

−nP

k=1

Fk

Fk+1Fk+2

�where Fk

denote the kth Fibonacci number.

Mihaly Bencze

PP30166. If x, y ∈ R and a ∈ (0, 1) and x0 = x, y0 = y,xn+1 = axn + (1− a) yn, yn+1 = (1− a)xn + ayn for all n ∈ N thenxn = x+y

2 + x−y2 (2a− 1)n and yn = x+y

2 + y−x2 (2a− 1)n for all n ∈ N.

Mihaly Bencze

PP30167. If ak > 0 (k = 1, 2, ..., n) then determine all λ > 0 for whichQ

cyclic

�λa1

a2+a3+...+an+ 1

�≥

�λ

n−1 + 1�n

.

Mihaly Bencze

PP30168. Prove that tg

� ∞Pn=1

arctg 1F2n+1

�= 1 when Fk denote the kth

Fibonacci number.

Mihaly Bencze


PP30169. If ak > 0 (k = 1, 2, ..., n) thennP

k=1

ank

1+an+1k

+

n�

k=1ak

1+n�

k=1

an+1k

≤ n+12 .

Mihaly Bencze


(x2+1)dx2x6+x3+3

≥ 29 ln

b3+1a3+1

.

Mihaly Bencze

PP30171. If 0 < a ≤ b then 29arctg

a3+b3

1−(ab)3≤

bRa

(x3+1)dx2x6+x3+3

.

Mihaly Bencze


k=1

12k+1 < ln (n+ 1) < 1

2

nPk=1

2k+1k(k+1) .

Mihaly Bencze


k=1

ln k(k+1)2k+1 < ln2 (n+ 1) < 1

2

nPk=1

(2k+1) ln k(k+1)k(k+1) .

Mihaly Bencze


k=1

(2k+1)(2k6+k3+3)(k+1)2(k3+1)(k6+1)

≤ 3n(n+2)

2(n+1)2.

Mihaly Bencze


2k6+k3+3(k3+1)(k6+1)

≤�π2

�2.

Mihaly Bencze

PP30176. Solve in Z the equation x2 + y2 + z2 + t2 = (x+ y)2(z + t)2.

Mihaly Bencze

PP30177. Solve in Z the equation x3 + y3 + z3 + t3 = (xyzt)3 .

Mihaly Bencze


PP30178. If xk > 0 (k = 1, 2, ..., n) thennP

k=1

x2k(2x

6k+x3

k+3)

(x3k+1)(x6

k+1)

≤ 3n2 .

Mihaly Bencze

PP30179. Let xk > 0 (k = 1, 2, ..., n) such that limn→∞

xn = 0 and∞Pn=1

xn = +∞. Prove that exist a subsequence (xnk)k≥1 of sequence (xn)n≥1

for which∞Pk=1

x2p+1nk

= +∞ and∞Pk=1

x2p+2nk

< +∞ when p ∈ N.

Mihaly Bencze


1≤i<j≤n

1

(ni)+(nj)

≥ n2(n−1)4(2n−1) .

Mihaly Bencze


1≤i<j≤n

1

(ni)2+(nj)

2 ≥ n2(n−1)

4((2nn )−1).

Mihaly Bencze

PP30182. We consider Fn : (a,+∞) → (a,+∞) when a > 0 such thatFn (x) = a+ (x− a)a

n

. Prove that Fm ◦ Fn = Fm+n for all m,n ∈ Z. IfG = {Fn|n ∈ Z} , then (G, ◦) is abelian group and (G, ◦) ∼= (Z,+) .

Mihaly Bencze

PP30183. If xk ∈ (0, 1) (k = 1, 2, ..., n), thennP

k=1

arctgxk >π

�

n�

k=1xk

�

2

�

n+n�

k=1x2k

� .

Mihaly Bencze

PP30184. If xk ∈�0,√3�(k = 1, 2, ..., n) , then

nPk=1

arctgxkarctg1xk

≤πn2

n�

k=1

xk

2

�

n2+

�

n�

k=1

xk

�2� .

Mihaly Bencze



A2 + 1�arctgAarctg 1

A ≤ π2

2 .

Mihaly Bencze


tg�

πA2

2(A2+π2)

�< π

2 .

Mihaly Bencze

PP30187. Determine all x, y > 0 for which

arctgxarctg 1y + arctgyarctg 1

x < π2

�x

x2+1+ y

y2+1

�.

Mihaly Bencze

PP30188. Denote pk the kth prime, and M =

�x ∈ R|x =

nPk=1

1prk, r ∈ N

�.

Show that M is neither an open set nor a closed set in R with the usualtopology.

Mihaly Bencze

PP30189. Compute2πR0

sinx cos 2x sin 4xdx

(1+sin2 x)(1+sin2 2x)(1+sin2 4x).

Mihaly Bencze

PP30190. Prove that

limx→0

1x2

��ax+bx

2

� 1x −

√ab− 1

8

�ln a

b

�2√abx

�= 1

128

�ln a

b

�4√ab for all a, b > 0.

Mihaly Bencze

PP30191. Determine the best constants c1, c2 > 0 such that in all triangleABC holds 2 (4R+ r)2 (2R− r) ≥ s2 (c1R− c2r) .

Mihaly Bencze


k=1

(k+1)2k−k2k

(k(k+1))k≥ 2 ln (n+1)n

n! .

Mihaly Bencze


PP30193. If Fk and Lk denote the kth Fibonacci, respective Lucasnumbers, then

1).nP

k=1

(Fk+1−Fk)Fk+2

FkFk+1≥ 2 lnFn+1

2).nP

k=1

(Lk+1−Lk)Lk+2

LkLk+1≥ 2 lnLn+1

Mihaly Bencze

PP30194. If a ≥ 1 thennP

k=1

�ak − 1

ak

�≥ n (n+ 1) ln a.

Mihaly Bencze

PP30195. In all triangle ABC holds s2+r2+4Rr2sr ≥ s

R + 2 ln 2R2

sr .

Mihaly Bencze


a41 +118 = a2

√3 +

�a23 − 1

�2+�√

2a4 −q

38

�2

a42 +118 = a3

√3 +

�a24 − 1

�2+�√

2a5 −q

38

�2

−−−−−−−−−−−−−−−−−−−a4n + 11

8 = a1√3 +

�a22 − 1

�2+�√

2a3 −q

38

�2

.

Mihaly Bencze

PP30197. If ak > 0 (k = 1, 2, ..., n) thenP

cyclic

a1�a42 − a2

√3 + 2

�≥ 5

8

nPk=1

ak.

Mihaly Bencze

PP30198. If ak > 0 (k = 1, 2, ..., n) thenP

cyclic

r(a31+a32)(a33+a34)

(a21+a1a2+a22)(a23+a3a4+a24)≥ 2

3

P√a1a3.

Mihaly Bencze

PP30199. If ak > 0 (k = 1, 2, ..., n) thennP

k=1

�a4k − ak

√3�+ 11n

8 ≥ 2nP

k=1

��a2k − 1

� �√2ak −

q38

�.��

Mihaly Bencze


PP30200. If x ∈�0, π2

�then

x6+(π2−x)

6

x2(π2−x)

2�

x4+x2(π2−x)

2+(π

2−x)

4� ≥ (sinx+cosx)(1−sinx cosx)

3 sin2 x cos2 x.

Mihaly Bencze

PP30201. If x ∈�0, π2

�and λ ≥ 0 then

x sinx+ (λ+ 1)x2 cosx ≤ (λ+ 2) sin2 x.

Mihaly Bencze

PP30202. If a, b, c > 0 thenP a3+(b+c)3

a2+b2+c2+ab+2bc+ca≥ a+ b+ c.

Mihaly Bencze

PP30203. In all acute triangle ABC holds�A2 +B2 + C2

�2 ≤ (s2−r(4R+r))(4s2r2−2(s2−r2−4Rr)(s2−r2−4Rr−4R2))8R2(s2−(2R+r)2)

2 .

Mihaly Bencze

PP30204. In all triangle ABC holds�A2 +B2 + C2

�2 ≤ 2(8R2+r2−s2)((4R+r)2−2s2)s2R2 .

Mihaly Bencze


1).P tgA

A2 ≥�s2+r2+Rr

2sr

�2− 4R

r

2).P ctgA

(π2−A)

2 ≥�

s2+r2−4R2

s2−(2R+r)2

�2− 8R(R+r)

s2−(2R+r)2

3).P �

sinAA

�2 ≥ s2−r2−4Rr2sr 4).

P�cosAπ2−A

�2≥ 2sr

s2−(2R+r)2

Mihaly Bencze


P A2

sinA ≤ 2srs2−(2R+r)2

2).P A2

tgA ≤ sR

3).P (π

2−A)

2

cosA ≤ s2−r(4R+r)2sr 4).

P (π2−A)

2

ctgA ≤ R+rR

Mihaly Bencze


PP30207. If x ∈�0, π2

�then

√sinxtgx+

√cosxctgx ≥ π

2 .

Mihaly Bencze

PP30208. Determine all x, y ∈�0, π2

�for which x2+ y2 ≤ sinxtgy+sin ytgx.

Mihaly Bencze

PP30209. If x ∈�0, π2

�then x2

�π2 − x

�2 ≤ sinx cosx.

Mihaly Bencze

PP30210. If x ∈�0, π2

�then 2x2 − πx+

�π2

�2 ≤ (sinx+cosx)(1−sinx cosx)sinx cosx .

Mihaly Bencze

PP30211. If x ≥ 0 then (arctgx)2 ≤ x2√1+x2

.

Mihaly Bencze

PP30212. In all acute triangle ABC holdsPAB ≤P sinAtgB +

PtgA sinB.

Mihaly Bencze

PP30213. If x ∈�0, π2

�then x2 sinx+ sin 2x

2 ≤ sinx+ x (1− cosx) .

Mihaly Bencze


1).P

A2 cosA ≤ s2−r(4R+r)2R2

2).P �

π2 −A

�2sinA ≤ 1− s2−(2R+r)2

2R2

Mihaly Bencze

PP30215. If x ∈�0, π2

�then�

π2

4 − (π − 1)x+ x2�sinx+

�x2 − x+ π

2

�cosx ≤ 2.

Mihaly Bencze

PP30216. If x ≥ 0 then xarctgx+ (arctgx)2 ≤ 2x2√1+x2

.

Mihaly Bencze



1).P

A sinA+P

A2 cosA ≤ s2−r(4R+r)R2

2).P �

π2 −A

�cosA+

P �π2 −A

�2sinA ≤ 2− s2−(2R+r)2

R2

Mihaly Bencze

PP30218. Let be a, b, c, d > 0 determine all function f : R×R → R forwhich (af (a, b+ cf (c, d))) (bf (a, b) + df (c, d)) ≥

�a2 + b2

� �b2 + d2

�.

Mihaly Bencze

PP30219. If a, b, c > 0 then1

2a+1 + 1a+c+1 + 1

2b+1 + 1b+c+1 ≤ 2 + 1

a+b+c+1 + 12a+2b+c+1 .

Mihaly Bencze


�

tgA2

3�

tgA2+2tgB

2tgC

2

�

tgC2

≤ 4R+r5r .

Mihaly Bencze


(rahb+sr 3√2s)2+(rbhc)

2+(sr 3√2s)2 ≤ 1

2(sr 3√2s)2 .

Mihaly Bencze

PP30222. Let ABC be a triangle. Prove thatP

tg2A2 tgB2 ≥ 2m where m is

a positive root of the equation 2m3 +�4R+r

s

�m− r

s = 0.

Mihaly Bencze

PP30223. In all triangle ABC holds s2+(4R+r)2

s2− 18R

4R+r ≥ 2 ln (4R+r)3

27Rs2.

Mihaly Bencze

PP30224. In all acute triangle ABC holdsP A

sinA +P

A2ctgA ≤ 2sR .

Mihaly Bencze

PP30225. In all triangle ABC holds s2+r2−8Rrr2

− 18R2R−r ≥ 2 ln 2(2R−r)3

27Rr2.

Mihaly Bencze



4R+r2R − 9s2

s2+(4R+r)2≤ 2 ln

16R2(s2+(4R+r)2)3

27s8.

Mihaly Bencze

PP30227. If xk ≥ 1 (k = 1, 2, ..., n) thennP

k=1

1xk

− n2

n�

k=1

xk

≤ 2 ln

�

1n

n�

k=1xk

�n

n�

k=1

xk

. If

xk ∈ (0, 1) (k = 1, 2, ..., n) , then holds the reverse inequality.

Mihaly Bencze

PP30228. In all triangle ABC holds sR − 18sr

s2+r2+4Rr≤ 2 ln

(s2+r2+4Rr)3

432s2r2R2 .

Mihaly Bencze

PP30229. Let ABCD be a tetrahedron inscribed in sphere with radius Rand denote r the radius of insphere. Prove that R− 3r ≥ 6Rr

R+3r lnR3r (A

refinement of Euler’s inequality.)

Mihaly Bencze

PP30230. In all triangle ABC holds Rr ≥ s2−3r2−4Rr

4Rr ≥�

a2c6sRr ≥ 2. (A

refinement of Eyler’s inequality.)

Mihaly Bencze


s2+r2−4R2

s2−(2R+r)2− 9R

R+r ≥ 2 ln 4(R+r)3

27R(s2−(2R+r)2).

Mihaly Bencze

PP30232. In all triangle ABC holds 2R−r2R − 9r2

s2+r2−8Rr≤ 2 ln

(s2+r2−8Rr)3

432R2r4.

Mihaly Bencze

PP30233. In all triangle ABC holds s2+r2+4Rr2sr − 9R

s ≥ 2 ln 2s2

27Rr .

Mihaly Bencze



R+rR − 9(s2−(2R+r)2)

s2+r2−4R2 ≤ 2 ln(s2+r2−4R2)

3

128R2(s2−(2R+r)2).

Mihaly Bencze

PP30235. In all triangle ABC holds1). R

r ≥ ln sr

2). s2+r2+4Rr2R ≥ 1 + 2 ln 2s2

Rr

Mihaly Bencze

PP30236. In all triangle ABC holds R− 2r ≥ 4RrR+2r ln

R2r . (A refinement of

Euler’s inequality).

Mihaly Bencze

PP30237. In all acute triangle ABC holdss2+r2−4R2

s2−(2R+r)2≥ R+r

R + 2 ln 4R2

s2−(2R+r)2.

Mihaly Bencze


1). s2+r2−8Rrr2

≥ 2R−r2R + 4 ln 4R

r

2).�4R+r

s

�2 ≥ 1 + r2R + 4 ln 4R

s

Mihaly Bencze


k=1

k(k2+1)k4+k2+1

≥ 12 ln

�n2 + n+ 1

�.

Mihaly Bencze

PP30240. Let ak > 0 (k = 1, 2, ..., n) and A = 1n

nPk=1

ak, G = n

snQ

k=1

ak,

H = nn�

k=1ak

. Prove that (A−G)(G−H)(lnA−lnG)(lnG−lnH) ≥

4AG2H(A+G)(G+H) .

Mihaly Bencze


PP30241. Let ABCD be a quadrilateral inscribed in a circle with radiusR = 1, and M ∈ Int (ABCD). Prove that6 ≤ 2 (AM + CM) +BM +DM ≤ 8.

Mihaly Bencze


k=1

2k+1k(k+1) ≥ 2 ln (n+ 1) .

Mihaly Bencze

PP30243. If A ∈ Mn (C) and Am = (−1)m In, m ∈ N,m ≥ 2 thendetermine all λk ∈ C (k = 1, 2, ..., n) for whichn ≤ rang (In + λ1A) + rang (In + λ2A) + ...+ rang (In + λnA) ≤ (m− 1)n.

Mihaly Bencze

PP30244. If a0 = 1 andnP

k=1

�nk

�an−k = 0 then study the convergence of

sequence bn = a01 + a1

1! +a22! + ...+ an

n! .

Mihaly Bencze

PP30245. Let be Hn =nP

k=1

and xn = 1H1

+ 1H2

+ ...+ 1Hn

+ ln 1Hn

. Study the

convergence of sequence (xn)n≥1 and compute its limit.

Mihaly Bencze

PP30246. If 0 < a ≤ b then 2 ln b3+1a3+1

+ arctg b3−a3

1+b3a3≤ 9

2 (b− a) .

Mihaly Bencze

PP30247. If xk > 0 (k = 1, 2, ..., n) thenP x1x2

2

x31+x3

2≤ n

2 .

Mihaly Bencze

PP30248. 1). If a, b, c, d > 0 thenQ

(a+ b+ c− d) ≤Q (a+ b)

2). If a, b, c, d > 0 then determine all λ ∈ R for whichQ(a+ b+ c+ λd) ≤Q (a+ b) .

Mihaly Bencze


PP30249. Prove that1). n

�n−2p−1

�+ n2

�n−2p−2

�= p2

�np

�

2). (p− 1)n�n+1p

�+ (n+ 1)

�np

�− n2

�n

p−1

�=

�np

�

Mihaly Bencze

PP30250. Solve in N the equation (n!) + (2n)! + (3n)! = m6.

Mihaly Bencze

PP30251. If b ≥ a ≥ 67 then ln�

2 ln b−12 ln a−1

�≤

bRa

π(x)x2 dx ≤ ln

�2 ln b−32 ln a−3

�.

Mihaly Bencze

PP30252. If En (x) =nP

k=0

xk

k! then limn→∞

n�

k=0k(ex−Ek(x))

n�

k=0(ex−Ek(x))

= x2 .

Mihaly Bencze

PP30253. 1). Prove that2nPk=1

tg kπ2n+1 tg

(k+1)π2n+1 = − (2n+ 1)

2). Compute3nPk=1

tg kπ3n+1 tg

(k+1)π3n+1 tg (k+2)π

3n+1

Mihaly Bencze

PP30254. Let ABCD be a convex quadrilateral inscribed in a circle withradius R, and denote r1, r2, r3, r4 the radii of incircles of triangles ABD,BDC, ADC, ABC. Prove that AB2 +AC2 +AD2 +BC2 +BD2 + CD2 ≥≥ 2 (R+ r1)

2 + 2 (R+ r2)2 + 2 (R+ r3)

2 + 2 (R+ r4)2 .

Mihaly Bencze


k=1

12k+1 < ln (n+ 1) < 1

3

nPk=1

(3k2+3k+1)(2k+1)

k(k+1)(2k2+2k+1).

Mihaly Bencze


cosnx(n+1)5

, when x ∈ [0, 2π] .

Mihaly Bencze


PP30257. Compute

1).∞Pn=0

1

Γ�

n+ pq

�

2).∞Pn=0

1

Γ�

n+ 1p

�

Γ�

1+ 1q

�

where p > q > 3 are given prime

Mihaly Bencze

PP30258. If λ ∈ (−∞, 0) ∪ (1,+∞) then

1 +�12

�2λ+�13

�2λ+ ...+

�1n

�2λ ≥ 1nλ−1

�π2

6 −∞Pk=0

Bk

nk+1

�λ

.

Mihaly Bencze

PP30259. In all triangle ABC holds1). (4R+ r)

�s2 + 48r2

�≥ 25s2r

2).�s2 + r2 + 4Rr

� �s2 + 24Rr2

�≥ 50s2Rr

Mihaly Bencze

PP30260. Denote pn the nth prime. Study the convergence of the sequence

xn =

��nP

k=1

1pk

��λ

− λ lnn where λ ∈ R, and {·} denote the fractional part.

Mihaly Bencze


Pha ≥ 791rhAhBhChD

6000r4+26hAhBhChD

20.P

rA ≥ 791rrArBrCrD750r4+52rArbrCrD

Mihaly Bencze

PP30262. Let f : [0, 1] → R be an increasing function. Prove that

n1R0

f (x) dx ≤nP

k=1

(k + 1)1R0

xkf (x) dx.

Mihaly Bencze


PP30263. If x > 0 and H (x) =∞Pn=0

xn

n!(n+1)!(n+2)! , then compute

limx→∞

xλH(x)

exp(µ√x)

where λ, µ > 0.

Mihaly Bencze

PP30264. Let p ≥ 13 be a prime. Determine all k ∈ N for whichp−12P

i=1

1ik

= AB , with (A,B) = 1 result that p divides A.

Mihaly Bencze

PP30265. If a1 = 2 and an+1 = 2an +p3 (a2n − 1) then compute

∞Pn=1

1an.

Mihaly Bencze


(sinx)2 sin�x2

�dx.

Mihaly Bencze

PP30267. Solve in Z the equation x(y+z)xn+1 + y(z+x)

yn+1 + z(x+y)zn+1 = 3.

Mihaly Bencze

PP30268. If x, a > 1 then∞Pk=0

xk

Γ(a+k) =(a−1)exx1−a(Γ(a−1)−Γ(a−1,x))

Γ(a) .

Mihaly Bencze

PP30269. Let f : [0, 1] → R be an integrable function such that1R0

f (x) dx = 0. Determine the best constants c1, c2 ≥ 0 such that

c1

�1R0

xf (x) dx

�2

≤1R0

f2 (x) dx ≤ c2

�1R0

xf (x) dx

�2

.

Mihaly Bencze

PP30270. If a > 1 thennQ

k=2

ln ak+1a+1

ln ak+1+12

≥ 2n(n+1) .

Mihaly Bencze


PP30271. Let be p ≥ 3 a prime. The numbers n, n+ 1, n+ 2 are p-healty ifand only if

�(n+ 2)

√p�+�n√p�− 2

�(n+ 1)

√p�= 1 where [·] denote the

integer part. What is the probability that three consecutive numbers to bep-healty?

Mihaly Bencze

PP30272. If n ≥ 3 then the equation xn + 2xn−1 + 3xn−2 + ...+ nx− 1 = 0

have a root xn ∈ (0, 1) . Compute∞Pn=1

1x2n.

Mihaly Bencze

PP30273. If ak ≥ 80, (k = 1, 2, ..., n) thennQ

k=1

π (a1a2...ak) ≥ (π (a1))n (π (a2))

n−1 ... (π (an−1))2 π (an) .

Mihaly Bencze

PP30274. In all triangle ABC holdstg πA

π2+A2 + tg πBπ2+2(A2+B2)

+ tg πCπ2+3(A2+B2+C2)

< 3.

Mihaly Bencze

PP30275. In all triangle ABC holdstg 2as

a2+4s2+ tg 2bs

42+2(a2+b2)+ tg cs

5s2−3r2−12Rr< 3.

Mihaly Bencze


1). tg rrar2+r2a

+ tg rr2arbr2ar

2b+2r2(r2a+r2

b)+ tg s2r

2rc(2s2−12Rr−3r2)< 3

2). tg rha

r2+h2a+ tg rh2

ahb

h2ah

2b+2r2(h2

a+h2b)

+ tg s2rhc(5s2−3r2−12Rr)

< 3

Mihaly Bencze


tg

�2R−r2R

sin2 A2

( 2R−r2R )

2+(sin A

2 )4

�+ tg

2R−r2R

sin2 B2

( 2R−r2R )

2+2

�

(sin A2 )

4+(sin B

2 )4�

!+

+tg

�2R(2R−r) sin2 C

232R2−8Rr+5r2−3s2

�< 3.

Mihaly Bencze


PP30278. In all triangle ABC holdstg s(s−a)

s2+(s−a)2+ tg s(s−b)

s2+2((s−a)2+(s−b)2)+ tg s(s−c)

4s2−6r2−24Rr< 3.

Mihaly Bencze

PP30279. In all triangle ABC holdstg (4R+r)ra

(4R+r)2+r2a+ tg (4R+r)rb

(4R+r)2+2(r2a+r2b)

+ tg (4R+r)rc4(4R+r)2−6s2

< 3.

Mihaly Bencze

PP30280. If xk > 0 (k = 1, 2, ..., n) then tg x1(x1+x2+...+xn)

(x1+x2+...+xn)2+x2

1

+

+tg x2(x1+x2+...+xn)

(x1+x2+...+xn)2+2(x2

1+x22)

+ ...+ tg xn(x1+x2+...+xn)

(x1+x2+...+xn)2+n(x2

1+x22+...+x2

n)< n.

Mihaly Bencze

PP30281. In all triangle ABC holds tg

�4R+r2R

cos2 A2

( 4R+r2R )

2+(cos A

2 )4

�+

+tg

4R+r2R

cos2 A2

( 4R+r2R )

2+2

�

(cos A2 )

4+(cos B

2 )4�

!+ tg

�4R(4R+r) cos2 C

2

5(4R+r)2−3s2

�< 3.

Mihaly Bencze


1). 2R−r2R + 3r2

s2+r2−8Rr≥ 4 3

q�r4R

�2

2). 4R+r2R + 3s2

s2+(4R+r)2≥ 3 3

q�s4R

�2

Mihaly Bencze

PP30283. If xk > 0 (k = 1, 2, ..., n) thennP

k=1

xk +n

n�

k=1

1xk

≥ (n+ 1) n

snQ

k=1

xk.

Mihaly Bencze

PP30284. In all triangle ABC holds1). s+ 6sRr

s2+r2+4Rr≥ 2 3

√4sRr

2). s(R+r)4R+r ≥ 3

√sr2

3). R+ r ≥ 3√s2r

Mihaly Bencze



k=1

1k+1

�e1+

12+...+ 1

k−c − k − 1

2

�< n

24(n+1) <nP

k=1

1k

�e1+

12+...+ 1

k−c − k − 1

2

�

when C is the Euler constant.

Mihaly Bencze


n

ln 2−

2n−1Pk=1

12n−1+k

!.

Mihaly Bencze

PP30287. Prove thatmP

n=1

nPk=0

(nk)2

n(2n−1k )

2 ≥ 4mm+1 .

Mihaly Bencze

PP30288. In all acute triangle ABC holds 2 <P �

π2 −A

�tgA < π2

2 .

Mihaly Bencze


π(π−A)2+π

2A2

A sinA ≤ 4π ≤P (π−A)2+A2

A sinA.

Mihaly Bencze


π(π−A)2−π

2A2

2π(π−A)2+π

2A2

≤ 1 + rR ≤P (π−A)2−A2

(π−A)2+A2.

Mihaly Bencze

PP30291. In all acute triangle ABC holds π ≤P (π −A) tgA2 ≤ π2

2 .

Mihaly Bencze

PP30292. In all triangle ABC holdsP(1 + sinA) arctg (sinA) ≤ πs

2R ≤P�π2 + sinA

�arctg (sinA) .

Mihaly Bencze

PP30293. In all acute triangle ABC holdsP(1 + cosA) arctg (cosA) ≤ π(R+r)

2R ≤P �

π2 + cosA

�arctg (cosA) .

Mihaly Bencze


PP30294. In all acute triangle ABC holds ln π−Aπ−B ≤

BRA

tg x2dx

x ≤ π2 ln

π−Aπ−B and

his permutations.

Mihaly Bencze

PP30295. Prove thatπ2

nPk=1

1�

π2+ 1

k(k+1)

�

arctg 1k(k+1)

≤ nn+1 ≤ π

2

nPk=1

1�

1+ 1k(k+1)

�

arctg 1k(k+1)

.

Mihaly Bencze

PP30296. If a1 = 1 and an+1 = 1 + 12

�1 + 1

n+1

�an for all n ≥ 1 then study

the convergence of the sequence bn = a1(n1)

+ a2(n2)

+ ...+ an(nn)

.

Mihaly Bencze

PP30297. If a1 = a2 = 1, a3 = 4 and an+3 = 2an+2 + 2an+1 − an for alln ≥ 1 then compute

1).nP

k=1

a3k 2).

n∞Pk=1

1ak

3).∞Pk=1

1a2k

Mihaly Bencze

PP30298. If a, b > 0 then determine all x ∈ R for which(1 + a2 − x)(1 + b2 − x) ≤ (1 + ab)2.

Mihaly Bencze

PP30299. If x, y ∈ (0, π2 ) then:

(sin2 x+sin2 y)sin2 + sin2 y ·(cos2 x+cos2 y)cos

2 x+cos2 y

(sinx)2 sin2 x·(sin y)2 sin2 y ·(cosx)2 cos2 x·(cos y)2 cos2 y≤ 4

Daniel Sitaru

PP30300. If a, b, c > 0; a+ b+ c = 3; 0 ≤ x ≤ 1 then:a�ba

�x+ b

�cb

�x+ c

�ac

�x+ b

�ab

�x+ c

�bc

�x+ a

�ca

�x ≤ 6

Daniel Sitaru

PP30301. Find: Ω =P∞

k=1

�P∞n=1n6=k

�1

n2−k2

��

Daniel Sitaru


PP30302. Find: Ω =R �P∞

n=1

�3n sinh3

�x3n

��dx

Daniel Sitaru

PP30303. If x, y, z, t > 0 then:

(xy + yz + zt+ tx)�

1x4 + 1

y4+ 1

z4+ 1

t4

�≥

�1x + 1

y + 1z + 1

t

�2

Daniel Sitaru

PP30304. If a, b, c, d > 0 then:(ab+ bc+ cd+ da)(a4 + b4 + c4 + d4) ≥ abcd(a+ b+ c+ d)2

Daniel Sitaru

PP30305. Let be: Ω(a, b, c) =P∞

n=1an2+bn+c

n! ; a, b, c > 0

Prove that: Ω(a, b, c) + Ω(b, c, a) + Ω(c, a, b) ≥ 3(4e− 1) 3√abc

Daniel Sitaru

PP30306. In ΔABC the following relationship holds:P

CyC(a,b,c) aq�

(b− c)2 + 4r2��(c− a)2 + 4r2

�≥ abc

Daniel Sitaru

PP30307. In acute ΔABC the following relationship holds:2sinA + ssinB + 2sinC + 2cosA + 2cosB + 2cosC > 9

Daniel Sitaru

PP30308. If a1, a2, ..., a8 ≥ 1 then: a41 + a42 + ...+ a48 ≤ (a1a2...a8)4 + 7

Daniel Sitaru

PP30309. Find: Ω = limn→∞ npΩn(a)− (a+ 1)!; a ∈ N ;

Ωn(a) =Pn

k=0(k2 − a2 + 1)(a+ k)!; n ∈ N

Daniel Sitaru

PP30310. Let be Ω(a) = limn→∞Pn

k=1

�1

3k sin3(3k sin a)

�Prove that if

a, b, c ∈ [0, π2 ] then: 4�bΩ(a) + cΩ(b) + aΩ(c)

�≤ 3(a2 + b2 + c2)

Daniel Sitaru


PP30311. If 0 < a ≤ b ≤ c then: 1(a−b+c)6

+ 1b6

≤ 1a6

+ 1c6

Daniel Sitaru

PP30312. If a, b > 0, a2 + b2 = 1 then: 1a + 1

b ≥ 2√2

Daniel Sitaru

PP30313. If x, y > 0; Ω(x, y) =P∞

n=12n2+(2x+2y+5)n+2xy+6x−y3n(n+y)(n+y+1)(n+y+2) then:

Ω(y, x) ≤ 19 3√xy

Daniel Sitaru

PP30314. If a, b, c, d > 0; a3 + b3 + d3 = 1 then: a+b+c+dabcd ≥ 16

Daniel Sitaru

PP30315. If a, b, c, > 0; a3 + b3 + c3 = 1 then: a+b+cabc ≥ 3 3

√9

Daniel Sitaru

PP30316. If a, b, c, d > 0; a2 + b2 + c2 + d2 = 1 then: a+b+c+dabcd ≥ 32

Daniel Sitaru

PP30317. If a, b, c > 0; a2 + b2 + c2 = 1 then: a+b+cabc ≥ 9

Daniel Sitaru

PP30318. If a, b, c > 0; x, y, z, t ∈ R then:(a+b+c)y2

a + (a+b+c)z2

b + (a+b+c)t2

c ≥ x(2y + 2z + 2t− x)

Daniel Sitaru

PP30319. If a, b > 0; x, y, z ∈ R then: (a+b)y2

a + (a+b)z2

b ≥ x(2y + 2z − x)

Daniel Sitaru

PP30320. In ΔABC the following relationship holds:3

qsinAsinB + 3

qsinBsinC + 3

qsinCsinA − 3

qsinAsinC − 3

qsinBsinA − 3

qsinCsinB < 1

Daniel Sitaru


PP30321. If ai > 0 (i = 1, 2, ..., n) and λ1,λ2, ...,λk > 0 such thatλ1 + λ2 + ...+ λk = 1 then

P λ1a2+λ2a3+...+λkak+1

a21≥

nPi=1

1ai

≥P 1λ1a1+λ2a2+...+λkak

.

Mihaly Bencze


P 1a2(b2+c2−a2)

≥ (s2+r2+4Rr)2

32s2R2r2(s2−r2−4Rr).

Mihaly Bencze

PP30323. Let be a1 = 2 and (n+ 2) an+1 = nan +�n2 + 2n+ 2

�n! for all

n ≥ 1. Compute∞Pk=1

1ak.

Mihaly Bencze


sin 2A ≥ s2+r2+4Rr2sr .

Mihaly Bencze


1).Q �

ctgA2

�a2 ≤�

2s(4R+r)s2−r2−4Rr

�2(s2−r2−4Rr)

2).Q

actgA2 ≤ (4r (4R+ r))

s2r

Mihaly Bencze


1).Q �

tgA2

�a3 ≤�s2(2R−3r)+r2(4R+r)

s2−3r2−6Rr

�2s(s2−3r2−6Rr)

2).Q

atgA2 ≤

�2s(s2(2R−3r)+r2(4R+r))

4R+r

� 4R+r3s

Mihaly Bencze



1).Q �

ctgA2

�a3 ≤�s2(2R+3r)−r(4R+r)2

s(s2−3r2−6Rr)

�2s(s2−3r2−6Rr)

2).Q

actgA2 ≤

�2r(s2(2R+3r)−r(4R+r)2)

s

� s3r

Mihaly Bencze


1).P

ctgA2 ≥ s

R−r 2).P

tgA2 ≥ s

4R+r

Mihaly Bencze

PP30329. Solve in Z the equation 1x(y+1) +

1(y+2)(z+3) =

1996 .

Mihaly Bencze


1).Q �

tgA2

�a2 ≤�

2s(R−r)s2−r2−4Rr

�2(s2−r2−4Rr)

2).Q

atgA2 ≤

�4s2(R−r)4R+r

� 4R+r2s

Mihaly Bencze


a ≤ min

�q2s(2R−r)

r ; (4R+ r)q

2s

�.

Mihaly Bencze


1).Q �

tgA2

�a ≤�2R−r

s

�2s2).

Qatg

A2 ≤

�2s(2R−r)4R+r

� 4R+r2s

Mihaly Bencze


1).Q �

ctgA2

�a ≤�4R+r

s

�2s

2).Q

actgA2 ≤

�2r(4R+r)

s

� sr

Mihaly Bencze



1).P 1

(a+1)(b+1)(a+b+1) ≤5√5−112Rr

2).P 1

(ra+1)(rb+1)(ra+rb+1) ≤(5

√5−11)(4R+r)

2s2r

3).P 1

(1+sin2 A2 )(1+sin2 B

2 )(1+sin2 A2+sin2 B

2 )≤ 4R(2R−r)(5

√5−11)

r2

4).P 1

(1+cos2 A2 )(1+cos2 B

2 )(1+cos2 A2+cos2 B

2 )≤ 4R(4R+r)(5

√5−11)

s2

Mihaly Bencze


1). 4R+r2 +

P r2arb+rc

≥r3�(4R+ r)2 − 2s2

�

2). 12 + r

P rarb(ra+rb)r2c

≥ 1s

p3 (s2 − 2r (4R+ r))

Mihaly Bencze

PP30336. If xi > 0 (i = 1, 2, ..., n) and k, p ∈ N∗ then

Pcyclic

xk1

x2+x3+...+xn+ 1

n−1

�nP

i=1xi

� 1p

≥ p+1n−1

p+1

snpp

nPi=1

xki .

Mihaly Bencze


Pma +

P m2a

mb+mc≥ 3

2

p2 (s2 − r2 − 4Rr).

Mihaly Bencze

PP30338. In all triangle ABC holdsP cos2 A

2

cos π−A4

cos B−C2

+P

cos A2 ≥

q6�4 + r

R

�.

Mihaly Bencze

PP30339. If ak > 0 (k = 1, 2, ..., n) thenP (5a1+a2)(a1+a2)

8a1+a2≥ 4

3

nPk=1

ak.


PP30340. If ak > 0 (k = 1, 2, ..., n) thenP (5a1+a2)(8a1+a2)

a1+a2≥ 27

nPk=1

ak.

Mihaly Bencze


PP30341. If ak > 0 (k = 1, 2, ..., n) andnQ

k=1

ak = 1 then

nPk=1

ak(1+a1)(1+a2)...(1+ak)

≥ 2n−12n .


PP30342. In triangle ABC, D ∈ (BC) , E ∈ (CA) , F ∈ (AB) andAD,BE,CF are Gergonne’s cevians. Prove that

r (4R+ r) +P

AD2 ≥P�b2(s−b)c2(s−c)

� 1a .

Mihaly Bencze

PP30343. If xi > 0 (i = 1, 2, ..., n) and k ∈ N, k ≥ 2 then

(k + 1)P

cyclic

√x2

x1+kx2≤ P

cyclic

��x21 − x1x2 + x22

�x2k+11 xk

2−22

� −1

2(k+1)2


PP30344. In triangle ABC, D ∈ (BC) , E ∈ (CA) , F ∈ (AB) andAD,BE,CF are the Gergonne’s cevians. Prove thatP

a ·AD2 = 2sr (2R+ 5r) .

Mihaly Bencze

PP30345. In all triangle ABC holdsP ma

(b+c)sa≥ s2+r2+4Rr

8sRr .

Mihaly Bencze

PP30346. In all triangle ABC holdsP (b+c)sa

ma≤ (s2+r2+4Rr)

2+8s2Rr

8s(s2+r2+2Rr).

Mihaly Bencze


k=1

sin

1

k2(k+1)2−1R0

e−t2dt

≤ n

n+1 .

Mihaly Bencze


PP30348. Let ABC be an acute triangle and M ∈ Int (ABC) . DenoteRa, Rb, Rc the circumradii of triangles BMC, CMA, AMB. Prove thatM ≡ H if and only if Ra = Rb = Rc where H is the orthocentrum of triangleABC.

Mihaly Bencze


k=1

cos

√

(k2+k)2−1R0

e−t2dt

≥ n

n+1 .

Mihaly Bencze

PP30350. If 0 < a ≤ b thenbRasin

�xR0

e−t2dt

�dx ≤ (b−a)(b+a)√

1+a2+√1+b2

.

Mihaly Bencze

PP30351. If 0 < a ≤ b thenbRax cos

�xR0

e−t2dt

�dx ≥ (b−a)(b+a)√

1+a2+√1+b2

.

Mihaly Bencze

PP30352. If x ∈�0, π2

�then 2 sin

tgxR0

e−t2dt

!sin

ctgxR0

e−t2dt

!≤ sin 2x.

Mihaly Bencze

PP30353. If x ∈�0, π2

�then 2 cos

tgxR0

e−t2dt

!cos

ctgxR0

e−t2dt

!≥ sin 2x.

Mihaly Bencze

PP30354. If x ∈�0, π2

�then cos2

tgxR0

e−t2dt

!+ cos2

ctgxR0

e−t2dt

!≥ 1.

Mihaly Bencze

PP30355. If x ∈�0, π2

�then sin2

tgxR0

e−t2dt

!+ sin2

ctgxR0

e−t2dt

!≤ 1.

Mihaly Bencze


PP30356. If x ∈ R thenp1 + sin4 x sin

sin2 xR0

e−t2dt

!+

√1 + cos4 x sin

cos2 xR0

e−t2dt

!≤ 1.

Mihaly Bencze

PP30357. If ak > 0 (k = 1, 2, ..., n) then

nPk=1

sin

�akR0

e−x2dx

�≤

nn�

k=1ak

�

n2+

�

n�

k=1ak

�2

.

Mihaly Bencze

PP30358. If ak > 0 (k = 1, 2, ..., n) thennP

k=1

cos

�akR0

e−x2dx

�≤ n2

�

n2+

�

n�

k=1ak

�2

.

Mihaly Bencze

PP30359. If ak > 0 (k = 1, 2, ..., n) thenP

cyclic

a1a2R0

e−x2dx ≤ π

2

nPk=1

ak


PP30360. Prove that

1).nP

k=1

tg

F 2kR

0

e−x2dx

!≤ FnFn+1

2).nP

k=1

tg

L2kR

0

e−x2dx

!≤ LnLn+1 − 2,

where Fk and Lk denote the kth Fibonacci respective Lucas numbers.

Mihaly Bencze

PP30361. If x > 0 then sin

�xR0

e−t2dt

�≤ x√

1+x2≤ x cos

�xR0

e−t2dt

�.

Mihaly Bencze



k=1

tg

1k(k+1)R0

e−x2dx

≤ n

n+1 .

Mihaly Bencze

PP30363. Prove that

1).nP

k=1

tg

FkR0

e−x2dx

!≤ Fn+2 − 1

2).nP

k=1

tg

LkR0

e−x2dx

!≤ Ln+2 − 3,

where Fk and Lk denote the kth Fibonacci respective Lucas numbers.

Mihaly Bencze


1).P 1√

ra

1√rbR0

e−x2dx ≤ π

q2r

1).P 1√

ha

1√hbR0

e−x2dx ≤ π√

r

Mihaly Bencze

PP30365. If ak > 0 (k = 1, 2, ..., n) then

Pcyclic

�a1R0

e−x2dx

�arctga2 ≤

nPk=1

arctg2ak.

Mihaly Bencze


�xR0

e−t2dt

�dx ≤ barctgb− aarctga− 1

2 ln1+b2

1+a2.

Mihaly Bencze


1).P√

a

√bR

0

e−x2dx < π

2

√6s

2).P√

ra

√rbR0

e−x2dx ≤ π

2

p3 (4R+ r)


3).P√

A

√BR

0

ex2dx < π

√3π2


PP30368. If x ∈ R then(1+2 cos2 x) cos4 x

1+sin2 x+

(1+2 sin2 x) sin4 x1+cos2 x

≤ 8 + 12 sin2 x cos2 x.


PP30369. Prove that

√3R

1√3

�xR0

e−t2dt

�dx ≤ 5π

6√3− 1

2 ln 3.

Mihaly Bencze

PP30370. Let ABCD be a tetrahedron and M ∈ Int (ABCD) . DenoteRa, Rb, Rc the radii of circumspheres of tetrahedron BCDM, CDAM, DABM,ABCM. Prove that M ≡ H if and only if Ra = Rb = Rc = Rd where H is theorthocentrum of tetrahedron ABCD.

Mihaly Bencze

PP30371. Let ABC be an acute triangle and H denote the orthocentrum.Prove that the circumradii of triangles AHB, BHC, CHA are equal with Rthe circumradii of triangle ABC.

Mihaly Bencze

PP30372. Find all n ∈ N for which 20162n−1 + 20172n + 20182n+1 is aperfect square.

Mihaly Bencze

PP30373. Let ABCD be a convex quadrilateral and G1, G2 the centroid oftriangles ABD and BCD. The parallel lines through a point M situated inthe plane of the triangle to the medians AA1, BB1, DD1

(A1 ∈ BD,B1 ∈ AD,D1 ∈ AB) and BB2, CC1, DD2

(B2 ∈ CD, C1 ∈ BD, D2 ∈ BC) intersect lines BD,DA,AB respectiveCD,DB,BC in points A2, B2, D2 respective B3, C2, D3. Prove thatA1A

22 +B1B

22 +D1D

22 +B1B

23 + C1C

22 +D1D

23 ≥ 3

4MG21 +

34MG2

2.

Mihaly Bencze


PP30374. Let ABC be a triangle such that

cos Aλ + cos B

λ + cos Cλ =

√λ2+1−1

λ . Determine all λ > 0 for which

max {A,B,C} > λ2πλ2+1

.

Mihaly Bencze

PP30375. Solve in Z the equation4�x2 + y2

�+ 4034 (x+ y)− 12xy − 4068285 = 0.

Mihaly Bencze

PP30376. Solve in Z the equationx+ y + z + (x− y)2 + (y − z)2 + (z − x)2 = xy + yz + zx.

Mihaly Bencze

PP30377. Let x be a real number such that xm�xn + xn−1 + ...+ x+ 1

�

and xn�xm + xm−1 + ...+ x+ 1

�are rational for some relatively prime

positive integers m and n. Prove that x is irrational.

Mihaly Bencze

PP30378. Let ABCD be a cyclic quadrilateral. Points E and F lie on thesides AB and BC, respectively, such that BFE∡ = λBDE. Determine allλ > 0 for which EF ·AE ·AD = AE · FC ·AD +AE2 · CD.

Mihaly Bencze

PP30379. In all triangle ABC holds 27π3 ≤ 1

ABC ≤ 27R2π3r

. (A refinement ofEuler’s inequality.)

Mihaly Bencze

PP30380. If xk ∈�0, π4

�(k = 1, 2, ..., n) then

nPk=1

ctg2xk ≥1−

n�

k=1cos2 xk

1−n�

k=1sin2 xk

.

Mihaly Bencze

PP30381. ComputeP

n,k≥0

2n+k

(22n+1)�

22k+1

� .

Mihaly Bencze


PP30382. Prove that∞Pn=1

7776n5−14256n4+9072n3−2412n2+216n−1n(6n−1)! = 6

�1− 1

e

�.

Mihaly Bencze

PP30383. Prove that the sequence FF

F11

1 + 1, FF

F22

2 + 1, ..., FFFnn

n + 1, ... andan arbitrary infinite increasing arithmetic sequence have aither infinitelymany terms in common or at most one term is common. Same question if wesubstitute Fn with Ln, when Fn and Ln denote the nth Fibonacci respectiveLucas numbers.

Mihaly Bencze

PP30384. Let 0 = a0 < a1 < ... < an ≤ an+1 such thatnP

k=1

aλk = 1 when

λ ≥ 1. Prove thata2λ−11

a2−a0+

a2λ−12

a3−a1+ ...+ a2λ−1

n

an+1−an−1≥ 1

anan+1.

Mihaly Bencze

PP30385. Let 0 = a0 < a1 < ... < an < an+1 = 1 such thatnP

k=1

aλk = 1 where

λ ≥ 1. Prove thatna2λ−1

1a1−a0

+(n−1)a2λ−1

2a3−a1

+ ...+2a2λ−1

n−1

an−an−2+ a2λ−1

n

an+1−an−1≥ n1+ 1

λ .

Mihaly Bencze

PP30386. If ak,λk > 0 (k = 1, 2, ..., n) , thenpλ1a21 + λ2a22 + ...+ λna2n+

+pλ1a22 + λ2a23 + ...+ λna21 + ...+

qλ1a2n + λ2a21 + ...+ λna2n−1 ≥

≥s

nPk=1

λk

�nP

k=1

ak

�(A generalization of problem J326 Mathematical

Reflections).

Mihaly Bencze

PP30387. If ak,λk > 0 (k = 1, 2, ..., n) and r ≥ 1 then

(λ1ar1 + λ2a

r2 + ...+ λna

rn)

1r + (λ1a

r2 + λ2a

r3 + ...+ λna

r1)

1r + ...

+�λ1a

rn + λ2a

r1 + ...+ λna

rn−1

� 1r ≥

�nP

k=1

λk

� 1r�

nPk=1

ak

�.

Mihaly Bencze


PP30388. If ak > 0 (k = 1, 2, ..., n) andnP

k=1

ak = n, then

Pcyclic

n−1

qan−11 + a2a3...an−1 ≤ n · 2

1n−1 .

Mihaly Bencze

PP30389. If a0 = 1 and an+1

�n2an + a2n + 1

�= an for all n ≥ 1 then

compute limn→∞

n�3− n3an

�.

Mihaly Bencze

PP30390. If a0 = 1 and an = an+1

�nkan + pa2n + 1

�for all n ≥ 1, where

k ∈ N∗, k ≥ 2 then limn→∞

nk+1an = k + 1.

Mihaly Bencze

PP30391. Solve in N the equation 1n

nPk=1

xk + n

snQ

k=1

xk = 2n.

Mihaly Bencze

PP30392. If a, b, c > 0 thenP bc

a+√a2+bc

≤ 12

P a2+bca+c .

Mihaly Bencze

PP30393. If ak > 0 (k = 1, 2, ..., n) thenP

n−1

qan−11 + a2...an ≤ 1

n−1

P an−11 +a2...ana2+...+an

+ n− 3 +nP

k=1

ak.

Mihaly Bencze

PP30394. If ak > 0 (k = 1, 2, ..., n) thenP

n−1

qan−11 + a2...an ≤ 1

n−1

P an−11 +a2...an

(a2+...+an)n−2 + (n− 2)

nPk=1

ak.

Mihaly Bencze

PP30395. If a, b, c, d > 0 and abcd+ (P

a)�P

a3�= 5 (

Pa)4 then

(P

abc) (P

a) ≤ abcd+ 3 (P

a)4 .

Mihaly Bencze


PP30396. If ak > 0 (k = 1, 2, ..., n) thenP

n−1

qan−11 + a2a3...an ≤ n

n−1

nPk=1

ak.

Mihaly Bencze

PP30397. If a, b, c, d > 0 thenP

a12 + 4 (abcd)3 ≥ 2P

(abc)4 .

Mihaly Bencze

PP30398. In all triangle ABC holdsPr�

1ra

+ 1hb

�2+�

1rb

+ 1hc

��1rc

+ 1ha

�≤ 3

r .

Mihaly Bencze

PP30399. If 0 = a0 < a1 < ... < an+1 such thata1a2 + a2a3 + ...+ an−1an = anan+1 thenn−1a23−a20

+ n−2a2n−a21

+ ...+ 2a2n−a2n−3

+ 1a2n+1−a2n−2

≥ (n−1)2

a21+a22+...+a2n−1.

Mihaly Bencze

PP30400. If xk > 0 (k = 1, 2, ..., n) then snP

k=1

x2k

! P

cyclic

qx21 + x22 + ...+ x2n−1

!+ 2

�1−

qn

n−1

� P1≤i<j≤n

xixj ≥

≥ (n− 1)nP

k=1

x2k.

Mihaly Bencze

PP30401. Each of the diagonals AD, BE, CF of the convex equiangularhexagon ABCDEF divides its area in half. Prove that�AB2 + CD2 + EF 2

� �AC2 + CE2 + FA2

�=

=�BC2 +DE2 + FA2

� �BD2 +DF 2 + FB2

�.

Mihaly Bencze

PP30402. Let ABC be a convex quadrilateral in which A∡ ≥ 60◦,D∡ ≥ 60◦ and E ∈ AB,F ∈ AC,M ∈ BD,N ∈ DC. Prove that10−8 cos A+D

2cos A−D

2BD ≤ 1

min{BE,EF,FC} + 1min{BM,MN,NC} .

Mihaly Bencze


PP30403. If a, b > 0 then�a+ 1

2

�2+�b+ 1

2

�2+ 23

2 ≥ 4 4

q4 (a+ b+ c)3.


PP30404. If a, b > 0 then (a+ 1)3 + (b+ 1)3 + 16 ≥ 4q(a+ b+ 2)3.



1). ch4x+ ch4y + 48 ≥ 8 4

q8 (ch2x+ ch2y)3

2). ch6x+ ch6y + 16 ≥ 4

q(ch2x+ ch2y)3


PP30406. Let ABCD be a convex quadrilateral such that A∡ = D∡ = 90◦.Let M,N be two points on segment AB. Lines through M and N tangent tothe circumcircle of triangles ADC and DEC intersect lines AB and BD inpoints M1,M2 respective N1, N2. Prove that AM1 + CN1 = BM2 +DN2.

Mihaly Bencze

PP30407. If x, y, z > 0 and x+ y > z, y + z > x, z + x > y then

min

�P (2z2+xy)2

2y2+2z2−x2 ;P (2z2+xy)

2

2z2+2x2−y2

�≥ 3

Px2.

Mihaly Bencze

PP30408. If ak > 0 (k = 1, 2, ..., n) then

Pcyclic

3

q2 (a1 + 3) (a2 + 3)

�a21 + 3

� �a22 + 3

�≥ 4

nPk=1

ak + 4n.

Mihaly Bencze

PP30409. Let be xk ∈ N (xi 6= xj) (i 6= j) (k = 1, 2, ..., n), n ≥ 5 andM = {xixjxkxr|1 ≤ i, j, k, r ≤ n} . Determine cardM.

Mihaly Bencze



1).P (8r2a+39)(8r2b+39)

(5r4a+34r2a+80)(5r4b+34r2b+80)

≤ 4R+r4s2r

2).P (8h2

a+39)(8h2b+39)

(5h4a+34h2

a+80)(5h4b+34h2

b+80)

≤ s2+r2+4Rr16s2r2

Mihaly Bencze


P (3r−ra)rbrn+1a (3r+rb)

≥ 0

2).P (3r−ha)hb

hn+1a (3r+hb)

≥ 0 for all n ∈ N

Mihaly Bencze

PP30412. Prove that exist infinitely many A ∈ M3 (Q) for which Ap = I3where p ≥ 5 is a prime.

Mihaly Bencze

PP30413. Solve in R the following inequation

8x2+395y4+34z2+80

+ 8y2+395z4+34x2+80

+ 8z2+395x4+34y2+80

≤ 12

�1x + 1

y + 1z

�.

Mihaly Bencze

PP30414. Determine all prime p such that (p+ 2)5 + (p+ 1)5 − 2p5 areperfect square.

Mihaly Bencze


arctg(3√3tg2x)

x2+1dx ≤ ln b2+1

a2+1for all 0 < a < b < π

4 .

Mihaly Bencze

PP30416. If x, y ∈�0, π4

�then�√

tg2x+√tg2y

�cosx cos y ≥

p3√3 sin (x+ y) .



PP30417. Determine all polynomials P ∈ R [x] for which1R0

(P (xm))1n dx =

1R0

(P (xm))1m dx, where m,n ∈ N∗

Mihaly Bencze


k=1

tg(2(k+1)π)Rtg(2kπ)

xdx

(1+x2)32 arctgx

< n2(n+1)π .

Mihaly Bencze

PP30419. If x, y ∈�0, π4

�then

2�x2 + y2

�≥ xarctg

�3√3tg2y

�+ yarctg

�3√3tg2x

�.

Mihaly Bencze

PP30420. In all triangle ABC holdsP 4R+r−rc

4R+r+2ra+rb≤ 1.

Mihaly Bencze

PP30421. Prove that 4(m−1)4m+1 +

mPn=2

1

nn�

k=2

k√

2k+1≤ m−1

2m +mP

n=2

1

(4n+1)n�

k=2

k√

2k+1

for all m ≥ 2,m ∈ N.

Mihaly Bencze


�x31 − 1

�4= x2 − 1�

x32 − 1�4

= x3 − 1−−−−−−−−�x3n − 1

�4= x1 − 1

.

Mihaly Bencze


1− sin 2x1 = 16�2−

√3�(cosx2)

6

1− sin 2x2 = 16�2−

√3�(cosx3)

6

−−−−−−−−−−−−−−−1− sin 2xn = 16

�2−

√3�(cosx1)

6

.

Mihaly Bencze


PP30424. In all tetrahedron ABCD holds 216√2R3 ≤

�P 1rA

��P 1r2A

�≤ 4

√2

r3.

Mihaly Bencze

PP30425. If x ∈�0, π2

�then�

x sinx+ (1 + sinx)�π2 − x

�� π2 − x

�cosx+ (1 + cosx)x

�≥

�π2

�2.

Mihaly Bencze

PP30426. Prove that�cos π

7 − cos 2π7 + cos 3π

7

�3= cos π

7 cos2π7 cos 3π

7 .

Mihaly Bencze

PP30427. In all tetrahedron ABCD holds32

√2R3

27 ≥ (P

rA)�P

r2A�≥ 32

√2r3.

Mihaly Bencze

PP30428. Determine all a, b ∈ R for which∞R0

�1

a+x2 − b cosx�

dxx = C when

C is the Euler constant.

Mihaly Bencze

PP30429. Determine all ak > 0 (k = 1, 2, ..., n) for which the function

f : (0,+∞) → R when f (x) =

�1n

nPk=1

axk

� 1x

is log− concave.

Mihaly Bencze

PP30430. Determine all z, v, w ∈ C for which |z| = |v| = |w| = 1 and��1 + z + v2 + w3�� ≥ 1;

��1 + v + w2 + z3�� ≥ 1;

��1 + w + z2 + v3�� ≥ 1.

Mihaly Bencze

PP30431. Let ABCD be a tetrahedron, denote r1, r2, r3, r4 the radii ofinspheres of tetrahedron GABC, GBCD, GCDA, GDAB where G is thecentroid of the given tetrahedron. Prove that16r3

R2

�1r1

+ 1r2

+ 1r3

+ 1r4

�≤ R2

r3(r1 + r2 + r3 + r4) .

Mihaly Bencze


PP30432. Denote P1, P2, ..., Pk, ... a sequence of convex polygons such thatfor all k ≥ 1 the edges of plygon Pk+1 dividet the sides of polygon Pk in sameraport. Prove that exist only one point in interior of all polygons.

Mihaly Bencze

PP30433. Determine all n, k ∈ N ∗ for which

tg nπ2k+1 tg

kπ2n+1 ≥

�π2 + 2tg (n−1)π

2k+1

��π2 + 2tg (k−1)π

2n+1

�.

Mihaly Bencze and Gyorgy Szollosy

PP30434. In all triangle ABC holdsQ(2λa+ b+ c) ≥ 2 (2λ+ 1)2 s

�s2 + r2 + 2Rr

�,

Q(2λra + rb + rc) ≥ 4 (2λ+ 1)2 s2R for all λ ≥ 1+

√5

2 .

Mihaly Bencze

PP30435. Let ABC be a triangle and m ≤ A,B,C ≤ M. Prove thatPtg

�πA2

2(A2+1)

�≤ (M+m)2

4mM .

Mihaly Bencze


tg�

π2(k2+1)

�< π2

6 .

Mihaly Bencze

PP30437. If 0 < a < b < 1 thenbRa

arctgxdxx > π

4 lnb2+1a2+1

.

Mihaly Bencze

PP30438. If b > a > 1 thenbRa

arctgxdxx < π

4 lnb2+1a2+1

.

Mihaly Bencze

PP30439. If b > a > 0 thenbRaarctgx · arctg 1

xdx < π4 ln

b2+1a2+1

.

Mihaly Bencze



k=1

1√k2+k−1

arctg√k2 + k − 1arctg 1√

k2+k−1< nπ

2(n+1)

Mihaly Bencze


A2 + 1�arctgAarctg 1

A < π2 .

Mihaly Bencze


arctgAarctg 1A

> 12π .

Mihaly Bencze

PP30443. If 0 < a < b then

(n+1)π2

bRa

�

x2

2+lnx

�n

dx

arctgxarctg 1x

>�b2

2 + ln b�n+1

−�a2

2 + ln a�n+1

for all n ∈ N∗.

Mihaly Bencze


tg πA2A+π < π <

Ptg π2A

2(πA+2) .

Mihaly Bencze

PP30445. In all acute triangle ABC holds4π

P Aπ−2A < 2sr

s2−(2R+r)2< π

P Aπ−2A .

Mihaly Bencze

PP30446. In all triangle ABC holds:P �A+ π

2

�arctgA > π2

2 >P �

A+ 2π

�arctgA.

Mihaly Bencze

PP30447. If 0 < a < b then π2 ln

2b+π2a+π <

bRa

arctgxx dx < π

2 lnπb+2πa+2 .

Mihaly Bencze

PP30448. If 0 < a < b then π4 ln

2b2+π2a2+π

<bRa

arctg(x2)dxx < π

2 lnπb2+2πa2+2

.

Mihaly Bencze



P Aπ−A ≤ 4R+r

s ≤P πA2(π−A)

2).P AB

(π−A)(π−B) ≤ 1 ≤P π2AB4(π−A)(π−B)

3). ABC(π−A)(π−B)(π−C) ≤

rs ≤ π3ABC

8(π−A)(π−B)(π−C)

Mihaly Bencze

PP30450. If 0 < a < b < 1 then π2 ln

2b+π2a+π ≤

bRa

arctgxx dx ≤ π

2 lnb+1a+1 .

Mihaly Bencze

PP30451. If 0 < a < b < 1 then π4 ln

2b2+π2a2+π

≤bRa

arctg(x2)dxx ≤ π ln b2+1

a2+1.

Mihaly Bencze


tg πA2A+π ≤ π ≤P tg πA

2(A+1) .

Mihaly Bencze

PP30453. If 0 < a < b < 1 then

13 ln

(2b+π)(π−a)(2a+π)(π−b) ≤

bRa

actgxtg x2

x2 dx ≤ π2

4 ln (b+1)(π−a)(a+1)(π−b) .

Mihaly Bencze


Ptg

λA2+(λ2−1)AA2+λA+λ2−1

≤ 2λ ≤P tgλ(λ2−1)A2+A

(λ2−1)A2+λA+1when λ = π

2 .

Mihaly Bencze


1). 2P 1

A − 6π ≤ s

r ≤ πP 1

A − 32). 4

π2

P �πA − 1

� �πB − 1

�≤ 4R+r

r ≤P�πA − 1

� �πB − 1

�

3). 8π3

�πA − 1

� �πB − 1

� �πC − 1

�≤ s

r ≤�πA − 1

� �πB − 1

� �πC − 1

�

Mihaly Bencze


�

λA−1+√

(λA−1)2+4A(λ2−1)(λ−A)��

λB−1+√

(λB−1)2+4B(λ2−1)(λ−B)�

4(λ2−1)2(λ−A)(λ−B)≤

≤ s2−r2−4Rrs2−(2R+r)2

≤


P�

λA+1−λ2+√

(λA+1−λ2)2+4A(λ2−1)(λ−A)��

λB+1−λ2+√

(λB+1−λ2)2+4B(λ2−1)(λ−B)�

4(λ−A)(λ−B) ,

where λ = π2 .

Mihaly Bencze


1).P λA−1+

√(λA−1)2+4A(λ2−1)(λ−A)2(λ2−1)(λ−A)

≤ 2srs2−(2R+r)2

≤

≤P λA+1−λ2+√

(λA+1−λ2)2+4A(λ2−1)(λ−A)2(λ−A) where λ = π

2 .

Mihaly Bencze

PP30458. Prove that the sequence xn = (−1)[3√n2+2017n+2018] where [·]

denote the integer part is not periodic.

Mihaly Bencze

PP30459. If a0 = a, a1 = b and a2 + b2 + c = kabc; an+2an = a2n+1 + c then

compute∞Pn=1

1a2n.

Mihaly Bencze

PP30460. In all acute triangle ABC holdsQ λA−1+

√(λA−1)2+4(λ2−1)(λ−A)2(λ2−1)(λ−A)

≤

≤ 2srs2−(2R+r)2

≤Q λA+1−λ2+√

(λA+1−λ2)2+4A(λ2−1)(λ−A)2(λ−A) where λ = π

2 .

Mihaly Bencze

PP30461. If a, b, c > 0 thenP a4+1

a3+1+P a4

b3+c3≥ 3

2 +P

a.

Mihaly Bencze


k=1

1

kn�

11n

+ 12n

+...+ 1(k−1)n

+(n!)r−1

knr + 1(k+1)n

+...+ 1nn

� ≥ 1 for

all −n+1n−1 ≤ r < 1 and n ≥ 2, n ∈ N.

Mihaly Bencze


PP30463. Determine all n, k ∈ N for which {n+ 1, n+ 2, ..., kn} can bepartitioned in four subsets A,B,C,D such the sum of element from A,B,C,Dare equals.

Mihaly Bencze

PP30464. Compute2πR0

sinx cos 2xdx(2−cos2 x)(2−cos2 2x)

dx.

Mihaly Bencze

PP30465. If λ = limn→∞

1n

nPk=1

2k−1√n2+8k2−8k+4

then compute

limn→∞

n

�λ− 1

n

nPk=1

2k−1√n2+8k2−8k+4

�.

Mihaly Bencze and Gyorgy Szollosy

PP30466. If a, b, c > 0 and abc = 1 thenP a3

a9+1≤ 3

2 .

Mihaly Bencze


(sinA)2k (cosA)2(n−k) ≤ 32(nk)

where

n ≥ 2, k ∈ {1, 2, ..., n− 1} .

Mihaly Bencze

PP30468. ComputenP

k=1

n�

1√π− (2n−1)!!

(2n)!!

qn+ 1

2

�.

Mihaly Bencze

PP30469. If a, b, c ≥ 1 thenP 1

a+1 ≥P 1√ab+1

+ 16

�Pr √ab−1

(1+a)(1+b)(1+√ab)

��√a−

√b��2

.

Mihaly Bencze

PP30470. If ai > 0 (i = 1, 2, ..., n) andnQ

i=1ak = 1 then

nPi=1

11+ai+a2i+...+aki

≥ nk+1 , where k ∈ N∗.

Mihaly Bencze



P�2

a3+1+ 1

b6+1

�≤ 3

2

��s2+r2+4Rr

4sRr

�2− 1

Rr

�.

Mihaly Bencze


2r3a+1

+ 1r6b+1

�≤ 3(s2−8Rr−2r2)

2s2r2.

Mihaly Bencze

PP30473. Determine all λ > 0 such thatλR0

�1x + 1

ln(λ−x)

�dx = C, where C

is the Euler constant.

Mihaly Bencze


k=1

1

1+(k(k+1))32+

nPk=1

11+k3(k+1)3

≤ 3n2(n+1) .

Mihaly Bencze


ln�A3 + 1

�+P

arctg�A3

�≤ 9π

2 .

Mihaly Bencze

PP30476. In all triangle ABC holds2P

ln�1 + r3a

�+P

arctg�r3a�≤ 9(4R+r)

2 .

Mihaly Bencze


2

1+(sin A2 )

6 + 1

1+(sin B2 )

12

�≤ 3(s2+r2−8Rr)

2r2.

Mihaly Bencze


2

1+(cos A2 )

6 + 1

1+(cos B2 )

12

�≤ 3

2

�1 +

�4R+r

s

�2�.

Mihaly Bencze



2 ln�

1+(cos A2 )

6�

+arctg�

(cos A2 )

6� ≥ 2

9

�1 +

�4R+r

s

�2�.

Mihaly Bencze

PP30480. If 0 < a < b then 2 ln b3+1a3+1

+ arctg b3−a3

1+b3a3≤ 9

2 (b− a) .

Mihaly Bencze


2 ln(1+h3a)+arctg(h3

a)≥ 2r

9 .

Mihaly Bencze


2 ln�

1+(sin A2 )

6�

+arctg�

(sin A2 )

6� ≥ 2(s2+r2−8Rr)

9r2.

Mihaly Bencze

PP30483. Solve on (0,+∞) the following system:2x2

1

x32+1

+x23

x64+1

=2x2

2

x33+1

+x24

x65+1

= ... = 2x2n

x31+1

+x22

x63+1

= 32 .

Mihaly Bencze

PP30484. If x, y ≥ 0 then�2 ln

�x3 + 1

�+ arctg

�y3��

2 ln�y3 + 1

�+ arctg

�x3

��≤ 81

4 xy.

Mihaly Bencze


3�x91 + x62 + x33 + 1

�= 2

�2x84 + x55 + 3x26

�

3�x92 + x63 + x34 + 1

�= 2

�2x85 + x56 + 3x27

�

−−−−−−−−−−−−−−−−−−3�x9n + x61 + x32 + 1

�= 2

�2x83 + x54 + 3x25

�.

Mihaly Bencze

PP30486. If a1 = 1 and (n+1)!an+1

= n!an

+ 2nn4+n2+1

for all n ≥ 1 then compute

limn→∞

1n3·n!

nPk=1

ak.

Mihaly Bencze and Lajos Longaver


PP30487. If x1 ≥ 1 and xn+1 = 1 + x1x2...xn for all n ≥ 1 then1x3n+ 1

(1−xn)3 + 1

(xn+1−1)3= 3

xn(1−xn)(xn+1−1) for all n ≥ 1.

Mihaly Bencze

PP30488. If x1 ≥ 1 and xn+1 = 1 + x1x2...xn for all n ≥ 1 then compute

limn→∞

n

�2x1

−nP

k=1

1xk

�.

Mihaly Bencze

PP30489. If x ∈�0, π2

�then�

(sinx)8 + (cosx)8 + 1��

1(sinx)4

+ 1(cosx)4

+ 1�2

≥ 7298 .

Mihaly Bencze

PP30490. Denote x1 and x2 the roots of the equation(2m+ 1)x2 −

�m2 − 1

�x−m2 − 2m− 2 = 0. Determine all m, α ∈ R for

which 1|x1| ≤ sinα and 1

|x2| ≤ cosα.

Mihaly Bencze and Lajos Longaver


k=1

(k+1)3

k(k+2)4≥ n

3(n+3) .

Mihaly Bencze

PP30492. If 0 < a ≤ b then

1). 3 ln b+3a+3 +

bRa

�x+1x+2

�3dx ≥ b− a

2). b− a+ 3 ln ba ≥

bRa

�x+2x+1

�3dx.

Mihaly Bencze

PP30493. Prove that 32arctg

2(b−a)4+ab +

bRa

�x2+2x2+3

�3dx ≥ b− a.

Mihaly Bencze

PP30494. Determine all x, y > 0 for which�x+1y+2

�3+�y+1x+2

�3≥ x

y+3 + yx+3 .

Mihaly Bencze



n

�1√e−�

nPk=1

1√n2+k

�n�.

Mihaly Bencze

PP30496. Solve the following system:

x21 + x1x2 + x22 = 3x3 − x4−−−−−−−−−−−x22 + x2x3 + x23 = 3x4 − x5x2n + xnx1 + x21 = 3x2 − x3

.

Mihaly Bencze

PP30497. Compute∞Pk=1

1

(4k−1)k2.

Mihaly Bencze

PP30498. Determine all a, b, c, d ∈ N for which all numbers2a + 3b + 5c + 7d; 2b + 3c + 5d + 7a; 2c + 3d + 5a + 7b; 2d + 3a + 5b + 7c aredivisible by 17.

Mihaly Bencze

PP30499. Determine all matrices Ak ∈ M2 (C) (k = 1, 2, ..., n) for which

P1≤i<j≤n

(AiAj)−1 =

P

1≤i<j≤nAiAj

!−1

.

Mihaly Bencze

PP30500. Prove thatmPk=1

sin πk(k+1)n ≥ m

m+1 sinπn where m,n ∈ N∗.

Mihaly Bencze

PP30501. Determine all xk ∈ C (k = 1, 2, ..., n) for whichnP

k=1

xk = 1,

nQk=1

xk = 1 and |xk| = 1 (k = 1, 2, ..., n) .

Mihaly Bencze

PP30502. In all acute triangle ABC holds sr�s2 − (2R+ r)2

�≤ R4.

Mihaly Bencze


PP30503. If a > 1 and G = (0,+∞),

x ∗ y = a loga

�1 +

�a

xa − 1

��a

ya − 1

��for all x, y ∈ G then (G, ∗) is abelian

group.

Mihaly Bencze


x = y + 2 {z}y = z + 3 {x}z = x+ 4 {y}

where {·}

denote the fractional part.

Mihaly Bencze


x2 = yz − 2 (2x+ z) {x}y2 = xz + {y}+ {z}z2 = xy + 2 (2y + z) {x}

where {·} denote the fractional part.

Mihaly Bencze

PP30506. Let ABC be a triangle and A1 ∈ (BC) , B1 ∈ (CA) , C1 ∈ (AB) .Determine the geometrical locus of points A1, B1, C1 for whichArea (A1B1C1) =

14Area (ABC) .

Mihaly Bencze


loga x+ loga+1 (y + 1) = loga+2

�z2 + 4x+ 4

�

loga y + loga+1 (z + 1) = loga+2

�x2 + 4y + 4

�

loga z + loga+1 (x+ 1) = loga+2

�y2 + 4z + 4

� where a > 1.

Mihaly Bencze

PP30508. If n ≥ 3, zk ∈ C (k = 1, 2, ..., n), zi 6= zj (i 6= j) then determine

all r ∈ N for which exist z ∈ C such that |z| = 1 andnP

k=1

|zr − zrk| ≥ n.

Mihaly Bencze

PP30509. Prove that exist a nonconstant geometrical progression formedby n natural numbers and which product is perfect n+ 1 power of a naturalnumber.

Mihaly Bencze


PP30510. Prove that exist a nonconstant arithmetical progression formedby n natural numbers and which product is a perfect n+ 1 power of anatural number.

Mihaly Bencze


52x+1 + (7y − 29) 5z + 3z2 − 17x = 152y+1 + (7z − 29) 5x + 3x2 − 17y = 152z+1 + (7x− 29) 5y + 3y2 − 17z = 1

.

Mihaly Bencze

PP30512. Prove that for all k ∈ N, k ≥ 2 exist points A,B,C on the graphicof the function f : (0,+∞) → R, f (x) = x2n+1 for which Area [ABC] = k2n.

Mihaly Bencze


n

e2

2 − n1+ 1

nR

−(1+ 1n)

x2n

1+ex dx

.

Mihaly Bencze

PP30514. If ak, bk, ck ∈ (0, 1) ∪ (1,+∞) (k = 1, 2, ..., n) then

3

snP

k=1

�logak bk

�3+ 3

snP

k=1

�logbk ck

�3+ 3

snP

k=1

�logck ak

�3 ≥ 3 3√n.

Mihaly Bencze

PP30515. Solve on R the equation(9x + 16x + 25x) (3−x + 4−x + 5−x) = 5

564 (3x + 4x + 5x)2 .

Mihaly Bencze

PP30516. Determine all ak ∈ (0,+∞) (k = 1, 2, ..., n) for which holds

[x] +n−1Pk=1

[x+ ak] = [anx] for all x ∈ R, where [·] denote the integer part.

Mihaly Bencze and Nicolae Papacu



nR0

dx

(1+x+x2+...+xk)(1+xn)where k ∈ {1, 2, ..., n} .

Mihaly Bencze

PP30518. If x, y, z > 0 andP x

y = 27 thenP x

√x√

x3+xy2+2y3≥

√32 .

Mihaly Bencze


2 + y3

3 + z4

4 = (z − 1) (x+ y)2 .

Mihaly Bencze

PP30520. If mk ∈ N∗ (k = 1, 2, ..., n) then

�n

nQk=1

mk

�sin π

n�

k=1mk

≥nP

k=1

mk sinπmk

.

Mihaly Bencze


k=1

sin πmk

sin πk

≥ 2nm(n+1) where n,m ∈ N∗.

Mihaly Bencze


1n

nPk=1

π2R0

cosx√k+sin 2x

dx.

Mihaly Bencze

PP30523. Let ABC be a triangle in which1

BC+λAB + 1BC+λAC = 1

λAB−BC + 1λAC−BC .

1). Determine all λ ∈ R for which ABC is rectangle2). Determine all λ ∈ R for which ABC is isoscele

Mihaly Bencze and Sandor Tamas

PP30524. If xk > 0 (k = 1, 2, ..., n) andnQ

k=1

xk ≤ 2n thennP

k=1

11+xk

≥ 1.

Mihaly Bencze


PP30525. If the equation x2 − ax+ b = 0 have integer roots, where a, b ∈ Rthen b3 + 2a3 − 6ab+ 9 and b4 + 4a4 − 16a2b+ 8b2 + 16 are compositenumbers.

Mihaly Bencze

PP30526. Let be Mn =nx ∈ R| [x]{x} = n ∈ N∗

owhere [·] and {·} denote the

integer respective fraction part. Compute∞Pk=1

1x21+x2

2+...+x2nk

where

x1, x2, ..., xnk∈ Mn.

Mihaly Bencze

PP30527. Determine all X ∈ M3 (R) for which xn + xn+1 =

2 2 20 2 20 0 2

where n ∈ N.

Mihaly Bencze

PP30528. If x1 ≥ 1 and xn+1 = 1 +nQ

k=1

xk for all n ≥ 1 then compute

∞Pn=1

1x2n.

Mihaly Bencze

PP30529. In all triangle ABC holdsP (ra+rb)

5+5r5c(ra+rb+4rc)r4c

≥ 32 + 32Rr .

Mihaly Bencze

PP30530. In all triangle ABC holdsP r10a +5r5

br5c

(r2a+2rbrc)r5br5c

≥ 3 + (4R+r)3

s2r− 12R

r .

Mihaly Bencze

PP30531. In all triangle ABC holdsP (s−a)10+5(s−b)5(s−c)5

((s−a)2+2(s−b)(s−c))(s−b)4(s−c)4≥ 3 + s2−12Rr

r2.

Mihaly Bencze


PP30532. In all triangle ABC holdsP (ha+hb)

5+160h5c

(ha+hb+4hc)h4c≥ 40 +

4(s2+r2)Rr .

Mihaly Bencze

PP30533. In all triangle ABC holdsP (a+b)5+160c5

(a+b+4c)c4≥ 40 +

4(s2+r2)Rr .

Mihaly Bencze

PP30534. In all triangle ABC holdsP a10+5b5c5

(a2+2bc)b4c4≥ s2−3r2

2Rr .

Mihaly Bencze

PP30535. In all triangle ABC holdsP (sin2 A

2+sin2 B

2 )5+160 sin10 C

2

(sin2 A2+sin2 B

2+2 sin2 C

2 ) sin8 C

2

≥ 96 +8((2R−r)2(s2+r2−8Rr)−2Rr2)

Rr2.

Mihaly Bencze

PP30536. In all triangle ABC holdsP (cos2 A

2+cos2 B

2 )5+160 cos10 C

2

(cos2 A2+cos2 B

2+2 cos2 C

2 ) cos8C2

≥ 96 +4((4R+r)3+s2(2R+r))

Rs2.

Mihaly Bencze

PP30537. In all acute triangle ABC holdsP (cosA+cosB)5+160 cos5 C

(cosA+cosB+2 cosC) cos4 C≥ 96 +

16((2R+r)2+s2r−2R(s2+r2+2Rr))R(s2−(2R+r)2)

.

Mihaly Bencze

PP30538. In all acute triangle ABC holdsP cos10 A+5 cos5 B cos5 C

(cos2 A+cosB cosC) cos4 B cos4 C≥ 9 +

3(s2+r2−4R2)((4R+r)2−3s2)16R4 .

Mihaly Bencze

PP30539. In all triangle ABC holdsP (ha+hb)

5+160h5c

(ha+hb+2hc)h4c≥ 80 +

8(s2+r2)Rr .

Mihaly Bencze

PP30540. In all triangle ABC holdsP (ra+rb)

5+160r5c(ra+rb+2rc)

≥ 80 + 32Rr .

Mihaly Bencze

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