Octogon Mathematical Magazine, Vol. 23, No.1, April 2015 … · 2016-05-29 · x,y∈ R for which 2...

204 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015

Proposed problems

PP24615. 24 Let a, b, c, d,α be a numbers from (0, 1) interval,f : (0, 1) → R a convex and decreasing function. Ten are true the followinginequality f

�1− a3

�+ f�1− b3

�+ f�1− c3

�+ f�1− d3

�≥

≥ f�1− a2b

�+ f�1− b2c

�+ f�1− c2a

�+ f�1− d2a

�.

Marius Dragan

PP24616. Find the best k ∈ Z such that�√n+

√n+ 1 +

√n+ 2 +

√n+ 3

�=�√

16n+ k�for each n ∈ N.

Marius Dragan

PP24617. Let x, y, z be an interior numbers. Find all the rest dividing thenumber x3yz + xy3z + xyz3 by 11.

Marius Dragan

PP24618. Let a, b, c be the positive numbers such that a = b+ c. then istrue the following inequality:�1 + 1

a2

�a2 �1− 1

a2−b2−c2

�≤�1 + 1

b2

�b2 �1 + 1

c2

�c2.

Marius Dragan

PP24619. Let a, b, c, d be a positive real numbers such that a ≤ b ≤ c. Thenis true the following inequality:�a2 + 14

� �b2 + 14

� �c2 + 14

�≥ 26 (3a+ 2b+ c+ 1)2 .

Marius Dragan

PP24620. Let A,B be two square matrices from C, a, b, c, d a strict naturalnumbers such that a < b, q the quotiend of dividing of b to a, such thatcq �= d and AaBc = AbBd = In. Then it exist a strictly integer number suchthat Bu = In.

Marius Dragan

24Solution should be mailed to editor until 30.12.2018. 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 205

PP24621. Let n be a positive integer. Prov that the equation

22n �

x8 + y8�2n

= z2 + t2 + w2 has the integer solutions.

Marius Dragan

PP24622. We consider the complex number Z such that |z| = 1. Re z ≥ 0,Im z > 0. If we denote x = π

2 arg z and we have��z[x] − 1��+��z[3x] − z[2x]

�� =��z[2x] − z[x]

��+��z[3x] − 1

�� . Then√2 ≥��z[x] − 1

�� ·��z[3x] − z[2x]

�� ·��z[2x] − z[x]

�� ·��z[3x] − 1

�� ≥ 1.

Marius Dragan

PP24623. Let be f : R → (0,+∞) be an increasing concave function. Prove

thatn�

k=1

f�

1(2k+1)

√2k−1

�≤ nf

�1n

�.

Mihaly Bencze

PP24624. Compute� � � dxdydz

(242x5−(y−1)5−(z+1)5)(242y5−(z−1)5−(x+1)5)(242z5−(x−1)5−(y+1)5)where

x, y, z > 12 .

Mihaly Bencze and Gyorgy Szollosy

PP24625. Solve in C the following system�x2 − 9

�(y − 2) (z + y) +

�y2 − 36

�(z − 4) (x+ 8) =

=�y9 − 9

�(z − 2) (x+ 4) +

�z2 − 36

�(x− 4) (y + 8) =

=�z9 − 9

�(x− 2) (y + 4) +

�x2 − 36

�(y − 4) (z + 8) = −153.

Mihaly Bencze and Ferenc Olosz

PP24626. If ak > 0 (k = 1, 2, ..., n) then

n+��a1

a2

�n≥� a1a2...an−1

an−1n

+� an−1

1a2a3...an

.

Mihaly Bencze

PP24627. If x0 ∈ R and xn+1

�2x2n − xn + 3

�= 3x2n − xn + 2 for all n ∈ N,

then compute limn→∞

n (1− xn) .

Mihaly Bencze

PP24628. Compute F (m) =∞�n=0

�n!m!(n+m)!(2n)!(2m)!

�2.

Mihaly Bencze


PP24629. If a ≥ 1 and f, g : [0, a] → [0, 1] are two increasing functions and

g is continuous, then compute limn→∞

n

�ag (0)−

a�ag (fn (x)) dx

�.

Mihaly Bencze

PP24630. If a, b, c > 0, thenab4 + b3c2 + a5c+ 3a2b2c ≥ c

�a2b2 + b3c+ a3c

�+

3√a2b2c2

�a3 + abc+ b3

�.

Mihaly Bencze

PP24631. If ak > 0 (k = 1, 2, ..., n) , then

3n�

k=1

a2k +�

cyclic

a21(a1+a2)4(a1+a3)

4+a22(a2+a1)4(a2+a3)

4+a23(a3+a1)4(a3+a2)

2

(a1+a2)4(a2+a3)

4(a3+a1)4 ≥ 3n

2 .

Mihaly Bencze

PP24632. If x > 1 then π2

6 > 1336 +

∞�n=2

�1− x−1

xn+1−1− (n−1)x

n+1

�2.

Mihaly Bencze

PP24633. If zk ∈ C∗ (k = 1, 2, ..., n) , thenn�

k=1

��zk + 1zk

��8≥ 16n

�1 + 2

n

n�k=1

Re�z2k��2

.

Mihaly Bencze

PP24634. Determine all x, y, z ∈ C for which3An = 2n (A+ I3) + (−1)n (2I3 −A) for all n ∈ N∗, where

A =

0 x yy 0 zz x 0

.

Mihaly Bencze

PP24635. Prove that 14

�π2

6 − 1�<

∞�n=2

�1�0

xndxx2+1

�2

< π2

24 .

Mihaly Bencze


PP24636. If a, b, c, d ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} , then one of equationsx3 − abcx2 + bcdx− cda = 0,x3 − bcdx2 + cdax− dab = 0,x3 − cdax2 + dabx− abc = 0,x3 − dabx2 + abcx− bcd = 0,have real solutions.

Mihaly Bencze

PP24637. In all scalene triangle ABC holds

1).� r2a

(ra−rb)2(ra−rc)

2 ≥ 1(4R+r)2−2s2

2).� AI4

(AI2−BI2)2(AI2−CI2)≥ 1

s2+r2−8Rr

Mihaly Bencze

PP24638. Let ABC be a triangle. Determine all n, p, k ∈ N for which��

a�

ama

�n+�

�

a�

ama

�p+�

�

a�

ama

�k≥

≥�

aamn

a+bmpb+cmk

c+

�

a

ampa+bmk

b+cmnc+

�

aamk

a+bmnb +cmp

c.

Mihaly Bencze

PP24639. If zk, wk ∈ C (k = 1, 2, ..., n) , n ≥ 2 and|z1| = |z2| = ... = |zn| = |w1| = |w2| = ... = |wn| ∈ (0, 1] then determine all

x, y ∈ R for which 2

��n�

k=1

zx+y2

k −n�

k=1

wx+y2

k

�� ≤n�

k=1

��zxk − wyk

��+n�

k=1

��zyk − wxk

�� .

Mihaly Bencze

PP24640. Let be (xn)n≥1 an increasing positive real numbers sequence and

λ = limn→∞

xn. Compute limn→∞

n

�λ− λ−xn

ln λxn

�.

Mihaly Bencze

PP24641. Let ABC be a triangle. Determine all λ ≥ 1 for which�

(sinA)λ

�

(cos Aλ )

λ ≥�

sinA�

cos Aλ

.

Mihaly Bencze


PP24642. Solve in R the equationn�

k=1

�x+kk+1

�= 2m, when [·] denote the

integer part.

Mihaly Bencze

PP24643. If ak > 0 (k = 1, 2, ..., n) , then

n

n�

k=1

1ak

n�

k=1

1ak

≥n�

k=1

a1akk .

Mihaly Bencze

PP24644. Determine all p ∈ N for which (pn)! =n�

k=1

� pk

pk−1

�pn−k

.

Mihaly Bencze

PP24645. Solve in R the following system:

4 arctan 1−x11+x2

= π − 4 (arctanx3)2

4 arctan 1−x21+x3

= π − 4 (arctanx4)2

−−−−−−−−−−−−−−−−4 arctan 1−xn

1+x1= π − 4 (arctanx2)

2

.

Mihaly Bencze

PP24646. Solve in C the following system:x61 − 8x52 + 18x43 − 6x34 − 12x25 + 2x6 = x62 − 8x53 + 18x44 − 6x35 − 12x26 + 2x7 =... = x6n − 8x51 + 18x42 − 6x33 − 12x24 + 2x5 = −1.

Mihaly Bencze

PP24647. Denote A (n) =Lpn−Ln

p when p is a prime and Ln denote the nth

Lucas numbers. Compute∞�n=1

11+A2(n)

.

Mihaly Bencze

PP24648. Solve in N the equationn�

k=1

(kr + 1) = pr, where r ∈ Z.

Mihaly Bencze


PP24649. In all triangle ABC holds 35 ≤� ln(ma)

ln(mam2bmc)

< 1.

Mihaly Bencze

PP24650. If S (n, p) =n�

k=1

kp then

t�r=1

�2t+12r

�S (n, 2r) = 1

2

�(n+ 1)2t+1 + n2t+1 − 2n− 1

�.

Mihaly Bencze

PP24651. Compute limn→∞

n

�lnΓ (x+ 1) + lnΓ (1− x)−

n�k=1

ζ (2k) x2k

k

�,

when x ∈ (−1, 1) .

Mihaly Bencze

PP24652. If x ∈�0, 12�, then

cosx (sinx)2n−1 − sinx (cosx)2n−1 ≥ 2x��

2x2�n−1 −

�1− 2x2

�n−1�for all

n ∈ N∗.

Mihaly Bencze

PP24653. In all triangle ABC holds256sRr ≤ (a+ 2s) (b+ 2s) (c+ 2s) ≤ 4sR(R+2r)

r2.

Mihaly Bencze


n

�ζ (λ)−

n�i=1

n�j=1

(i−1)!(j−1)!(i+j)!

�.

Mihaly Bencze

PP24655. In all triangle ABC holds�

cos2�

tgA2tgB

2

1−tgA2tgB

2

�> 3

2 .

Mihaly Bencze

PP24656. In all triangle ABC holds 272 ≤� 1

tgA2tgB

2 (1−tgA2tgB

2 )≤ s2

2r2.

Mihaly Bencze



n

�3e4

256 −n�

k=2

4e2

�1 + 1

k

�2k+1 (k−1)(k+1)(2k−1)(2k+1)

�.

Mihaly Bencze

PP24658. If a, b, c ∈ C∗ then 6561�|a|2 + |b|2 + |c|2

�−1·

·�

1|a2| +

1|b2| +

1|c2|

�−1· |abc|2 ≤

�3 |a+ b+ c|2 +� |2a− b− c|2

�·

·�3 |ab+ bc+ ca|2 +� |2bc− ac− ab|2

�.

Mihaly Bencze

PP24659. Determine all λ ∈ R for which

��n�

k=0

(−1)kλ

�� ≤�nλ�when [·]

denote the integer part.

Mihaly Bencze

PP24660. If a, b, c > 0 then� √

a+√b

a+b+c+3√ab

≥ 1√a+b+c

.

Mihaly Bencze

PP24661. In all triangle ABC holds� (tgA

2 )x

(tgA2 )

y+(tgB

2 )y ≥ 3

2

�1√3

�x−yfor all

x > y ≥ 0.

Mihaly Bencze

PP24662. Compute1�0

lnk�x

1k − (1− x)

1k

�dx when k ∈ N∗

Ovidiu Furdui and Mihaly Bencze

PP24663. Compute

limn→∞

n

�arctan

�tan�

π√2

�√2− 1

�cth�

π√2

�√2 + 1

�− π

8 −n�

k=1

arctan�

1k2+1

��.

Mihaly Bencze


n

�1−

n�k=1

kk−1e−k

k!

�.

Mihaly Bencze


PP24665. If x > y > 0 and a > 1 then�a−1a

�2loga

xy < x−y

xy .

Mihaly Bencze

PP24666. If P2 = 4�2−

√2, P3 = 8

�2−�2 +

√2,

P4 = 16

�2−�2 +�2 +

√2 etc, then compute

limn→∞

n�π3

8 − 4n (Pn − Pn−1)�.

Mihaly Bencze

PP24667. In all triangle ABC holds�� a+b+c

−a+b+c

� 2λa−a+b+c ≥ s2

3λ−1r2for all

λ ≥ 1.

Mihaly Bencze

PP24668. If a > 1 thenn(an+1−1)

n+1 > n(n+1)4

�a−1a

�n+

n�k=1

ak−1k .

Mihaly Bencze

PP24669. Compute∞�0

dxx+ex+e2x

.

Mihaly Bencze

PP24670. Compute

√3�

0

ln(1+√3x)dx

cos2 x.

Mihaly Bencze


n�1 + (n+1)n+1

nn − (n+2)n+2

(n+1)n+1

�.

Mihaly Bencze

PP24672. Compute Sk =1�0

dxxk(1+lnx)

.

Mihaly Bencze


PP24673. Compute S (k, p) =n�

i=1

�nk

pi

�where [·] denote the integer part.

Mihaly Bencze

PP24674. Compute

√22�0

(1−x2) lnxdx√x(1−2x2)

.

Mihaly Bencze

PP24675. If x0 = 0 and xn+1 = λxn + xn−1 for all n ≥ 1, then determine allλ ∈ R for which xn =

�n1

�+ λ�n3

�+ λ2

�n5

�+ λ3

�n7

�+ ...

Mihaly Bencze

PP24676. Compute1�0

ln�

1+x1+

√1−x2

�√1−x2

dx.

Mihaly Bencze

PP24677. If P0 (x) = 1 and Pn (x) =dn

2nn!·dxn

�x2 − 1

�nthen

n�k=1

cos2 (kx)P 2n−k (cosx) ≥ n2P 2

n (cosx) for all n,m ∈ N, m < n and all

x ∈ R.

Mihaly Bencze

PP24678. Compute∞�n=1

nn

(n!en)2.

Mihaly Bencze

PP24679. Computeb�a

ln(b+1−x)dxln(b+1−x)+ln(x−a+1) where 0 < a < b.

Mihaly Bencze

PP24680. Determine all λ ∈ R for which limn→∞

n�k=0

1n+λk > 1

λ .

Mihaly Bencze


PP24681. Let a, b, c ∈ N∗ and x1 = a, xn+1 = xn + bn+ c. Compute∞�n=0

1xn+1

.

Mihaly Bencze

PP24682. If λ = limn→∞

nlnn

n−1�k=1

1k(n−k) , then compute

limn→∞

n

�λ− n

lnn

n−1�k=1

1k(n−k)

�.

Mihaly Bencze

PP24683. In all acute triangle ABC holds�s2+r2+Rr

2sr

�2+�

s2+r2−4R2

s2−(2R+r)2

�2≥ 12 + 4R

r + 8R(R+r)

s2−(2R+r)2.

Mihaly Bencze

PP24684. Let p ≥ 3 be a prime. Determine all r ∈ N for whichp−1�k=0

�kp−1k

�n

is divisible by p if n is even and not divisible by p if n is odd.

Mihaly Bencze

PP24685. Determine all p ∈ N for whichn�

k=1

(−1)k−1 kp = (−1)n Tm when

Tm is a triangular number.

Mihaly Bencze

PP24686. In all triangle ABC holds� (a+b)2

c2≤ 4sR

r .

Mihaly Bencze

PP24687. In all triangle ABC holds� cos 2A+cos 2B

sinC ≤ s2+r2

sr + 2�2− 3

√3�.

Mihaly Bencze

PP24688. Determine all prime p for which L22n ≡ (p− 1)mod p and

L22n+1 = − (p− 1)mod p when Ln is the nth Lucas number.

Mihaly Bencze


PP24689. If z, w �= 1 and zn = wn = 1, then��(n− 2) z2 − 2 (n− 1)w + n��+��(n− 2)w2 − 2 (n− 1) z + n

�� ≤≤ (n−1)(2n−1)

6

�|z − 1|3 + |w − 1|3

�for all n ∈ N, n ≥ 2.

Mihaly Bencze and Jose Luis Dıaz-Barrero


sinx1 + cosx2 = sinx2 + cosx3 = ... = sinxn = cosx1 =12

�3 +

√5.

Mihaly Bencze

PP24691. If sinx+ sin y = k (cos y − cosx)2 for x, y ∈ R, then determine allk ∈ R for which sin

�k2x�+ sin

�k2y�= 0.

Mihaly Bencze

PP24692. Determine all a, b, c, d ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} for

�abcd�bcda

+�bcda�cdab

+�cdab�dabc

+�dabc�abcd

is a perfect square.

Mihaly Bencze

PP24693. Compute Ak,p =∞�n=1

(−1)n−1 nk

pn .

Mihaly Bencze

PP24694. Determine all xk ∈ R (k = 1, ..., n) for which

[x1]3x2+{x3} + {x2}

3x3+[x4]≥ 4

15[x2]

3x3+{x4} + {x3}3x4+[x5]

≥ 415

−−−−−−−−−−−−[xn]

3x1+{x2} + {x1}3x2+[x3]

≥ 415

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

denote the fractional part.


PP24695. Compute� (cosx−sin2 x) cosxdx

sinx cosx(sinx−cosx)+(1+cos x) cosx .

Mihaly Bencze


PP24696. If Φ = 1+√5

2 then determine all a, b ∈ R for which the series∞�n=1

1na|sin(nbπΦ)| converges.

Mihaly Bencze

PP24697. Find all polynomials P for whichP (x− 2)P (x− 1)P (x) = P

�x3�for all x ∈ R.

Mihaly Bencze

PP24698. Determine all a, b, c, d ∈ N for which F2an+2b − 2cn− 2d isdivisible by 5 for all n ∈ N , when Fk denote the kth Fibonacci number.

Mihaly Bencze

PP24699. Denote xa, xb, xc the distance of orthocenter H to the sides

BC,CA and AB respectively. Prove that xa + xb + xc ≤ 6

�9s2R3r√

3.

Mihaly Bencze


(p (n− 1) + k)xn−1 if |x| < 1 and p, k ∈ Z.

Mihaly Bencze

PP24701. Find the volume of the region bounded by the surfacex2n+1 + y2n+1 + z2n+1 = (xyz)n when n ∈ N∗.

Mihaly Bencze

PP24702. Determine all k,m ∈ N for which the sum of (m+ k) (kn+ 1)consecutive terms of the Fibonacci sequence is divisible by the(kn+m+ 1)st Lucas number.

Mihaly Bencze

PP24703. In all triangle ABC holds:2(s2−r2−Rr)s2+r2+2Rr

+ 3 3√4Rsr4s ≥ 2.

Mihaly Bencze


PP24704. Evaluate the sumn�

k=0

(−1)k�nk

�3�3n3k

�−1.

Mihaly Bencze

PP24705. If x ∈�0, π2�then 1 < 1

sin2 x cos2 x+ 1

ln(sin2 x)+ 1

ln(cos2)x< 2.

Mihaly Bencze

PP24706. If x ∈ (0, 1) then n2 < xn−1

(x−1)xn +n�

k=1

1ln(1−xk)

< n.

Mihaly Bencze

PP24707. If x0 = 0, xk > 0 (k = 1, 2, ..., n) andn�

k=1

xk = 1. Determine all

λ ∈ R for which 1 ≤n�

k=1

xk

(1+x0+x1+...+xk)λ(xk+xk+1+...+xn)

λ ≤ π2 .

Mihaly Bencze

PP24708. If ak ∈ (0, 1] (k = 1, 2, ..., n) , then�

cyclic

1a21+a22

≥ 2[n2 ]+1

[n2 ]+1+n�

k=1ak

.

Mihaly Bencze

PP24709. Let be 1kλ

+ 1(k+1)λ

+ ...+ 1nλ = p

q when k, n,λ ∈ N∗. Determine

all λ ∈ N∗ for which p is odd.

Mihaly Bencze

PP24710. If x, y, z > 0 then

(x+ y + z)�

1x + 1

y + 1z

�≥ 3 + 2 (x+ y + z)

�(xxyyzz)

1x+y+z

xyz .

Mihaly Bencze

PP24711. Computeπ�0

sin2n+1 x sin (kx) dx, when n, k ∈ N.

Mihaly Bencze


PP24712. Let Z × Z × Z → R be a function which satisfies f (0, 0, 0) = 1and f (n, p, k) + f (n+ 1, p, k) + f (n, p+ 1, k) + f (n, p, k + 1) +f (n+ 1, p+ 1, k) + f (n+ 1, p, k + 1)++f (n, p+ 1, k + 1) + f (n+ 1, p+ 1, k + 1) = 0 for all n, p, k ∈ Z. Prove that|f (n, p, k)| ≥ 1

4 for infinitely many integers n, p, k.

Mihaly Bencze

PP24713. Let Tn =�n+12

�for all n ≥ 1, be the nth triangular number. Show

that for all integers k ≥ 2, the sequence�T kn

�n≥1

does not contain anyinfinite subsequence with all terms in geometric progression.

Mihaly Bencze

PP24714. Solve in N the equation�nk

��n+1k+1

��n+2k+2

�=�3n+33k+3

�.

Mihaly Bencze

PP24715. Find the closed form of�p

�1 + 2

p2+ 4

p4

�−1when the product is

over all primes.

Mihaly Bencze

PP24716. Let ABCD be an equilateral tetrahedron with edge length dinscribed in a sphere. Let P be a point of minor semisphere ABC. LetPA = a, PB = b, PC = c. Is it possible for a, b, c and d to all be distinctpositive integers?

Mihaly Bencze

PP24717. The unsigned Stirling number of the first kind�nk

�is the number

of permutation of [n] that have k cycles. Express (−1)n�k

�n3k

�(−1)k ;

(−1)n�k

�n

3k+1

�(−1)k , (−1)n

�k

�n

3k+2

�(−1)k in function of combinatorial

expressions.

Mihaly Bencze



n1�0

�12

�nxdx, then compute

limn→∞

n

�λ− n

1�0

�12

�nxdx

�.

Mihaly Bencze

PP24719. Solve in Z the following equation(x+ 3)

�x2 + 3

� �x3 + 3

�= y2 + 15.

Mihaly Bencze

PP24720. Determine all a, b ∈ R for which1�0

(ax+ b)2k+1 dx =1�0

(ax+ b)2p+1 dx when k, p ∈ N.

Mihaly Bencze

PP24721. Compute

limn→∞

n

�2√3π9 − 1−

n�k=1

1(b−a)2n+1

b�a(x− a)n (b− x)n dx

�when 0 < a < b and

n ∈ N.

Mihaly Bencze

PP24722. If z1, z2, z3 ∈ C such that |z1| = |z2| = |z3| = a andz1 + z2 + z3 = b. Determine all a ∈ R and b ∈ C such that

|z − z1|2 + |z − z2|2 + |z − z3|2 = 3�1 + |z|2

�for all z ∈ C.

Mihaly Bencze

PP24723. If ak > 0 (k = 1, 2, ..., n) , thenn�

k=1

ak

ak+nn

n−1≤ n

n+1 .

Mihaly Bencze

PP24724. Solve in R the following systemarctg3x1 · arctg3x2 = arctg3x2 · arctg3x3 = ... = arctg3xn · arctg3x1 = π2

18 .

Mihaly Bencze


PP24725. Solve in R the following system

�√2 + 2 sinx = 2

54 sin y +

�√2− 2 sin z�√

2 + 2 sin y = 254 sin z +

�√2− 2 sinx�√

2 + 2 sin z = 254 sinx+

�√2− 2 sin y

Ionel Tudor and Mihaly Bencze

PP24726. Solve in N the equation(a+ b)b+c + (b+ c)a+b = (a+ b+ c)a+b+c .

Mihaly Bencze

PP24727. If xk > 1 (k = 1, 2, ..., n) andn�

k=1

xk = n, then

�logx1

�xn−12 + xn−2

2 + ...+ x2 + 1�≥ n(3n−1)

2

Mihaly Bencze

PP24728. Solve in N the equation ab + bc + ca = ba! + cb! + ac!.

Mihaly Bencze

PP24729. If xk > 1√n(k = 1, 2, ..., n) , then

�lognx1x2

�(n− 1)x1 +

x2x1

�≥ 1 + n(n−1)

log3

�

3n�

n�

k=1xk

�n−1� .

Mihaly Bencze

PP24730. If A,B ∈ Mn (Z) and AB = BA, detA = detB = 0, thendetermine all n ∈ N for which det (An +Bn) = xn + yn when x, y ∈ Z.

Mihaly Bencze

PP24731. If x, y, z > 0 and xyz = 1 then determine all λ ∈ R for which� 1+xλn

1+xn ≥ 1 for all n ∈ N.

Mihaly Bencze

PP24732. If ak > 0 (k = 1, 2, ..., n) , then

1n

m�i=1

ai1+ai2+...+ainn√

ai1ai2...a

in

+nm n

√a1a2...an

a1+a2+...+an≥ 2m.

Mihaly Bencze


PP24733. Let A1A2...An be a convex polygon and S =n�

k=1

ak. Prove that

�n

�S−a2−a3S−2a1

≥ n.

Mihaly Bencze

PP24734. If a, b, c > 0 and ab +

bc +

ca ≤ 11 then

ab+ bc+ ca+ (a+ b+ c)�√

ab+√bc+

√ca�≤ 6 (a+ b+ c)2 .

Mihaly Bencze

PP24735. Solve in (0,+∞) the following

system

(1 +√x)

y= 1 + 2y

√z + x2�

1 +√y�z

= 1 + 2z√x+ y2

(1 +√z)

x= 1 + 2x

√y + z2

.

Mihaly Bencze

PP24736. Solve the following system

x4 + 12y2 + 6 = 6�z2 + 2t

�

y4 + 12z2 + 6 = 6�t2 + 2x

�

z4 + 12t2 + 6 = 6�x2 + 2y

�

t4 + 12x2 + 6 = 6�y2 + 2z

�

Mihaly Bencze and Ionel Tudor

PP24737. In all triangle ABC holds

1).� ma

ha≤

√3(s2−r2−4Rr)

2sr

2).��ma

ha

�2≤ (s2−r2−4Rr)

2

4s2r2

Mihaly Bencze

PP24738. Solve in R the following system:729x+6 ·27y+8 ·27−z = 729y+6 ·27z+8 ·27−x = 729z+6 ·27x+8 ·27−y = 15.

Mihaly Bencze

PP24739. In all triangle ABC holds1).� tgA

2+tg2A≤ R+r

R√3

2).� ctgA

2+ctg2A≤ s

R√3

Mihaly Bencze



1).� tg2A

(2+tg2A)(2tg2A+1)≤ s

3R

2).� ctg2A

(2+ctg2A)(2ctg2A+1)≤ R+r

3R

Mihaly Bencze

PP24741. If xk > 0 (k = 1, 2, ..., n) andn�

k=1

1chxk

= 3 then� sh2xk

2+sh4xk≤ 1.

Mihaly Bencze

PP24742. Prove thatn�

k=1

k2(k+1)2

(k2+2)2(k2+2k+3)2≤ n

18(n+1) .

Mihaly Bencze

PP24743. Compute∞�

n1,n2,...,nk=1

1n21+n2

2+...+n2k.

Mihaly Bencze


xy2+2

≤ 1√3(z+1)

yz2+2

≤ 1√3(x+1)

zx2+2

≤ 1√3(y+1)

.

Mihaly Bencze

PP24745. If a, b, c > 0 and 2abc+�

ab ≤ 1 then� ab

(2a2+1)(2b2+1)≤ 1

3 .

Mihaly Bencze

PP24746. Prove that∞�n=0

�n

n2+2

�4< π2

36 .

Mihaly Bencze

PP24747. Solve in N the following system

a�b2 + 3c+ 2

�= x2 + y2

b�c2 + 3a+ 2

�= y2 + z2

c�a2 + 3b+ 2

�= z2 + x2

.

Mihaly Bencze


PP24748. Let ABC be a triangle. Determine all M ∈ Int (ABC) for whichBMC∡−A∡; AMC∡−B∡; AMB∡− C∡ are the angle of a triangle.

Mihaly Bencze

PP24749. If f : R → R is convex, then�

af�

b1+a+ab

�≥ (a+ b+ c) f (1)

when a, b, c > 0 and abc = 1.

Mihaly Bencze

PP24750. Solve in Z the following system:|x− 2015|+ y2 − 3z = |y − 2015|+ z2 − 3x = |z − 2015|+ x2 − 3y = 4.

Mihaly Bencze

PP24751. Solve in N the equation x2 + y2 = 169n when n ∈ N.

Mihaly Bencze

PP24752. Determine all xk ∈ C (k = 1, 2, ..., n) such that� x1x2...xn−1

1+x1+x1x2+...+x1x2...xn−1= 1.

Mihaly Bencze

PP24753. Compute

� �3x2 − 2y + 5

� �3y2 − 2z + 5

� �3z2 − 2x+ 5

�lnxyzdxdydz.

Mihaly Bencze

PP24754. In all triangle ABC holds12sr ≤� (b+ c)wa ≤ 2s

�3 (s2 + r2 − 8Rr) ≤

√3�

a2.

Mihaly Bencze

PP24755. Determine all f : [0, 1] → (0,+∞) of class C1 for which�1�0

f (x) dx

�n

≥1�0

fn+1 (x) dx when n ∈ N.

Mihaly Bencze


PP24756. Let f : [0,+∞) → R be a continuous function. Compute

limn→∞

n

�1−

1�0

f (nx) dx

�if lim

x→∞f (x) = 1.

Mihaly Bencze

PP24757. Determine all functions f : [0, 1] → (0,+∞) of class C1 for which

f (e) = 1 and1�0

f2 (x) dx+1�0

dx(f ′(x))2

≤ 2.

Mihaly Bencze

PP24758. Determine all functions f : (0,+∞) → (0,+∞) for which

f ′ (x) f�1y

�= 1

z

f ′ (y) f�1z

�= 1

xf ′ (z) f

�1x

�= 1

y

, for all x, y, z ∈ (0,+∞) .

Mihaly Bencze

PP24759. If ai > 0 (i = 1, 2, ..., n) and k ∈ {1, 2, ..., n} thenn�

i=1ai

n

�

n�

i=1ai

+kn

�

i=1ai

�

cyclic(a1+a2+...+ak)

≥ n+ 1. (A generalization of problem 5.343

Mathematical Reflections).

Mihaly Bencze

PP24760. Prove that∞�n=0

5n�

35n+1−5·35n+4

�

7295n−2435

n−35n+1

= 12 .

Mihaly Bencze

PP24761. If a0 = 0, a1 = 1, a2 = 2, a3 = 6 andan+4 = 2an+3 + an+2 − 2an+1 − an for all n ≥ 0 then denote α (n) =

�ann2

�

when [·] denote the integer part. Compute∞�n=1

1α2(n)

.

Mihaly Bencze

PP24762. In all triangle ABC holds�(|a− c|+ |b− c|)2 ≤ 8

�s2 − 3r2 − 12Rr

�.

Mihaly Bencze



n

�1− (n+ 1)!

�e−

n�k=0

1k!

��.

Mihaly Bencze


√2x − 1 +

√3y − 2x +

√2z+1 − 3y = 2x + 1√

2y − 1 +√3z − 2y +

√2x+1 − 3z = 2y + 1√

2z − 1 +√3x − 2z +

√2y+1 − 3x = 2z + 1√

2x − 1 +√3y − 2x +

√2z+1 − 3y = 2x + 1

.

Longaver Lajos and Mihaly Bencze

PP24765. Solve in R the following system(3x + 8 log3 y + 8z) (3y − 8 log3 z + 8x) == (3y + 8 log3 z + 8x) (3z − 8 log3 x+ 8y) == (3z + 8 log3 x+ 8y) (3x − 8 log3 y + 8z) = 121.

Mihaly Bencze and Bela Kovacs

PP24766. In all scalene triangle ABC holds1).� (b−a)rc

mb−ma< 2 (4R+ r)

2).� (b−a)hahb

mb−ma< 4s2r

R

Mihaly Bencze

PP24767. Compute� � (1−cos 2x−sin 2y)(1−cos 2y−sin 2x)dx

(1+sin 2x)(1+sin 2y) .

Mihaly Bencze

PP24768. In all triangle ABC holds� tgA

2tgB

2

9tg3 A2tg2 B

2+tgC

2

≤ s6r .

Mihaly Bencze


n

�ln 2−

π2�π6

(1−(sinx)2n)ctgxdx1+(sinx)2n

�.



�x2 − 2y

�+ 2 [z] = [t]2�

y2 − 2z�+ 2 [t] = [x]2�

z2 − 2t�+ 2 [x] = [y]2�

t2 − 2x�+ 2 [y] = [z]2�

x2 − 2y�+ 2 [z] = [t]2

,


when [·] denote the integer part.

Mihaly Bencze

PP24771. Determine the left-most digit of the decimal expansion of20172000.

Mihaly Bencze

PP24772. If ak ∈ [−3, 3] (k = 1, 2, ..., n) then determine the maximum of�cyclic

��a21 − a2a3 + 1�� .

Mihaly Bencze

PP24773. In all triangle ABC holds2r�5s2 + r2 + 4Rr

�≤ 3 (R+ r)

�s2 + r2 + 2Rr

�.

Mihaly Bencze

PP24774. In all triangle ABC holds� (ctgA

2 )6

ctg2 A2+ctg2 B

2

≥ s2

2r2.

Mihaly Bencze

PP24775. If xk > 0 (k = 1, 2, ..., n) andn�

k=1

xk = 1, then

n−1�

cyclicx1x2...xn−1

− 1n�

k=1xk

≤ n−2n .

Mihaly Bencze

PP24776. In all triangle ABC holds��3r

s

�3rs ctg

A2 ctg

A2 + 2

�3rs ctg

A2 + 2

�≤ 216.

Mihaly Bencze

PP24777. Prove that exist infinitely many triangles ABC such that

�� 1m

rba

�� 1mrc

b

�� 1mra

c

�≥ 1

34R+r−1

�R2r2

�4R+r.

Mihaly Bencze


PP24778. Determine all a, b, c ≥ 3 for which(a−1)2(b−1)

(c−2)a−2 + (b−1)2(c−1)

(a−2)b−2 + (c−1)2(a−1)

(b−2)c−2 ≤ ab + bc + ca.

Mihaly Bencze

PP24779. Find all functions f : R → R such that f (0) ∈ Q andf�x+ f2

�y3�+ f3

�z2��

= f6 (x+ y + z) for all x, y, z ∈ R.

Mihaly Bencze

PP24780. Solve in N the equation (2a + b)�2b + a

�= a!b!.

Mihaly Bencze

PP24781. Find all functions f : R → R such that(f ◦ f) (x− y) f

�x2 + xy + y2

�= x3 − yf

�y2�for all x, y ∈ R.

Mihaly Bencze


n

2−

1�

0(x2−x−2)

ndx

1�

0

(4x2−2x−2)ndx

.

Mihaly Bencze


n

2−

1�

0(2x2−5x−1)

ndx

1�

0

(x2−4x−1)ndx

.

Mihaly Bencze


n

2−

n+1�

k=1k! csc

�

π

2k

�

n�

k=1k!scs

�

π

2k

� + 2n

.

Mihaly Bencze

PP24785. In all triangle ABC holds 1s2+r2+2Rr

≥ 18R2 .

Mihaly Bencze



� (tgA2 )

n−1(tgB

2 )n−2

tgC2

ctgA2+ctgB

2

≥ 22n−2·3n−1rs for all n ∈ N, n ≥ 2.

Mihaly Bencze

PP24787. In all triangle ABC holds� ctgA

2

1+9tg2 A2tg2 B

2

≥ 12 .

Mihaly Bencze


n

�lnπ − 3

2 −n�

k=2

�k2 ln

�1− 1

k2

�+ 1��

Ovidiu Furdul and Mihaly Bencze

PP24789. In all triangle ABC holds� a3

a2+ab+b2≥ 9Rr

s .

Mihaly Bencze


k=1

�e2 + π2k

�< 2ne

n+π(π2−1)e(π−1) .

Mihaly Bencze


k=1

�e2 +

�e+ 1

k

�2�< 2ne2n+

en(n+1)2 .

Mihaly Bencze

PP24792. In all triangle ABC holds� ma

a ≥ s2−4Rr−r2

2sR .

Mihaly Bencze

PP24793. In all triangle ABC holds��

tgA2 tg

B2

�4 ≥�rs

�2.

Mihaly Bencze

PP24794. In all triangle ABC holds�� s3+27r3(ctgA

2 )3

s2+9r2(ctgA2 )

2

�2

≥ 9r (4R+ r) .

Mihaly Bencze


PP24795. If ak > 0 (k = 1, 2, ..., n) andn�

k=1

ak = n, then

n�k=1

�1+a3k1+a2k

�2≥

n�k=1

a2k.

Mihaly Bencze

PP24796. Determine all convex pentagon ABCDE for whichPerimeter (ACE) ≥ 2BD.

Mihaly Bencze

PP24797. Let a, b ∈ Q such that a1k + b

1k + (ab)

1k is also rational.

Determine all k ∈ N∗ for which a1k and b

1k must be rationals.

Mihaly Bencze

PP24798. Let be p ≥ k ≥ 3 prime numbers andA =

�k≤i1<i2<...<ik≤p−1

i1i2...ik. Determine all k ∈ N∗ for which A+ 1 is

divisible by p.

Mihaly Bencze

PP24799. Let ABC be an acute-angled triangle with AC �= BC, letM ∈ Int (ABC) and F the foot of the altitude through C. Furthermore, letP and K be the feet of the perpendiculars dropped from A and Brespectively to the extension of C. The line FM intersects the circumcircleof FPK a second time at L. Determine all M for which ML < MF.

Mihaly Bencze

PP24800. The first n primes are p1 = 2, p2 = 3, ... etc. Set K = pp11 pp22 ...ppnn .Find all positive integers a, b such that K

a(a+1)...(a+b−1) is even, andK

a(a−1)...(a+b−1) has exactly a (a+ 1) ... (a+ b− 1) divisors.

Mihaly Bencze

PP24801. Let (d) a line through point A, and ABC be a triangle. The line(d) intersects the line BC at P . Let Q,R be the symmetrical of P withrespect to the lines AB respectively AC. Determine all (d) for whichBC ⊥ QR.

Mihaly Bencze


PP24802. Find all non-constant polynomialsP (x) = xn + an−1x

n−1 + ...+ a1x+ a0 with rational coefficients whose rootsare exactly the numbers 1

a0, 1a1, ..., 1

an−1with the same multiplicity.

Mihaly Bencze

PP24803. Find all non-decreasing functions f : R → R for whichf�f�x3�+ y + f

�z2��

= x3 + f (y) + f2 (z) for all x, y, z ∈ R.

Mihaly Bencze

PP24804. Prove thatn−1�k=0

�n−1k

� �(x+ k)k−1 (y + n− k)n−k−1 + (y + k)k−1 (x+ n− k)n−k−1

�=

= (x+y)(x+y+n)n−1

xy for all x, y ∈ C∗.

Mihaly Bencze

PP24805. Solve in Z the equation (x+ y + z)�

1x+1 + 1

y+1 + 1z+1

�= 9.

Mihaly Bencze

PP24806. Consider a triangle ABC with an inscribed circle with centre Iand radius r. Let CA, CB and CC be circles internal to ABC, tangent to itssides and tangent to the inscribed circle with corresponding radii rA, rB andrC . Show that rλA + rλB + rλC ≥ rλ

3λ−1 for all λ ∈ (−∞, 0] ∪ [1,+∞) . Whathappent if λ ∈ (0, 1)?

Mihaly Bencze

PP24807. If xk > 0 (k = 1, 2, ..., n) then

�cyclic

x21

x2≥ �

cyclic

(x1 − x2)2 +

�n

n�k=1

x2k.

Mihaly Bencze

PP24808. Determine all a, b ∈ N for which for each number(n!)a + 2b; (n!)a + 3b; ...; (n!)a + nb there exist a prime divisor that does notdivide any other number from this set.

Mihaly Bencze



ar2a ≥ 6sr (2R− r) .

Mihaly Bencze

PP24810. Let H0 = 0 and Hn =n�

k=1

1k . Determine all p ∈ N∗ for which

n�k=1

(−1)n−k �nk

�pkHk = pHn −H�

np

�, when [·] denote the integer part.

Mihaly Bencze

PP24811. If ak ∈ R (k = 1, 2, ..., n) then2n�k=1

|sin ak|+2n�k=1

|cos ak|+��sin�

2n�k=1

ak

��+��cos�

2n�k=1

ak

�� ≤ 2 (2n+ 1) .

Mihaly Bencze

PP24812. In all triangle ABC holds� |cos 3A| ≤ 3

��2 +

√2− s

R

�.

Mihaly Bencze

PP24813. Compute min�ab + bc + ca

�where a, b, c > 0 and abc = 1.

Mihaly Bencze

PP24814. Find all injective functions f : Z → Z that satisfy|f (x)− f (y)|+

��f2 (y)− f2 (z)��+��f3 (z)− f3 (x)

�� ≤≤ |x− y|+

��y2 − z2��+��z3 − x3

�� for all x, y, z ∈ Z.

Mihaly Bencze

PP24815. Denote d (n) the number of different divisors of n, and letan+1 = an + kd (n) where a1 ∈ N∗. Determine all k ∈ N∗ for which doesthere exist an a1 such that k consecutive numbers of the sequence are perfectk powers of natural numbers?

Mihaly Bencze

PP24816. Let be ai, bi, ci ∈ C (i = 1, 2, ..., n) , |ai| = |bi| = |ci| = 1

(i = 1, 2, ..., n) and a = 1n

n�i=1

ai, b =1n

n�i=1

bi, c =1n

n�i=1

ci and


di = abci + abic+ aibc− 2aibici. Prove thatn�

i=1|di| ≤ n

Mihaly Bencze


n

�1

128 −n�

k=2

k2+1k3(k2−1)4

�.

Mihaly Bencze

PP24818. Determine all prime numbers Pk (k = 1, 2, ..., n+ 1) for which

P 2n+1 =

n�k=1

P 2k .

Mihaly Bencze

PP24819. If a, b, c > 0 then 67 +� a+b

2(a+b)+3c ≥ 2� a+b+2c

5(a+b)+4c .

Mihaly Bencze

PP24820. If xk ∈ [−5, 3] (k = 1, 2, ..., n) , then�cyclic

√3x1 − 5x2 − x1x2 + 15 ≤ 4n. When holds the equality?


PP24821. If λ ∈ [0, 1] thenn�

k=1

�1

(2k+1)√2k−1

�λ≤ n1−λ

�1− 1√

2n+1

�λ.

Mihaly Bencze

PP24822. If A,B ∈ Mn (R) such that AB = On. Determine all m ∈ N for

which det

�In +

m�k=1

�Ak +Bk

��≥ 0.

Mihaly Bencze


3x1 + 2x2 = 9log2 x3 + x243x2 + 2x3 = 9log2 x4 + x25−−−−−−−−−−−3xn + 2x1 = 9log2 x2 + x23

Mihaly Bencze and Lajos Longaver



x212x22 + 2 log2 x3 = x4 + 2x5+2x6

x222x23 + 2 log2 x4 = x5 + 2x6+2x7

−−−−−−−−−−−−−−−x2n2

x21 + 2 log2 x2 = x3 + 2x4+2x5


PP24825. Determine all f : R → R for which(x+ 1)2 f (x+ 1)− (x+ 2)2 f (−x− 1) = (x+ 1)

�2x2 + 6x+ 5

�for all

x ∈ R.

Mihaly Bencze

PP24826. Determine all n ∈ N∗ for which kn is a perfect k power,pn is aperfect p power and rn is a perfect r power, when k < p < r are positiveintegers.

Mihaly Bencze

PP24827. 1). Let ÷a1, a2, ..., an be an arithmetical progression. Determine

the inverse of the matrice A =

a1 a2 ... an−1 anan a1 ... an−2 an−1

− − − −− −−a2 a3 ... an a1

2). What happens when a1, a2, ..., an is a geometrical progression?

Mihaly Bencze

PP24828. If a = 1+√2

2 then determine all n, k ∈ N for which an + a−k ∈ N.

Mihaly Bencze


2 + tgx1 =√3 + 2 [tgx2]

2 + tgx2 =√3 + 2 [tgx3]

−−−−−−−−−−−−2 + tgxn =

√3 + 2 [tgx1]

when [·] denote the integer part.




�x21 − x22 − 1

� �x22 − x23 + 1

�= 4x1x2�

x22 − x23 − 1� �

x23 − x24 + 1�= 4x2x3

−−−−−−−−−−−−−−−−�x2n − x21 − 1

� �x21 − x22 + 1

�= 4xnx1

.


PP24831. If ak > 0 (k = 1, 2, ..., n) andn�

k=1

ak = 1 then

√5 ≤

n�k=1

√2ak + 1 ≤

�n (n+ 2).

Mihaly Bencze

PP24832. Determine all p, n, k ∈ N for which the function

f (x) =

�xn sin

�x−k�− x−p cos

�x−k�if x ∈ (0, 1)

0 if x = 0have primitive

functions.

Mihaly Bencze

PP24833. If xk > 1 (k = 1, 2, ..., n) , then determine all n ∈ N for whichx21

x2−1 +x22

x3−1 + ...+ x2n

x1−1 ≥ 2n+1.

Mihaly Bencze

PP24834. If ak > 0 (k = 1, 2, ..., n) , thenn�

k=1

�a2k + ak + 1

�≥ 3n

n�k=1

ak.

Mihaly Bencze

PP24835. Prove that 2√6n

n+1 ≤n�

k=1

3( 23)

k+2( 3

2)k

k(k+1) ≤ 5nn+1 .

Mihaly Bencze

PP24836. If a, b, x > 0 then axm+n + b ≥ xn (m+ n)��

an

�n � bm

�m� 1n+m

for

all n, n ∈ N∗.

Mihaly Bencze


PP24837. Prove thatn(n+1)

2 ≤n�

k=1

50k(k+1)2k−45ke(k2+k)k−11e2kk+1

25k(k+1)2k−25ke(k2+k)2−6e2kk+1≤ 11n(n+1)

12 .

Mihaly Bencze

PP24838. Prove that 2n9(n+3) ≤

n�k=1

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

≤ n3(n+3) .

Mihaly Bencze

PP24839. Prove that∞�k=1

k2+2k√k2+1

> π2

3 .

Mihaly Bencze


k=1

k2(k4+2k3+3k2+2k2+2k+2)k2+k+1

≥ n(n+1)(2n+1)3 .

Mihaly Bencze

PP24841. Solve in R the following

system

√x2 + x− 1 +

�y − y2 + 1 = z2 − z + 2�

y2 + y − 1 +√z − z2 + 1 = x2 − x+ 2√

z2 + x− 1 +√x− x2 + 1 = y2 − y + 2

.

Mihaly Bencze


x2

3+√

9−y2+ 1

4(3−√9−z2)

= y2

3+√9−z2

+ 14(3−

√9−x2)

= z2

3+√9−x2

+ 1

4�

3−√

9−y2� = 1.

Mihaly Bencze


�x2 + 3

�4= 256 (2y − 1)�

y2 + 3�4

= 256 (2z − 1)�z2 + 3

�4= 256 (2x− 1)

.

Mihaly Bencze



√x2 + 1 +

�y3 − 1 = z2 + z + 1�

y2 + 1 +√z3 − 1 = x2 + x+ 1√

z2 + 1 +√x3 − 1 = y2 + y + 1

.

Mihaly Bencze


n

�1− 2

n�k=1

2k2−14k4+1

�.

Mihaly Bencze


x2 + 13y + 4 = 6 (z + 2)√t

y2 + 13z + 4 = 6 (t+ 2)√x

z2 + 13t+ 4 = 6 (x+ 2)√y

t2 + 13x+ 4 = 6 (y + 2)√z

.

Mihaly Bencze

PP24847. Solve in R the following system�x+ 2 + 2

√y + 1 +

�y + 2− 2

√z + 1 =

�y + 2 + 2

√z + 1+

+�z + 2− 2

√x+ 1 =

�z + 2 + 2

√x+ 1 +

�x+ 2− 2

√y + 1 = 2.

Mihaly Bencze


n

�23 −

n�k=2

k3−1k3+1

�.

Mihaly Bencze


√x+ 4 +

√y − 4 = 2

�z2 +

√t2 − 16− 6

�

√y + 4 +

√z − 4 = 2

�t2 +

√x2 − 16− 6

�

√z + 4 +

√t− 4 = 2

�x2 +

�y2 − 16− 6

�

√t+ 4 +

√x− 4 = 2

�y2 +

√z2 − 16− 6

�.

Mihaly Bencze

PP24850. Solve in R the following system1

x−√

y2−y− 1

y+√z2−x

= 1y−

√z2−z

− 1z+

√x2−x

= 1z−

√x2−x

− 1

x+√

y2−y=

√3.

Mihaly Bencze



√x− 2 +

√4− y = z2 − 6z + 11√

y − 2 +√4− z = x2 − 6x+ 11√

z − 2 +√4− x = y2 − 6y + 11

.

Mihaly Bencze

PP24852. Solve in R the following system√x2 + 32−2 4

�y2 + 32 =

�y2 + 32−2 4

√z2 + 32 =

√z2 + 32−2 4

√x2 + 32 = 3.

Mihaly Bencze

PP24853. Solve in R the following system√x2 + 9−

�y2 − 7 =

�y2 + 9−

√z2 − 7 =

√z2 + 9−

√x2 − 7 = 2.

Mihaly Bencze

PP24854. Solve in R the following systemx2 + 3y + 4

√z2 + 3t− 6 = y2 + 3z + 4

√t2 + 3x− 6 =

= z2 + 3x+ 4�x2 + 3y − 6 = t2 + 3y + 4

�y2 + 3z − 6 = 28.

Mihaly Bencze

PP24855. Solve in R the following system�3x2 − 2y + 15 +

�3y2 − 2z + 8 =

�3y2 − 2z + 15 +

√3z2 − 2x+ 8 =

=√3z2 − 2x+ 15 +

�3x2 − 2y + 8 = 7.

Mihaly Bencze

PP24856. Solve in R the following system√15− x+

√3− y =

√15− y +

√3− z =

√15− z +

√3− x = 6.

Mihaly Bencze

PP24857. Solve in R the following system3x

y2−4y+1− 2y

z2+z+1= 3y

z2−4z+1− 2z

x2+x+1= 3z

x2−4x+1− 2x

y2+y+1= 8

3 .

Mihaly Bencze

PP24858. Solve in R the following system3x2−1

y + 5y3z2−z−1

= 3y2−1z + 5z

3x2−x−1= 3z2−1

x + 5x3y2−y−1

= 1198 .

Mihaly Bencze



(x+ 1)2 = (y − 3)3

(y + 1)2 = (z − 3)3

(z + 1)2 = (x− 3)3.

Mihaly Bencze

PP24860. Solve in R the following system2x

2y2−5y+3+ 13y

2z2+z+3= 2y

2z2−5z+3+ 13z

2x2+x−3= 2z

2x2−5x+3+ 13x

2y2+y−3= 6.

Mihaly Bencze


�3x2 + 5y + 8 = 1 +

�3y2 + 5z + 1�

3y2 + 5z + 8 = 1 +√3z2 + 5x+ 1√

3z2 + 5x+ 8 = 1 +�3x2 + 5y + 1

Mihaly Bencze


�x− 1 + 2

√y − 2 = 1 +

�y − 1− 2

√z − 2�

y − 1 + 2√z − 2 = 1 +

�z − 1− 2

√x− 2�

z − 1 + 2√x− 2 = 1 +

�x− 1− 2

√y − 2

.

Mihaly Bencze


xy2−y+1

+ 2yz2+z+1

= 1y

z2−z+1+ 2z

x2+x+1= 1

zx2−x+1

+ 2xy2+y+1

= 1

.

Mihaly Bencze

PP24864. Solve in C the following system

2x2 − 8y + 12 =�y2 − 4z − 6

�2

2y2 − 8z + 12 =�z2 − 4x− 6

�2

2z2 − 8x+ 12 =�x2 − 4y − 6

�2.

Mihaly Bencze

PP24865. Solve in Z∗\ {−1} the equation 1x(y2+1)

+ 1(x+1)(y+1) +

1y(x2+1)

= 1.

Mihaly Bencze


PP24866. Solve in Z∗ the equation 1x + 1

y + 1z = 1

9 + 1xyz .

Mihaly Bencze

PP24867. If x ∈ R\ {−1, 1} thenn�

k=1

�x2+1−1x−1

�k+

n�k=1

�x2n+1+1

x+1

�k> 2− 1

2n−1 .

Mihaly Bencze

PP24868. Solve in Z∗ the equation 1x3 + 1

x2y+ 1

xy2+ 1

y3= 1.

Mihaly Bencze


x12 + y4 + 1 ≥ z9 + ty12 + z4 + 1 ≥ t9 + xz12 + t4 + 1 ≥ x9 + yt12 + x4 + 1 ≥ y9 + z

.

Mihaly Bencze


k=1

2007

�k2014+k2008

k4+1≥ n(n+1)

2 .

Mihaly Bencze


k=1

k12+k4+1k8+1

≥ n(n+1)2 .

Mihaly Bencze


2x4 + 1 ≥ 2y3 + z2

2y4 + 1 ≥ 2z3 + x2

2z4 + 1 ≥ 2x3 + y2.

Mihaly Bencze


k=1

2k4+1k+2 ≥ n(n+1)(2n+1)

6 .

Mihaly Bencze

PP24874. Prove that∞�k=1

kk+1 ≥ π2

12 .

Mihaly Bencze



�2x3 ≤ y + 12y3 ≤ x+ 1

.

Mihaly Bencze

PP24876. Prove that n(n−1)2 ≤

n�k=1

4√k4 + k3 + k2 + k + 1 ≤ n(n+3)

2 .

Mihaly Bencze

PP24877. Prove that n(n+1)2 ≥

n�k=1

2m�2m (k − 1) + 1 when m ∈ N∗.

Mihaly Bencze


13x1 + 1

5x2 + 115x3 = 2

1+15x4 + 23x5+25x6 + 2

5x7+9x81

3x2 + 15x3 + 1

15x4 = 21+15x5 + 2

3x6+25x7 + 25x8+9x9

−−−−−−−−−−−−−−−−−−−−−1

3xn + 15x1 + 1

15x2 = 21+15x3 + 2

3x4+25x5 + 25x6+9x7

.

Mihaly Bencze


|z1 − 1|+ |z2 + 1| =√2 |z3 − 1|

|z2 − 1|+ |z3 + 1| =√2 |z4 − 1|

−−−−−−−−−−−−−−|zn − 1|+ |z1 + 1| =

√2 |z2 − 1|

.

Dorin Andrica and

Mihaly Bencze


x21 + x22 + 16x23 = 9x24 + 1x22 + x23 + 16x24 = 9x25 + 1−−−−−−−−−−−−x2n + x21 + 16x22 = 9x23 + 1

.

Mihaly Bencze



2�x21 + 4 =

�x22 + 1 +

�x23 + 9

2�x22 + 4 =

�x23 + 1 +

�x24 + 9

−−−−−−−−−−−−−2�x2n + 4 =

�x21 + 1 +

�x22 + 9

.

Mihaly Bencze


x10 + 20y4 = 32 + 10z6 + 21t5

y10 + 20z4 = 32 + 10t6 + 21x5

z10 + 20t4 = 32 + 10x6 + 21y5

t10 + 20x4 = 32 + 10y6 + 21z5

.



4 cos π2x =

√y + tg π

2z4 cos π

2y =√z + tg π

2x

4 cos π2z =

√x+ tg π

2y



x6 + 3y5 + 3z = 1 + 5t3

y6 + 3z5 + 3t = 1 + 5x3

z6 + 3t5 + 3x = 1 + 5y3

t6 + 3x5 + 3y = 1 + 5z3

.


PP24885. Let n ∈ N and a, b, c, d ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} ; a �= 0 such thatabcd+ cd = 2n − 20. Prove that n ≥ 10 and determine all n ∈ N and abcd forwhich holds the given equality.

Ionel Tudor

PP24886. Solve in R the equation64 cos6 x+ 96 cos5 x− 40 cos3 x+ 6 cosx− 1 = 0.

Ionel Tudor

PP24887. a). If n ∈ N, n ≥ 3 then exist an increasing arithmeticalprogression x1, x2, ..., xn ∈ (−1, 1) such that 2x1 + 2x2 + ...+ 2xn = n


b). If n = 3 then determine one arithmetical progression which verify a).

Ionel Tudor

PP24888. Prove that the function f : [8,+∞) → R wheref (x) = 4 cos π

2x − tg π2x −√

x is increasing. Solve in N the equation4 cos π

2n =√n+ tg π

2n .

Ionel Tudor

PP24889. If x ∈ R thenx8 + x7 + 3x6 + 5x5 + x4 − 10x3 + 12x2 − 8x+ 16 > 0. Solve in R theequation x10 − 10x6 − 21x5 + 20x4 − 32 = 0.

Ionel Tudor

PP24890. If�1 + 1

x

�x= e

�1−

∞�k=1

bk(x+1)k

�when x > 0 then compute

∞�k=1

11+b2k

.

Mihaly Bencze

PP24891. If�1 + 1

x

�x= e

�1−

∞�k=1

dk

( 1112

+x)k

�, then compute

∞�k=1

11+d2k

.

Mihaly Bencze

PP24892. If

An = 2n−1

8 0 0 04n 4n+ 8 4n 4n

n2 − n n2 − n n2 − 5n+ 8 n2 − 5n−n2 − 3n −n2 − 3n −n2 + n −n2 + n+ 8

, then

compute∞�n=1

11+(det(An))2

.

Mihaly Bencze


x+ [y] = {z}+ 54y + [z] = {x}+ 54z + [x] = {y}+ 54

, when [·]

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

Mihaly Bencze



�2x+ 11− 6

�2 (y + 1) < z+9

2�2y + 11− 6

�2 (z + 1) < x+9

2�2z + 11− 6

�2 (x+ 1) < y+9

2

.

Mihaly Bencze

PP24895. If x ∈ R then

32 + sinx+ cosx < 2 sin2 x

1+sin2 x+ 2 cos2 x

1+cos2 x+�

1+sin4 x2 +

�1+cos4 x

2 .

Mihaly Bencze

PP24896. If x ∈ R then�

1+sin4 x2 +

�1+cos4 x

2 + sinx+ cosx < 3.

Mihaly Bencze


�1 +

n�k=1

(4k−1)!(4k)!−(4k−2)!

�then compute

limn→∞

n

�λ− 1−

n�k=1

(4k−1)!(4k)!−(4k−2)!

�.

Mihaly Bencze

PP24898. If ak > 0 (k = 1, 2, ..., n) , thenn�

k=1

a2k ≥ �cyclic

a1a2 +1

2r+1

�(|a1 − a2|+ |a1 − a3|+ ...+ |a1 − ar|)2 when

r ∈ {2, 3, ..., n− 1} .

Mihaly Bencze


x4 + 12y2 + 32 = 12z2 + 16ty4 + 12z2 + 32 = 12t2 + 16xz4 + 12t2 + 32 = 12x2 + 16yt4 + 12x2 + 32 = 12y2 + 16z

.

Mihaly Bencze



x√3 + y +

√39− 3z = 4

√3 + t2

y√3 + z +

√39− 3t = 4

√3 + x2

z√3 + x+

√39− 3x = 4

�3 + y2

t√3 + y +

√39− 3y = 4

√3 + z2

Mihaly Bencze

PP24901. Solve in C the following equation

(3x − 2)2x + (4x − 2)

2x + ...+ (nx − 2)

2x = (n−2)(2n−3)(n−1)

6 .

Mihaly Bencze

PP24902. Solve in N the following system

(x!)3 = xy + y + z

(y!)3 = yz + z + x

(z!)3 = zx+ x+ y

.

Mihaly Bencze

PP24903. Determine all n ∈ N for which�

cyclic

n

�xn+xy

xn−xy+yn ≤ 3 n√2 for all

x, y, z > 0.

Mihaly Bencze


k=1

�1√k+�

k2+12k2

�≤ n+ 1.

Mihaly Bencze


x3 + 5y = z2 − 5 + 4 3

�(5t2 + t− 5)2 + 6 3

√5x2 + x− 5

y3 + 5z = t2 − 5 + 4 3

�(5x2 + x− 5)2 + 6 3

�5y2 + y − 5

z3 + 5t = x2 − 5 + 4 3

�(5y2 + y − 5)2 + 6 3

√5z2z +−5

t3 + 5x = y2 − 5 + 4 3

�(5z2 + z − 5)2 + 6 3

√5t2 + t− 5

.

Mihaly Bencze


PP24906. Prove that

3 <

�4 sin4 x1+sin4 x

+�2�1 + sin8 x

�+�

4 cos4 x1+cos4 x

+�2 (1 + cos8 x) for all x ∈ R.

Mihaly Bencze

PP24907. Let ABC be a triangle and M1,M2,M3 ∈ Int (ABC) . Provethat 2 (AM1 +AM2 +AM3) +BM1 +BM2 +BM3 +CM1 +CM2 +CM3 <

< 3 (2BC +AB +AC) .

Mihaly Bencze

PP24908. Let A1A2A3...An be a convex polygon for whichA1A2 = A2A3 = An−1An = AnA1. If A1∡ ≥ A2∡ ≥ ... ≥ An∡ then thepolygon is regular.

Mihaly Bencze

PP24909. In all acute triangle ABC holds

exp

�A�0

tgxx dx+

B�0

tgxx dx+

C�0

tgxx dx

�>�

π3

(π−2A)(π−2B)(π−2C)

� 2π.

Mihaly Bencze

PP24910. Solve in M2 (N) the following system:A2

1 − 8A2 + 7I2 = A22 − 8A3 + 7I2 = ... = A2

n − 8A1 + 7I2 = O2.

Mihaly Bencze

PP24911. Prove that 15 + 1

7 + ...+ 12n+3 < ln

�n+22 .

Mihaly Bencze


log2�x31 + 2x2

�= 1 + logx3

(2 + x4)log2�x32 + 2x3

�= 1 + logx4

(2 + x5)−−−−−−−−−−−−−−−−−log2�x3n + 2x1

�= 1 + logx2

(2 + x3)

Mihaly Bencze



log9 (4x1 + 1) = log4 (9

x2 − 1)log9 (4

x2 + 1) = log4 (9x3 − 1)

−−−−−−−−−−−−−log9 (4

xn + 1) = log4 (9x1 − 1)

.

Mihaly Bencze


3�x1

√x2 − 7

�x3�x4

√x5 = 6 12

�x6

√x7

3�x2

√x3 − 7

�x4�x5

√x6 = 6 12

�x7

√x8

−−−−−−−−−−−−−−−−−−3�xn

√x1 − 7

�x2�x3

√x4 = 6 12

�x5

√x6

.

Mihaly Bencze

PP24915. If a1 > 0 and�n+ a4n

�a2n+1 = na2n for all n ≥ 1 and λ = lim

n→∞an

then compute limn→∞

n (λ− an) .

Mihaly Bencze

PP24916. Solve in R the following system�3x1 +

4x2+1 + 8

�

2(1+x23)

��3x2 +

4x3+1 + 8

�

2(1+x24)

�=

=

�3x2 +

4x3+1 + 8

�

2(1+x24)

��3x3 +

4x4+1 + 8

�

2(1+x25)

�= ...

=

�3xn + 4

x1+1 + 8�

2(1+x22)

��3x1 +

4x2+1 + 8

�

2(1+x23)

�= 81.

Mihaly Bencze


n

�2e

π2 − n

�n�

k=1

�1 + n2

k2

��.

Mihaly Bencze

PP24918. Solve in R the following system[x1] +

�xk2�= [x2] +

�xk3�= ... = [xn] +

�xk1�= k when k ∈ N∗ is given, and [·]



Mihaly Bencze

PP24919. In all triangle ABC holds�1 + sin A

2

� �1 + sin B

2

� �1 + sin C

2

�≤ (a+3b−c)(b+3c−a)(c+3a−b)

8abc .

Mihaly Bencze

PP24920. If x0 = x1 = 1 and n (n+ 1)�xn+1

√xn−1 − xn

√xn�=

√xn−1xn

for all n ≥ 1 then compute limn→∞

n (4− xn) .

Mihaly Bencze

PP24921. Solve in R the following system:�x21+x2

22 + x3+x4

2 =

�x22+x2

32 + x4+x5

2 = ... =

�x2n+x2

12 + x2+x3

2 = 9.

Mihaly Bencze

PP24922. Let ABC be a triangle, N ∈ (BC) ,M ∈ (CA) ,K ∈ (AB) suchthat NAC∡ = α, MBA∡ = β, KCB∡ = γ. Prove that�3 (NC2 +MA2 +KB2) ≥� ab sinα

b sinα+c sin(A−α) .

Mihaly Bencze


n�k=1

e1

n+k

n+k then compute limn→∞

n

�λ−

n�k=1

e1

n+k

n+k

�.

Mihaly Bencze


log3 2 log2 x1 = log2 (x2 − 1)log3 2 log2 x2 = log2 (x3 − 1)−−−−−−−−−−−−−log3 2 log2 xn = log2 (x1 − 1)

.

Mihaly Bencze

PP24925. Prove that (n!)2 <n�

k=1

�1

arctg 1k

�2<�(n+1)(2n+1)

6 + 1�n

.

Mihaly Bencze


PP24926. If ak =

�1

(arctg 1k )

2

�when [·] denote the integer part, then

n�k=1

2k+1aka2k+1

= n(n+2)

(n+1)2.

Mihaly Bencze

PP24927. Let p ∈ {1, 2, ..., n− 1} a given number. Determine all n ∈ N for

which

�n�

k=1

�1

arctg 1√k

�2�= p+ n(n+1)

2 when [·] denote the integer part.

Mihaly Bencze

PP24928. If a, b,λ > 0 thenb�a

√lnxdx ≤ (λ−1)(b−a)

2√λ

+ 12√λln bb

aa .

Mihaly Bencze

PP24929. If x, y, z > 1 then�

log x+y2

�13

�x�≤ 27

�

(x+y)2

8xyz(�

x)3

�logx

x+y2 .

Mihaly Bencze

PP24930. If Fn denote the nth Fibonacci number, then�n�

k=1

1Fk

�2

≥n�

k=1

k+1F 2k.

Mihaly Bencze


k=1

�arctg 1

kk

�2> n2

n2+1.

Mihaly Bencze

PP24932. If ak ∈ (0, 1) (k = 1, 2, ..., 2n+ 1) then

2n+1�k=1

logak

�1

2n+1

2n+1�k=1

ak

�≥

(2n+1)2n+12n+1�

k=1ak

�

2n+1�

k=1ak

�2n+1 .

Mihaly Bencze


PP24933. If x > 0 (k = 1, 2, ..., n) andn�

k=1

xk = n, then

n�k=1

xn−1k +

n�k=1

xk + n− 1 ≥ 2�

x1x2...xn−1.

Mihaly Bencze

PP24934. If A,B ∈ Mn (C) such that xBA− yAB = In. Determine allx, y ∈ C for which det (AB −BA) = 0.

Mihaly Bencze

PP24935. If a > 1 and x, t > 0 then ax

x − ax+t

x+t ≤ tx(x+t) .

Mihaly Bencze

PP24936. In all triangle ABC holds�� rb+rc

ra

�λ≥ 3�Rr

�λfor all λ ≥ 1.

Mihaly Bencze


(13− 7x1)�1 + 3

1x2

�= 16 (2 + 7x3)

(13− 7x2)�1 + 3

1x3

�= 16 (2 + 7x4)

−−−−−−−−−−−−−−−−−(13− 7xn)

�1 + 3

1x1

�= 16 (2 + 7x2)


PP24938. If x0, x1 > 1 and xn+1 = log3 (1 + xn+1 + xn) then computelimn→∞

n (1− xn) .

Mihaly Bencze

PP24939. In all triangle ABC holds� r3a

r2a+2rbrc≥ 4R+r

3 .

Mihaly Bencze

PP24940. In all triangle ABC holds� 1

(tgA2+tgB

2 )(tgB2+tgC

2 )≤ 9

4 .

Mihaly Bencze



11a − 2b = c2

11b − 2c = a2

11c − 2a = b2

Gheorghe Stoica and Mihaly Bencze

PP24942. If b1 ≥ 2b2 ≥ ... ≥ nbn ≥ 0 then b1 +

�n�

k=1

√bk

�2

≥n�

k=1

(k + 1) bk.

Mihaly Bencze

PP24943. Let ABC be a triangle, A1 ∈ (BC) , B1 ∈ (CA) , C1 ∈ (AB) .The medians AA1, BB1, CC1 intersect second time the circumcircle in pointsD,E and F. Prove that

s2 + r2 + 4Rr +BD ·DC + CE ·AE +AF · FB ≤ AD2 +BE2 + CF 2.

Mihaly Bencze


r3b+r3c≥ 3(4R+r)

2(4R+r)2−4s2.

Mihaly Bencze

PP24945. Determine all n ∈ N for which

�n�

k=1

1

(arctg 1k )

2

�= n(n+1)(2n+1)

6 + p

when p ∈ {1, 2, ..., n− 1} is given.

Mihaly Bencze

PP24946. If p ∈ N∗ is given, then compute the integer part ofn�

k=1

p

�p+ 1

k! .

Mihaly Bencze


n

�e−1 −

n�p=1

pap

�n−1�

p=1(p+1)ap

where

ap = 1 +n�

k=1

f ′k (0) and fk (x) = kx (1− x) (2− x) ... (k − x) .

Mihaly Bencze


PP24948. If x0 ∈ [0,π] and (n+ 1)xn =n�

k=0

sinxk then compute

limn→∞

n�√

3−√lnnxn+1

�.

Mihaly Bencze

PP24949. Solve in M3 (R) the equation x3 + x2 + x =

1 1 11 1 11 1 1

.

Mihaly Bencze

PP24950. If a ∈ (0, 1) ∪ (1,+∞) then solve in R the following system

2ax+1 = (a− 1)2 y2 +�a2 − 1

�z + 2a

2ay+1 = (a− 1)2 z2 +�a2 − 1

�x+ 2a

2az+1 = (a− 1)2 x2 +�a2 − 1

�y + 2a


PP24951. If x, y, z ∈ R and z + y + z = (2k + 1)π, k ∈ Z then determineall a, b, c ∈ R for which

�sin ax sin (by − cz) =

�sin ax (sin by − sin cz)

Mihaly Bencze


(wa)b+c ≤

�√3(s2−r2−4Rr)

2s

�4s

.

Mihaly Bencze

PP24953. In all triangle ABC holds�� cos2 B

2+cos2 C

2

cos2 A2

≤�

2�(4R+r)3+s2(2R+r)

2Rs2+ 1�.

Mihaly Bencze


(4R+r)ra+3r2b≥ 1

2 .

Mihaly Bencze

PP24955. Determine all a, b ∈ N for which�7a + 51b − 2

� �7b + 51a − 2

�is

divisible by 64.

Mihaly Bencze



(n!)a((n+2)!)b

((2n+3)!)cif a+ b = c; a, b, c ∈ N.

Mihaly Bencze

PP24957. Prove that�

n�k=0

�nk

�2ak�− 1

n

+

�n�

k=0

�nk

�1ak

�− 1n

=

[n2 ]�k=0

�n2k

��2kk

�ak

(1+n)2k

− 1n

for all a > 0

and n ∈ N∗.

Gyorgy Szollosy

PP24958. In all triangle ABC holds�� sin2 B

2+sin2 C

2

sin2 A2

≥√2 +�

(2R−r)(s2+r2−8Rr)−2Rr2

8Rr2.

Mihaly Bencze

PP24959. Solve in R the equation√60− x+

√65− x+

√156− x = 15.

Gyorgy Szollosy

PP24960. If a, b, c > 0 and a ≥ b+ c then solve the equation√bc+ x+

√ca+ x+

√ab+ x = 3a−b−c

2 .

Mihaly Bencze

PP24961. In all triangle ABC holds�� rb+rc

ra≤ 3�

Rr .

Mihaly Bencze

PP24962. If ak ∈ (0, 1) (k = 1, 2, ..., n− 1) such thatn−1�k=1

ak > 1 then solve

in R the following system:

ax11 + ax2

2 + ...+ axn−1

n−1 = (a1 + a2 + ...+ an)xn

ax21 + ax3

2 + ...+ axnn−1 = (a1 + a2 + ...+ an)

x1

−−−−−−−−−−−−−−−−−−−−axn1 + ax1

2 + ...+ axn−2

n−1 = (a1 + a2 + ...+ an)xn−1

.

Mihaly Bencze


PP24963. If xk ≥ 1 (k = 1, 2, ..., n) , then

�cyclic

x71+6

x62−3x5

2+5x42−5x3

2+3x22−x2+1

≥ 7n.

Mihaly Bencze


arccos x12 + arccos 2x2 = arccos x2

2 + arccos 2x3 = ... =arccos xn

2 + arccos 2x1 =2π3 .

Mihaly Bencze

PP24965. If ak > 0 (k = 1, 2, ..., n) , then�

cyclic

a41+1

2a22−3a2+2≥ 2n n

�n�

k=1

ak.

Mihaly Bencze

PP24966. If ak > 0 (k = 1, 2, ..., n) , then

�cyclic

a61+1

6a42−15a32+20a22−15a2+6≥ n n

�n�

k=1

ak.

Mihaly Bencze

PP24967. If x > 0 and λ ∈ (−∞, 0) ∪ [1,+∞) thenn�

k=1

��x+ k−1

k

� 1λ

�≤�nλ−1 [nx]

� 1λ where [·] denote the integer part.

Mihaly Bencze

PP24968. If ak ≥ 1 (k = 1, 2, ..., n) then� a51+4

a42−2a32+2a23−a3+1≥ 5n.

Mihaly Bencze

PP24969. Let ABC be a triangle. Determine all points M in the plane ofthe triangle such that (

�MA) (

�MA ·MB) ≥�MA ·BC2.

Mihaly Bencze


PP24970. If a (x, y) = x+y2 , a (x, y) =

√xy and h (x, y) = 2

1x+ 1

y

then solve in

R+ the following system

a (x1, x2) + h (x2, x3) = 2g (x4, x5)a (x2, x3) + h (x3, x4) = 2g (x5, x6)−−−−−−−−−−−−−−−−a (xn, x1) + h (x1, x2) = 2g (x3, x4)

.

Mihaly Bencze

PP24971. Let f : R → R be a convex function. Determine all a, b > 0 anda+ b = 1 such that f (x) + f (y) + f (z) +

�f (ax+ by) ≥ 2

�f (bx+ ay)

for all x, y, z ∈ R.

Mihaly Bencze

PP24972. Let f : R → R be a convex function. Determine all a, b, c > 0and a+ b+ c = 1 such thatf (x)+ f (y)+ f (z)+ f (ax+ by + cz)+ f (ay + bz + cx)+ f (az + bx+ cy) ≥≥ 2f (2bx+ 2cy) + 2f (2by + 2cz) + 2f (2bz + 2cx) for all x, y, z ∈ R.

Mihaly Bencze

PP24973. Solve in R the following system(3x1+x2 + 4x3 + 5x4) (3x2+x3 + 4x4 + 5x5) == (3x2+x3 + 4x4 + 5x5) (3x3+x4 + 4x5 + 5x6) = ...= (3xn+x1 + 4x2 + 5x3) (3x1+x2 + 4x3 + 5x4) = 324.

Mihaly Bencze

PP24974. Determine all functions f : R → R for whichf (x) + 2f (y) + f (z) ≥ xf (y) + yf (z) for all x, y, z ∈ R.

Mihaly Bencze

PP24975. Solve in Z the equation x2 + y3 + z4 + t6 = u5.

Mihaly Bencze

PP24976. Let f : R → R be a convex function. Determine all a, b > 0 anda+ b = 1 such that f (x) + f (y) + f (z) + 3f

�x+y+z3

�≥

≥� f (ax+ by) +�

f (bx+ ay) for all x, y, z ∈ R.

Mihaly Bencze


PP24977. Solve in R the following system(2x1 + 3x2) (5− x3) = (2x2 + 3x3) (5− x4) = ... = (2xn + 3x1) (5− x2) = 20.

Mihaly Bencze

PP24978. Determine all n, k ∈ N for which (n+ 1)3 + k3 is prime.

Mihaly Bencze

PP24979. If a, b, c > 1 then solve in R the following system:(ax1 + bx2 + cx3) (a+ b+ c+ x4 + x5 − 2x6) == (ax2 + bx3 + cx4) (a+ b+ c+ x5 + x6 − 2x7) = ...= (axn + bx1 + cx2) (a+ b+ c+ x3 + x4 − 2x5) = (a+ b+ c)2 .

Mihaly Bencze

PP24980. If a, b, c > 0 and a+ b+ c = 1, then� a4

1−a3+ 1

3

� (2a+b)4

27−(2a+b)3≥ 2

3

� (a+2b)4

27−(a+2b)3.

Mihaly Bencze

PP24981. If a, b, c > 0 and a+ b+ 1 = 1, then� a4

1−a3+ 1

4

� (a+1)4

64−(a+1)3≥� (1−a)4

4−(1−a)3.

Mihaly Bencze

PP24982. If a, b, c > 0 and a+ b+ c = 1 then 126 +

� a4

1−a3≥ 1

2

� (1−a)4

3−2a−a2.

Mihaly Bencze

PP24983. If a, b, c > 0 and a+ b+ c = 1, then126 +

� a4

1−a3≥ 1

3

� (2a+b)4

27−(2a+b)3+ 1

3

� (a+2b)4

27−(a+2b)3.

Mihaly Bencze

PP24984. Let be A ∈ Mn (R) where n ≥ 2 such that Ak +Ak+1 = On.Determine all k ∈ N for which In −A (A+ In) is invertable.

Mihaly Bencze


PP24985. Determine all functions f : R → [0,+∞) for which fk isdifferentiable and

�fk�′= f where k ∈ N, k ≥ 2.

Mihaly Bencze

PP24986. Solve in R the equation�6x + 13

1x + 4x+

1x

��13x + 6

1x + 4x+

1x

�= 1225.

Mihaly Bencze

PP24987. If a, b, c > 0 then� a+1

(2a+b+c)2≤ 1

16

� a+1a2

.

Mihaly Bencze

PP24988. Let f : N∗ → N∗ a bijective function. Compute

limn→∞

n�1− f(n)

n

�.

Mihaly Bencze

PP24989. Solve in R the equation4x + 6x + 8x + 2016x = 5x + 7x + 9x + 2013x.

Mihaly Bencze

PP24990. Sove in N the equation 2x+y + (x+ y)2 = 2z+t + (z + t)2 .

Mihaly Bencze

PP24991. Compute

π2�0

x sin 2xdxsin8 x−2 sin6 x+sin4 x−1

.

Mihaly Bencze

PP24992. Compute In =� (2x3+15x2+41x+40)dx

(x2+5x+7)nwhen n ∈ N.

Mihaly Bencze

PP24993. If x1 = 1 and (n+ 1)xn+1 = xn + 1n for all n ≥ 1 then compute

limn→∞

n�1− n2xn

�.

Mihaly Bencze


PP24994. If 0 < a1 < a2 < ... < a2n then solve the equation

(ax1 + ax2n) +�ax3 + ax2n−2

�+ ... =

�ax2 + ax2n−1

�+�ax4 + ax2n−3

�+ ...

Mihaly Bencze

PP24995. If Ak ∈ Mn (C) (k = 1, 2, ..., n) such that A1 +A2 + ...+Am−1 == TAm ;A2 +A3 + ...+Am = TA1 ; ...;Am +A1 + ...+Am−2 = TAm−1 thenm�k=1

Ak = On.

Mihaly Bencze

PP24996. If a, b, c ∈ C, |a| = |b| = |c| = 1 and |a+ b+ c| ≤ 1 then��a2 + bc��+��b2 + ca

��+��c2 + ab

�� ≥ |a+ b|+ |b+ c|+ |c+ a| .

Mihaly Bencze

PP24997. If a, b > 1 then solve in R the following system:

abx1 + bax2 = abx2 + b

ax3 = ... = abxn + b

ax1 = ab + ba.

Mihaly Bencze

PP24998. If xk > 0 (k = 1, 2, ..., n) , then

a� x2+x3+...+xn

x21

+ b�

x21

�1x2

+ 1x3

+ ... 1xn

�≥ 2 (n− 1)

√ab for all a, b > 0.

Mihaly Bencze

PP24999. If xk > 0 (k = 1, 2, ..., n) andn�

k=1

xk = 1, thenn�

k=1

x4k

1−x3k≥ 1

n3−1

(A generalization of problem IX.378, RMT)

Mihaly Bencze

PP25000. If a1 = 2 and (n− 1)xn = (n+ 1) (x1 + x2 + ...+ xn−1) for alln ≥ 2 then prove that

1).n�

k=1

xk21−k = n(n+3)

2 2).∞�n=1

4n−1

x2n

= π2

6 − 1

3).n�

k=1

2k−1

kxk= n

n+1 4).n�

k=1

xkk+1 = 2n − 1

Mihaly Bencze


PP25001. If x1 = a ∈ R and xn+1 = 4xn (1− xn) for all n ≥ 1 thendetermine all a for which x2016 = 0.

Mihaly Bencze

PP25002. If x1 = 2 and (n− 1)xn = (n+ 1) (x1 + x2 + ...+ xn−1) for alln ≥ 2 then compute

1).∞�n=1

1xn

2).∞�n=1

1xnxn+1

Mihaly Bencze

PP25003. Let ABC be a triangle. Solve in R the following system:�x

(x2+1)ma= y

(y2+1)mb= z

(z2+1)mc

xy + yz + zx = 1.

Mihaly Bencze

PP25004. Determine the number of maximal of nonzero terms of the sum�i,j=1

|f (i)− f (j)| for the all possible f : {1, 2, ..., n} → {a1, a2, ..., ak} when

1 ≤ k ≤ n− 1 and a1, a2, ..., ak ∈ R are given.

Mihaly Bencze


n

�2(π+2)

π − 1n2

n�k=1

(n+ k) cos (n+1−k)π2n+1

�

Mihaly Bencze


�5x21 + 3x2 + 2

� �5x22 − 3x3 + 2

�= 31x1x2�

5x22 + 3x3 + 2� �

5x23 − 3x4 + 2�= 31x2x3

−−−−−−−−−−−−−−−−−−−�5x2n + 3x1 + 2

� �5x21 − 3x2 + 2

�= 31xnx1

.

Mihaly Bencze

PP25007. Let ABC be a triangle. Determine all points A1, B1, C1 in theplane of the given triangle such thatMA1 +MB1 +MC1 ≤ 1

2 (MA+MB +MC) + 32MG for all M in the plane

of the given triangle.

Mihaly Bencze


PP25008. If a, b, c > 1 then solve in R the following system:

ax1 + bx2 + cx3 = (a+ b+ c− 3)x4 + 3ax2 + bx3 + cx4 = (a+ b+ c− 3)x5 + 3−−−−−−−−−−−−−−−−−−axn + bx1 + cx2 = (a+ b+ c− 3)x3 + 3

.

Mihaly Bencze

PP25009. If a, b ∈ C are given, then solve the following system(x+ a) y + b = (y + a) z + b = (z + a) t+ b = (t+ a)x+ b = 0.

Mihaly Bencze

PP25010. If xk > 0 (k = 1, 2, ..., n) then

1).�

cyclic

x21

x2≥

n�k=1

xk +1n

��

cyclic

|x1−x2|√x1

�2

2). nn�

k=1

x2k ≥n�

k=1

x2k +2

n(n−1)

��

1≤i<j≤n|xi − xj |

�2

Mihaly Bencze

PP25011. Let ABCDE be a convex pentagon in which DC = DE andBCD∡ = DEA∡ = 90◦. Determine all points F ∈ (AB) such thatFCE∡ = ADE∡ and FEC∡ = BDC∡.

Mihaly Bencze

PP25012. If x0, x1 > 1 and xn+1 = log3 (1 + xn−1 + xn) for all n ≥ 1 thencompute lim

n→∞n (1− xn) .

Mihaly Bencze

PP25013. Let ABC be an acute triangle, AI,BI, CI intersect thecircumcircle in points A1, B1, C1 and AH,BH,CH intersect the circumcirclein points A2, B2, C2. Prove that

�(AI)a (AH)tgA ≤� (A1I)

a (A2H)tgA .

Mihaly Bencze

PP25014. Compute�

1≤i<j≤n

(2i)!(2j)!i!j!(i+j)! .

Mihaly Bencze



�1≤i<j≤n

�i!j!(i+j)!(2i)!(2j)!

�3.

Mihaly Bencze

PP25016. Let ABC be a triangle, the medians AD,BE and CF intersect

the circumcircle in points M,N and P. Prove that (AM+BN+CP )2

AD+BE+CF ≥ 16√3s

9 .

Mihaly Bencze


�� HAHA1

�ctgA≤�

4s2r2

(s2−(2R+r)2)(s2−4Rr−r2)

� s2−4Rr−r2

2sr

when AH,BH,CH

intersect the circumcircle in points A1, B1, C1.

Mihaly Bencze

PP25018. Prove that n! < exp�n(n+1)

2e

�.

Mihaly Bencze

PP25019. Determine all prime pk (k = 1, 2, ..., n+ 1) for which

1p2n+1

=n�

k=1

1p2k.

Mihaly Bencze


n�k=2

�k ln�2k+32k−3

�− 3�, then compute

limn→∞

n

�λ−

n�k=2

�k ln�2k+32k−3

�− 3��

.

Mihaly Bencze

PP25021. Solve in Z the equation�x+ y

2

� �2y + z

3

� �3z + x

4

�= 91

8 .

Mihaly Bencze

PP25022. Determine all prime p for which exist k, n ∈ N such thatnk + k + 1 and (n+ 1)k + k + 1 is divisible by p.

Mihaly Bencze


PP25023. If xk > 0 (k = 1, 2, ..., n) andn�

k=1

xk = 1 then

n�k=1

x3k

x2k+xk+1

≥ 1n2+n+1

.

Mihaly Bencze


k=1

k1k < ne

1e .

Mihaly Bencze


k=1

eke > n(n+1)

2 .

Mihaly Bencze

PP25026. If a, b, c > 0 then�� a3

a2+1

�2+ 1

16

�� (2a+b+c)3

(2a+b+c)2+16

�2≥ 1

2

�� (a+b)3

(a+b)2+4

�2.

Mihaly Bencze

PP25027. If a, b, c, d are the sides of a convex quadrilateral then�� (a+b+c−d)3

(a+b+c−d)2+1

�2≥�

�(a+b)3

(a+b)2+1

�2.

Mihaly Bencze

PP25028. If a, b, c > 0 and a+ b+ c = 3, then34 +�� a3

a2+1

�2≥ 1

9

�� (2a+b)3

(2a+b)2+9

�2+ 1

9

�� (a+2b)3

(a+2b)2+9

�2.

Mihaly Bencze

PP25029. If a, b, c > 0 then�� a3

a2+1

�2+ 1

9

�� (2a+b)3

(2a+b)2+9

�2≥ 2

9

�� (a+2b)3

(a+2b)2+9

�2.

Mihaly Bencze

PP25030. If a, b, c > 0 and a+ b+ c = 3 then34 +�� a3

a2+1

�2≥ 1

2

�� (a+b)3

(a+b)2+4

�2.

Mihaly Bencze


PP25031. Solve in N∗ the following equation�1 + 1

x

� �1 + 2

y

� �1 + 3

z

�= 8.

Determine all solutions in Z∗.


PP25032. If x > 1 thenn�

k=1

1k2

�x+ x2

22+ ...+ xk

k2

�> n

n+1 .

Mihaly Bencze

PP25033. In all triangle ABC holds1). (�

ma)3 ≥�m3

a + 3mambmc

2).��

cos A2

�3 ≥� cos3 A2 + 3s

4R

Mihaly Bencze

PP25034. If a > 0 and a �= 1 thena2n+2−1a2−1

+ a(an−1)a−1 + n+ 1 > an+1−1

a−1 + (n+ 1) a.

Mihaly Bencze

PP25035. Prove that�1 + 6

π

�π �1 + 6

e

�e> 576.

Mihaly Bencze

PP25036. If ak ∈ R (k = 1, 2, ..., n) , n ≥ 3 and a12 < a2

3 < ... < ann+1 then

n�k=1

a1+a2+...+akk2(k+3)ak

< n2(n+1) .

Mihaly Bencze

PP25037. In all triangle ABC holds12

��wa

ma−wa+�

hawa−ha

+�

mama−ha

�≥ wa

ma+ ha

wa+ ma

2ma−haand his

permutations.

Mihaly Bencze


k=1

(k + 1)��

1 + 1k

�k − 1�< n+ 2n−1

6n + 11n(n+1)12 .

Mihaly Bencze


PP25039. Prove that 3n−12·3n−1 +

n�k=1

�1 + 1

3k

�3k−1

< 3n+ 32n−18·32n for all n ∈ N∗.

Mihaly Bencze

PP25040. Determine all x, y ∈ R for which

e2√2 ln a ln b ≤ asinx+sin ybcosx+cos y ≤ e2

√ln2 a+ln2 b when a, b > 0.

Mihaly Bencze


k=1

loga�ak − 1

�loga�ak + 1

�loga�a2k + 1

�< 16n2(n+1)2

27 when a > 1.

Mihaly Bencze

PP25042. If xk > 1 (k = 1, 2, ..., n) , thenn�

k=1

lgn�

n�k=1

xk

�≥

m�k=1

lg xk.

Mihaly Bencze


k=1

�k√1! + k

√2! + ...+ k

√k!�≤ n(2n2+21n+13)

6 .

Mihaly Bencze

PP25044. Compute λ = limn→∞

1

(2nn )

�n�

k=0

(nk)2

n−k+1

�2

and

limn→∞

n

�λ− 1

(2nn )

�n�

k=0

(nk)n−k+1

�2�.

Mihaly Bencze


k=1

�3√1 · 2 + 3

√2 · 3 + ...+ 3

�k (k + 1)

�≤ n(n+1)(n+5)

9 .

Mihaly Bencze


n

�γ − ln 2−

�n�

k=1

(−1)k−1

k

�nk

�− ln 2k

��when

γ = 0, 92 is the Euler constant.

Mihaly Bencze



5− x =�5− y2

�2

5− y =�5− z2

�2

5− z =�5− x2

�2.

Mihaly Bencze

PP25048. Prove that√1 +

√2 + ...+

√n2 − 1 ≥ 2n(n2−1)

3 .

Mihaly Bencze


�n2�arctg(n+1)

n+1 − arctgnn

�− π

2

�.

Mihaly Bencze


n

�ln 2−

n�k=1

sinh 1n+k

�.

Mihaly Bencze


n

�1−

�

1≤i<j≤n

1ij

�

1≤i<j≤n

1(i+1)(j+1)

�.

Mihaly Bencze

PP25052. Solve in M3 (C)×M3 (C) the following system:�X3 + Y 3 = O3

X−1 + Y −1 = I3.

Mihaly Bencze


argzn where zn =�i+ 12

� �i+ 22

�...�i+ n2

�

and limn→∞

n (λ− arg zn) .

Mihaly Bencze

PP25054. Determine all a > 0 for whicha2 ≤ (a− 1)

�a+1a−1

�x+ (a+ 1)

�a−1a+1

�x≤ a3 for all x ∈ [0, 1] .

Mihaly Bencze



k=1

�k

�k�k√k ≤ n(15n+17)

2 .

Mihaly Bencze

PP25056. If A,B,C ∈ Mn (z) , n ≥ 3 are nonsingular matrices such that(A∗B∗)∗ = BA, (B∗C∗)∗ = CB, (C∗A∗)∗ = AC, then

detA+ detB + detC <√52 .

Mihaly Bencze

PP25057. Let A ∈ Mn (R) such that TrA = Tr (A∗) = 0. Determine alln ∈ N for which det (An + nIn) det (A

n + In) ≥ 0.

Mihaly Bencze


1 + 6x2+y2+z2

= 1x + 1

y + 1z

1 + 6y2+z2+t2

= 1y + 1

z + 1t

1 + 6z2+t2+x2 = 1

z + 1t +

1x

1 + 6t2+x2+y2

= 1t +

1x + 1

y

.

Mihaly Bencze

PP25059. Solve in Z the equation 1 + 6x2+y2+z2

= 1x + 1

y + 1z .

Mihaly Bencze

PP25060. Solve in Z the equation 1 + n(n+1)n�

k=1

x2k

=n�

k=1

1xk.

Mihaly Bencze

PP25061. Prove that�

1≤i<j≤n

3

�(n+1)2

n2ij(i+1)(j+1)≤ (n−1)(n+2)

6 .

Mihaly Bencze


w2λa

≥ 3�

2s2−r2−4Rr

�λfor all

λ ∈ (−∞, 0] ∪ [1,+∞) .

Mihaly Bencze


PP25063. In all triangle ABC holds�� a

b+c−a ≥ 4(s2−r2−Rr)s2+r2+2Rr

.

Mihaly Bencze

PP25064. In all triangle ABC holds� ab2

a+b ≤ s2 − r2 − 4Rr.

Mihaly Bencze


�n�

k=1

1|sin k|

�≥ 2016 when [· ]


Mihaly Bencze


�n�

k=1

1|cos k|

�≥ 2016 when [· ]


Mihaly Bencze


�n�

k=1

tg 1k

�≥ 2016 when [· ] denote

the integer part.

Mihaly Bencze


�n�

k=1

ctg 1k

�≥ 2016 when [· ]


Mihaly Bencze

PP25069. Determine all n ∈ N∗ for which ne < (n+1)n+1

nn < (n+ 1) e.

Mihaly Bencze

PP25070. Denote pn the nth prime and xn = p1p1−1 · p2

p2−1 ...pn

pn−1 . Determineall n ∈ N for which xn < 2016.

Mihaly Bencze


PP25071. If xk ∈ [0, 1] (k = 1, 2, ..., n) thenn�

k=1

�xk (1− xk) +

n�k=1

xk ≤ (1+√10)n

2 .

Mihaly Bencze

PP25072. In all triangle ABC holds� 1√

2ab+√3bc+2ca

≥ 98(s2−r2−4Rr)

.

Mihaly Bencze

PP25073. If ak > 0 (k = 1, 2, ..., n) and a12 < a2

3 < ... < ann+1 then

n�k=1

(a1 + a2 + ...+ ak) >n!(n+3)!an13·22n+1 .

Mihaly Bencze

PP25074. In all triangle ABC holds��a2mbmc

bcma+ bcma

mbmc− a�≥ 4Rsr.

Mihaly Bencze

PP25075. If a, b, c > 0 then��a2+b2

a+b

�3≥ 3abc.

Mihaly Bencze

PP25076. If a, b, c > 1, then��a2

bc + bc− a�3

≥ 3abc.

Mihaly Bencze

PP25077. Solve in Z the equation�x2

yz + yz − x��

y2

zx + zx− y��

z2

xy + xy − z�= xyz.

Mihaly Bencze

PP25078. If xk > 0 (k = 1, 2, ..., n) andn�

k=1

xk = 1 then�

1≤i<j≤nxixj ≤ n−1

2n .

Mihaly Bencze

PP25079. If xj > 0 (j = 1, 2, ..., n) andn�

k=1

xk = 1 compute

max�

1≤i1<...<ik≤nxi1xi2 ...xik .

Mihaly Bencze



k=1

k8+k5+1k3+1

≥ n(n+1)2 .

Mihaly Bencze

PP25081. If xk ∈�0, π2�(k = 1, 2, ..., n) , then

�cyclic

11−sinx1 sinx2

≤n�

k=1

1cos2 xk

.

Mihaly Bencze

PP25082. In all triangle ABC holds�� r2a

a

�λ≥ 33λ+1

�r2

2s

�λfor all

λ ∈ (−∞, 0] ∪ [1,+∞) .

Mihaly Bencze

PP25083. In all triangle ABC holds�� a2b

a3+b

�λ≤ sλ

3λ−1 for all λ ∈ [0, 1] .

Mihaly Bencze

PP25084. In all triangle ABC holds��2a2 + b2 + c2

�≥ 4s2

�s2 + r2 + 2Rr

�2.

Mihaly Bencze

PP25085. Prove that 119 +

π1�0

x2tgxdx ≤ 3 ln 22 − 3π2

32 +

π2�0

tg (sinx) dx.

Mihaly Bencze

PP25086. If n ∈ N∗ and f : [0, 1] → R is a continuous function then∞�k=0

�1�0

x2k(k+1)+nf (x) dx

�2

≤ π2

8

1�0

x2nf2 (x) dx.

Mihaly Bencze

PP25087. In all acute triangle ABC holds�

AtgA ≥ 2sπr3(s2−(2R+r)2)

.

Mihaly Bencze


AtgA2 ≥ π(4R+r)

3s .

Mihaly Bencze



Ama�

ma+

�

Aha�

ha+

�

Awa�

wa≤ π.

Mihaly Bencze

PP25090. Determine all continuous functions f : R∗ → R∗ for which2�1

f (x) dx+ e52�1

dxf(x) ≤ e4 + e.

Mihaly Bencze


n

�3 ln 3− π√

3− 2

n�k=1

1k(3k−2)

�.

Mihaly Bencze

PP25092. Solve in C the following system 1x + 2

y−1 + 3z−2 + 4

t−3 =

= 1y + 2

z−1 + 3t−2 + 4

x−3 = 1z + 2

t−1 + 3x−2 + 4

y−3 = 1t +

2x−1 + 3

y−2 + 4z−3 = 1.

Mihaly Bencze


�

�

tgA2+�

tgC2

�

�

tgB2

1+3tgB2

�

tgA2tgC

2

≥√3.

Mihaly Bencze

PP25094. In all triangle ABC holds 272 ≤� ctgA

2ctgB

2

1−tgA2tgB

2

≤ s2

2r2.

Mihaly Bencze

PP25095. If xk ≥ 0 (k = 1, 2, ..., n) andn�

k=1

xk = n then

n�k=1

exk+2e2xk+exk+1

≥ n(e+2)e2+e+1

.

Mihaly Bencze

PP25096. If a, b ∈ [0, 1] then

a+2a2+a+1

+ b+2b2+b+1

≥ 1 + ab+2(ab)2+ab+1

+(1−a)(1−b)(1−ab)(

√a+2

√b)(

√b+2

√a)

√ab

(a2+a+1)(b2+b+1)((ab)2+ab+1).

Mihaly Bencze


PP25097. If ak > 0 (k = 1, 2, ..., n) then�

cyclic

a1an−11 +(n−1)a2a3...an

≤

�

n�

k=1ak

�2

n2n�

k=1ak

.

Mihaly Bencze


sin2 A cos2 A2

≥ 163 .

Mihaly Bencze

PP25099. If ak > 0 (k = 1, 2, ..., n) then� a1an−12 (a2+...+an)

n ≥ nn

(n−1)n(�

a1a2)n−1 .

Mihaly Bencze

PP25100. If ak > 0 (k = 1, 2, ..., n) then��

cyclic

a1a2

�2

≥ 2

�n�

k=1

ak

��n�

k=1

1ak

�− n2.

Mihaly Bencze

PP25101. If 1 ≥ b ≥ 0 and a1 = 0 and 2an+1 = b+ a2n for all n ≥ 1 thencompute lim

n→∞n�1−

√1− b− xn

�.

Mihaly Bencze

PP25102. If ak > 1 (k = 1, 2, ..., n) then�

ax1 loga1a2a3...an =

�n�

k=1

ak

�x

.

Mihaly Bencze

PP25103. If xk > 0 (k = 1, 2, ..., n) and

fa (x1, x2, ..., xn) =(xa

1+xa2+...+xa

k)(xa2+xa

3+...+xak+1)...(x

an+xa

1+...+xak−1)

n�

k=1aak

when

k ∈ {1, 2, ..., n} then fa (x1, x2, ..., xn) ≥ fb (x1, x2, ..., xn) for all a ≥ b > 0.

Mihaly Bencze

PP25104. If ak > 0 (k = 1, 2, ..., n) then� an1

a2a3...an≥

n�k=1

ak.

Mihaly Bencze


PP25105. If ap > 0 (p = 1, 2, ..., n) and k ∈ {1, 2, ..., n− 1} , then�

cyclic

(a1a2...ak)n+1

(a1a2...an)k ≥� a1...ak.

Mihaly Bencze

PP25106. In all triangle ABC holds2(s2−r2−Rr)s2+r2+2Rr

+ 3

�2Rr

s2+r2+2Rr≥ 2.

Mihaly Bencze

PP25107. In all triangle ABC holds s2+r2

2Rr + 3�

rR ≥ 8.

Mihaly Bencze

PP25108. Solve in Z the equation x3

y+z2+ y3

z+x2 + z3

x+y2= 3a2

1+a when

a ∈ Z ∈ \ {−1} is given.

Mihaly Bencze

PP25109. In all acute triangle ABC holdsR(s2+8Rr+7r2)−2r(s2−r2)

r(s2+2Rr+r2)+ 3

�R(s2−(2R+r)2)

r(s√2+r

√2+2Rr)

≥ 2.

Mihaly Bencze

PP25110. If x, y, z > 1 then�

logxy z +3

��logxy z ≥ 2.

Mihaly Bencze

PP25111. If x, y, z > 0 then� x2+yz

x(y+z) + 2 3

�xyz

(x+y)(y+z)(z+x) ≥ 4.

Mihaly Bencze

PP25112. If xk > 0 (k = 1, 2, ..., n) then

�cyclic

x1x2+...+xn

+n

��n�

k=1xk

�

cyclic(x1+...+xn−1)

≥ n+1n−1 .

Mihaly Bencze

PP25113. If a ≥ 1 then 2�a9 + 2a

�≥ 3a3 + 2

�a6 − 1

�√a6 − 1.

Mihaly Bencze



x+ 12 3√y ≥ 3

2 + (z − 1)√t− 1

y + 12 3√z

≥ 32 + (t− 1)

√x− 1

z + 12 3√t

≥ 32 + (x− 1)

√y − 1

t+ 1

2√

3√x≥ 3

2 + (y − 1)√z − 1

.

Mihaly Bencze


k=2

√k−1

2k52−3k

32+k

76≤ n−1

2n

Mihaly Bencze

PP25116. If ak > 1 (k = 1, 2, ..., n) then

4n�

k=1

1x2k+

n�k=1

�2x8

k−3x2k+4

x6k−1

�2≥ 4

n3

�n�

k=1

ak

�4

.

Mihaly Bencze

PP25117. Prove that 4n+n�

k=2

�2k3−3k+4k

13

k2−1

�2

≥ 2n(n+1)(2n+1)3 .

Mihaly Bencze

PP25118. If x, y, z > 0 then(x+ y) (y + z) (z + x) ≥ 8xyz + 1

3 (�√

x |y − z|)2 .

Mihaly Bencze

PP25119. If x, y, z > 0 then�� x

y+z

�2+ 14xyz

(x+y)(y+z)(z+x) ≥ 4.

Mihaly Bencze

PP25120. If ak > 1 (k = 1, 2, ..., n) and λ ≥ 1 thenn�

k=1

ak +1

n�

k=1

aλk

≥n�

k=1

1aλ−1k

.

Mihaly Bencze


PP25121. If a ∈�0, π2�then

3

(1+sin2 a)(1+cos2 a)+ 2 sin (sin a+ cos a) cos (sin a− cos a) < 1

sin2 a cos2 a.

Mihaly Bencze

PP25122. Determine all x, y, z ∈ Z for which

sin 2x < 1y2(2−z2)

sin 2y < 1z2(2−x2)

sin 2z < 1x2(2−y2)

.

Mihaly Bencze

PP25123. If xk > 0 (k = 1, 2, ..., n) , thenn�

k=1

x2n−3k ≥ 1

2n−2 |(x1 − x2) (x2 − x3) ... (xn − x1)|+

+ n−1

��xn−11 x2...xn−1

� ��x1x

n−12 x3...xn−1

�...��

x1x2...xn−2xn−1n−1

�.

Mihaly Bencze


(tgA2+tgC

2 )2tg2 B

2

≥ (4R+r)3−12s2R36 .

Mihaly Bencze

PP25125. In all triangle ABC holds� ra

2rb+rc+ s2

3((4R+r)2−2s2)≥ 4

3 .

Mihaly Bencze

PP25126. In all triangle ABC holds� −a+b+c

3a−b+c + r(4R+r)3(s2−2r2−8Rr)

≥ 43 .

Mihaly Bencze

PP25127. If ak ∈ (0, 1] (k = 1, 2, ..., n) then determine all λ ∈ R for which

n�k=1

(1−ak)λ+1

aλk≤

�

1−n�

k=1ak

�λ+1

n�

k=1aλk

.

Mihaly Bencze

PP25128. If a, b > 0 then�a5 + 4a2b3 + b6

� �b5 + 4b2a3 + a6

�≥�

ab�a3 + 2b3

�+ b2

�2a3 + b3

�� ab�b3 + 2a3

�+ a2

�2b3 + a3

��.

Mihaly Bencze


PP25129. If x ∈�0, π2�then

sinn x1+sinx+...+sinn x + cosn x

1+cosx+...+cosn x + (n−1)(sin x+cosx)(n+1) sinx cosx ≥ 2n

n+1 .

Mihaly Bencze


x21 + 2 {x2}2 = 5 [x3]2

x22 + 2 {x3}2 = 5 [x4]2

−−−−−−−−−x2n + 2 {x1}2 = 5 [x2]

2

where

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

Mihaly Bencze


1�0

1�0

ln (xn + ey) ln (yn + ex) dxdy.

Mihaly Bencze

PP25132. Determine all a, b, c ∈ N for which xm,n = (am)!(bn)!m!n!(m+n)! and

xm,n−1 + xm−1,n = cxm−1,n−1 for all m,n ∈ N∗.

Mihaly Bencze

PP25133. If n ∈ N , n ≥ 2 and 1 = d1 < d2 < ... < dk = n are the divisorsof n, then determine all r ∈ {1, 2, ..., n} for which n is prime if and only ifd1d2...dr + d2d3...dr+1 + ...+ dnd1...dr−1 = nk (r − 1) .

Mihaly Bencze

PP25134. In all scalene triangle ABC holds� m2a

(ma−mb)2(ma−mc)

2 ≥ 23(s2−r2−4Rr)

Mihaly Bencze

PP25135. Determine all a, b, c ∈ Z for which the equation�a2 + b2

�x2 − 2

�b2 + c2

�x−�c2 + a2

�= 0 have rational roots.

Mihaly Bencze


tgA2 tg

B2 + 3

√3�

ctgB2

�ctgA

2 ctgC2 ≥ 28

√3.

Mihaly Bencze


PP25137. Solve in Z the equation x1x2(1+x3)

+ x2x3(1+x4)

+ ...+ xnx1(1+x2)

= nk+1

when k ∈ Z\ {−1} is given.

Mihaly Bencze


√n (2n−1)!!

(2n)!! then compute limn→∞

n�λ−√

n · (2n−1)!!(2n)!!

�.

Mihaly Bencze

PP25139. If xk ∈�0, π2�(k = 1, 2, ..., n) then

n�k=1

11−(sinxk)

n−1 ≥ �cyclic

11−sinx1 sinx2... sinxn−1

.

Mihaly Bencze


k=1

��1

3√

k(k+1)− sin 1

3√

k(k+1)

�� <6nn+1 .

Mihaly Bencze

PP25141. In all triangle ABC holds�� b+c

2

�2cos2 A

2 ≤ 6�s2 − r2 − 4Rr

�.

Mihaly Bencze

PP25142. If xk > 0 (k = 1, 2, ..., n) then�

cyclic

�x1 −

√x1x2 + x2

�2 ≥n�

k=1

x2k.

Mihaly Bencze

PP25143. If xk > 0 (k = 1, 2, ..., n) then determine all p ∈ N for whichn�

k=1

xk ≥ p n

�1n

n�k=1

xnk + (n− p) n

�n�

k=1

xk.

Mihaly Bencze

PP25144. If λ > 0 and zk ∈ C∗ (k = 1, 2, ..., n) , thenn�

k=1

��zk + λzk

��4≥ 4nλ2 + 8λ

n�k=1

Re�z2k�.

Mihaly Bencze


PP25145. Determine all a, b, n,m ∈ N for which(an)! + (bm)! ≤ (n!)a+1 + (m!)b+1 .

Mihaly Bencze

PP25146. Determine all m,nk ∈ N (k = 1, 2, ..., n) for whichm�k=1

1((nk)!)

3 ≤ m�

n�

k=1nk

�

!.

Mihaly Bencze

PP25147. Let AkBkCk (k = 1, 2, ..., n) be triangles. Prove that

�wa1wa2 ...wan ≥ 3

�3�

cyclic

sa1ra1 .

Mihaly Bencze

PP25148. If ak ∈ [0, 1] (k = 1, 2, ..., n) then

�cyclic

√a1a2 +

�cyclic

�a1+a21+a1a2

≤ 2n�

k=1

√ak.

Mihaly Bencze

PP25149. In all triangle ABC holds� ra√

r2a+2(rb+rc)2≥ 1.

Mihaly Bencze

PP25150. If xk > 0 (k = 1, 2, ..., n) , then��{x1}2

x2+ [x1]

2

x3

�≥

n�

k=1x2k

n�

k=1xk

when

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

Mihaly Bencze

PP25151. Determine all polynomials P for which the zeros of P are inarithmetical progression, and the zeroes of P ′ are too in arithmeticalprogression.

Mihaly Bencze


PP25152. Determine all polynomials P for which the zeros of P are ingeometrical progression, and the zeroes of P ′ are too in geometricalprogression.

Mihaly Bencze

PP25153. Determine all postve integers k, n for which the numberk (k − 1) ...10k (k − 1) ...10...k (k − 1) ...1 is prime.

Mihaly Bencze

PP25154. Let x be a nonzero real number sarisfying the equationa0x

n + a1xn−1 + ...+ an−1x+ an = 0, where a0, a1, ..., an ∈ Z such that

|a0|+ |a1|+ ...+ |an| > 1. Determine all n ∈ N∗ for which|x| > 1

|a0|+|a1|+...+|an| .

Mihaly Bencze

PP25155. Let ABC be a triangle. Determine all λ ∈ R for which�a2�

λra

− rrbrc

�= 4 (R+ (λ+ 1) r) .

Mihaly Bencze

PP25156. Consider n distinct points in the plane, n ≥ 3 arranged such thatthe number r (n) of segments of length d is maximed. Prove that

π2

2 − 154 <

∞�n=3

1r(n) .

Mihaly Bencze


rstgB

2+2tgA

2tgC

2+2(tgA

2tgC

2 )2 ≥ s2

r(4R+r) .

Mihaly Bencze

PP25158. Solve in Z the following system:

�x+ y2

� �y2 + z

�= (z + x)3�

y + z2� �

z2 + x�= (x+ y)3�

z + x2� �

x2 + y�= (y + z)3

.

Mihaly Bencze



x4 − y + 1 = z2

y4 − z + 1 = x2

z4 − x+ 1 = y2.

Mihaly Bencze

PP25160. Determine all xk ∈ N∗ (k = 1, 2, ..., n) for which nx1x2...xn−1−11+x1x2...xn

isinteger.

Mihaly Bencze

PP25161. If a, b, c > 0 then� 1√

a3+2b3+6abc≤ 1√

abc.

Mihaly Bencze

PP25162. Denote pn the nth prime number. If n ≥ k (k + 1) then pn > knfor all k ∈ N∗, k ≥ 2.

Mihaly Bencze

PP25163. If a, b, c > 0 and� a2+1

a = 4 then

a√bc+ b

�c�1a + 1

b +1c

�+ c�a�1a + 1

b +1c

�+�1a + 1

b +1c

�√ab ≤

≤ 2�1 +�abc�1a + 1

b +1c

��.

Mihaly Bencze

PP25164. If xk ∈ R thenn�

k=1

|cosxk|+�

cyclic

|cos (x1 − x2)| ≥ n− 1.

Mihaly Bencze


n

�14 − ln

�12n

n�k=1

�2 + k

n2

��.

Mihaly Bencze

PP25166. In all triangle ABC holds� sin A

2

(cos A2 )

2 ≥ 2.

Mihaly Bencze


PP25167. If b1, b2, ..., bn is a rearangement of numbers a1, a2, ..., an > 0 then

�1≤i<j≤n

aibjai+bj

≤ n

2n�

k=1ak

��

1≤i<j≤naibj

�.

Mihaly Bencze

PP25168. Let ABC be a triangle. Determine all λ > 0 for which� (a+b−c)λ

aλ+bλ−cλ≤ 3.

Mihaly Bencze


x21 +�x22 + 32 =

�x23 + 96

x22 +�x23 + 32 =

�x24 + 96

−−−−−−−−−−−−x2n +

�x21 + 32 =

�x22 + 96

Mihaly Bencze

PP25170. In all triangle ABC holds��a(b+c)

a2+bc≤�

2(5s2+r2+4Rr)s2+r2+2Rr

.

Mihaly Bencze

PP25171. Let n ≥ 2 be an integer. Find the number of integers k with0 ≤ k < n and such that k3 leaves a remainder of 1 when divided by n.

Mihaly Bencze


�x2 − 2y + 6 log3 (9− 2z) = x�y2 − 2z + 6 log3 (9− 2x) = y√z2 − 2x+ 6 log3 (9− 2y) = z

.

Mihaly Bencze

PP25173. If λ ∈ (0, 1) then determine all x ∈ R for which

�(a+ x) (b+ x) = x+

�aλ+bλ

2

� 1λ.

Mihaly Bencze


PP25174. If an+1 = [an] {an} for all n ≥ 0 when [·] and {·} denote theinteger part respectively the fractional part. Determine all a0 ∈ R for whichthe sequence (an)n≥0 is periodical.

Mihaly Bencze

PP25175. If M is a set for which cardM = 2n, then the number ofpartitions of M formed by 2-subsets of M is (2n)!

2n·n! .

Mihaly Bencze

PP25176. Let ABC be a triangle. Determine the best constant λ > 0 for

which�

R2r ≥ λ3a2b2c2

�

(λa2−(b−c)2).

Mihaly Bencze

PP25177. Prove that�2k+1

2k

�−�

2k

2k−1

�is divisible by 8k for all k ∈ N.

Mihaly Bencze

PP25178. Determine all functions f : (0,+∞) → (0,+∞) for which(f (x) + f (y) + 2xf (xy)) f (x+ y) = f (xy) for all x, y > 0.

Mihaly Bencze

PP25179. Solve in R the equationn�

k=1

�k3

x

�= n2(n+1)2

4 when [·] denote the

integer part.

Mihaly Bencze

PP25180. If x ∈�0, π2�then�

n�k=1

1+(sinx)4kn

1+(sinx)8kn

��n�

k=1

1+(cosx)4kn

1+(cosx)8kn

�< 1

sin2 x cos2 x.

Mihaly Bencze

PP25181. Solve in R the following equationn�

k=1

�kx

�= n(n+1)

2 when [·]denote the integer part.

Mihaly Bencze


PP25182. Solve in R the following equationn�

k=1

�k2

x

�= n(n+1)(2n+1)

6 when

[·] denote the integer part.

Mihaly Bencze


�n�

k=1

arctgkk2+1

− 12arctg

2n

�and

limn→∞

n

�λ+ 1

2arctg2n−

n�k=1

arctgkk2+1

�.


PP25184. In all triangle ABC holds� √

sinA+√sinB

�

cos A−B2

cos C2+�

|sin A−B2 | sin C

2

≤ 6.

Mihaly Bencze

PP25185. In all triangle ABC holds−195

8 ≤ 3�

cos 2A− 8�

sinA cosB + 5(2R−s+2r)R ≤ 60.


PP25186. Ifn�

k=1

|shkx| ≥ n(n+1)2 |shx| .

Mihaly Bencze

PP25187. If xk > 0 (k = 1, 2, ..., n) andn�

k=1

xk = 1 thenn�

k=1

xarctgxkk ≥ 1.

Mihaly Bencze

PP25188. In all acute triangle ABC holds��√tgA+ 4ctgB +

√tgB + 4ctgA+

√tgA+ 9ctgB +

√tgB + 9ctgA

�≥

≥ 21√2.

Mihaly Bencze


2�

tg2A2 tg2B2 ≥ 1 +

�tg2B2

�12

�tg4A2 + tg4C2

�.

Mihaly Bencze


PP25190. Solve the following system

x41 + 35x22 + 24 = 10x3�x24 + 5

�

x42 + 35x23 + 24 = 10x4�x25 + 5

�

−−−−−−−−−−−−−−x4n + 35x21 + 24 = 10x2

�x23 + 5

�.

Mihaly Bencze

PP25191. If x ∈ R then 2ex

e2x+1+ e−x

�e4x+1

2 ≤ (ex+1)2

2ex +((x−1)e2x+x+1)

2

2x2ex(e2x+1).

Mihaly Bencze


k=1

k

(k+2)�

k√1·2+ k√2·3+...+ k√

k(k+1)� ≥ n

n+3 .

Mihaly Bencze


1).� 1

1−sinA sinB ≤�

s2+r2−4R2

s2−(2R+r)2

�2− 8R(R+r)

s2−(2R+r)2

2).� 1

1−cosA cosB ≤�s2+r2+Rr

2sr

�2− 4R

r

Mihaly Bencze

PP25194. If ak > 0 (k = 1, 2, ..., n) andn�

k=1

ak = 1 then

n�k=1

�1 + 1

ak

�≥ (n+ 1)n .

Mihaly Bencze

PP25195. Determine all a, b, c > 0 for which (a+ b+ c)�1a + 1

b +1c

�≤ 81

8 .

Mihaly Bencze

PP25196. Let ABCD be a convex quadrilateral. Determine min and maxof the following expression

�sin A

2 cos B2 tg

C4 .

Mihaly Bencze

PP25197. If ak, bk, ck ∈ N\ {0, 1} thenn�

k=1

�1− 1

a2k

�k �1− 1

b2k

�k �1− 1

c2k

�k> 8n−1

7·8n .

Mihaly Bencze



R+rR <

� 1A2

�2A�A

sinxx dx

��2A�A

xdxsinx

�< s2+r2−4R2

2(s2−(2R+r)2).

Mihaly Bencze

PP25199. Solve in N the equation xy + yx = (x− y)2(x+y) .

Mihaly Bencze


�1 + 3

1x

�(13− 7y) = 16 (2 + 7z)�

1 + 31y

�(13− 7z) = 16 (2 + 7x)�

1 + 31x

�(13− 7x) = 16 (2 + 7y)

.



k=1

(1 + xk) = ynn�

k=1

xk.

Mihaly Bencze


2x2 + 3 ≥ 3y + 2√z2 − z + 1

2y2 + 3 ≥ 3z + 2√x2 − x+ 1

2z2 + 3 ≥ 3x+ 2�y2 − y + 1

.

Mihaly Bencze

PP25203. If zp ∈ C∗ (p = 1, 2, ..., n) such that |z1| = |z2| = ... = |zn| anda1 =

z1+z2z1−z2

, a2 =z2+z3z2−z3

, ..., an = zn+z1zn−z1

then�a1a2...a2k +

�a1a2...a2k−2 + ...+ 1 = 0 if n = 2k + 1 and�

a1a2...a2k−1 +�

a1a2...a2k−3 + ...+�

a1 = 0 if n = 2k.

Mihaly Bencze

PP25204. If a > 1 and xk > 0 (k = 1, 2, ..., n) then

n

�n�

k=1

loga xk ≤ loga

�1n

n�k=1

xk

�.

Mihaly Bencze


PP25205. Solve in Z the equationn�

k=1

xk−1xk+1 = n(m−1)

m+1 when m ∈ Z\ {−1} is

given.

Mihaly Bencze


1

[ 3√n3+2n+1]2 when [·] denote the integer part.

Mihaly Bencze

PP25207. If x0 = x1 = 1 and n (n+ 1)�xn+1

√xn−1 − xn

√xn�=

√xn−1xn

for all n ≥ 1 thenn�

k=1

√xk

k(2k−1) ≤n

n+1 .

Mihaly Bencze


c�1 + sin A

2

�2 �1 + sin B

2

�2 ≤≤ 3(s2−3r2−12Rr)(3s2+3r2−4Rr)+s2(−9s2+36r2+144Rr)

4sRr ≤≤ 3(s2−3r2−12Rr)(3s2+3r2−4Rr)+9s2(−s2+4r2+16Rr)

4sRr .

Mihaly Bencze

PP25209. If x0 = x1 = 1 and n (n+ 1)�xn+1

√xn+1 − xn

√xn�=

√xn−1xn

for all n ≥ 1 thenn�

k=1

�(k + 1)

√xk�3 ≤ n2

�2n2 − 1

�.

Mihaly Bencze

PP25210. If a, b, c, d > 0 then��

1 + 1a

�a�2 ≥ 3��

1 + 1a

�2a.

Mihaly Bencze

PP25211. If a, b > 0 thenn�

k=1

�a

1k b

k−1k + a

k−1k b

1k

�≤ n (a+ b) .

Mihaly Bencze

PP25212. If a, b, c > 0 then��a+b

2

�n ≥��√

ab�n

+ 13·2n��

√an −

√bn��2

for all n ∈ N.

Mihaly Bencze


PP25213. If ak > 0 (k = 1, 2, ..., n) then�

cyclic

�a21 + a22 ≤

√2

n�k=1

ak +√24

�cyclic

(a1−a2)2

a1+a2.

Mihaly Bencze

PP25214. If ak > 0 (k = 1, 2, ..., n) , then determine all n ∈ N for which�

cyclic(a1−a2)

2

2n+1 max{a1,a2,...,an} ≤n�

k=1

ak −�

cyclic

√a1a2 ≤

�

cyclic(a1−a2)

2

2n+1 min{a1,a2,...,an} .

Mihaly Bencze

PP25215. If a, b, c > 0 then determine all λ1,λ2,λ3 > 0 for whicha2

λ1a+λ2b+λ3c+ b2

λ1b+λ2c+λ3a+ c2

λ1c+λ2a+λ3b≥ a+b+c

λ1+λ2+λ3.

Mihaly Bencze

PP25216. If 1 ≤ xk ≤ 2 (k = 1, 2, ..., n) , thenn2(n+1)2

8 ≤n�

k=1

(x1 + x2 + ...+ xk)�

1x1

+ 1x2

+ ...+ 1xk

�2≤ n2(n+1)2

4 .

Mihaly Bencze

PP25217. If a, b, c > 0 then determine all λ ∈ R for which� a

a+λb+c ≥ λ+1λ+2 .

Mihaly Bencze

PP25218. If x, y, z > 0 and x+ y + z = 1 then� x1−x4 + 81

80 ≥ 16� 1−x

16−(1−x)4.

Mihaly Bencze

PP25219. If x, y, z > 0 and x+ y + z = 1, then� x1−x4 + 81

80 ≥ 27� 2x+y

81−(2x+y)4+ 27

� x+2y

81−(x+2y)4.

Mihaly Bencze


� 2x+y

81−(2x+y)4≥ 54

� x+2y

81−(x+2y)4.

Mihaly Bencze



� 1+x256−(1+x)4

≥ 16� x+y

16−(x+y)4.

Mihaly Bencze

PP25222. Solve in R the following system:� √x4 + 1 +

�y4 + 1 +

√z4 + 1 = 3

√2

xy + yz + zx = 1

Mihaly Bencze

PP25223. If a, b, c ∈ C then

3�|a|2 + |b|2 + |c|2 + |a+ b+ c|2

�≥ (|a+ b|+ |b+ c|+ |c+ a|)2 .

Mihaly Bencze

PP25224. If a, b, c ∈ C then

4�|a+ b|2 + |b+ c|2 + |c+ a|2

�≥ (|a|+ |b|+ |c|+ |a+ b+ c|)2 .

Mihaly Bencze

PP25225. If ak ∈ [0, 1] (k = 1, 2, ..., n) then�

cyclic

a11+a2+a3+...+an

+n�

k=1

(1− ak) ≤ 1.

Mihaly Bencze

PP25226. If x, y, z > 0 and λ ≥ 1 then��

1 + x1λ

�λ+ 3

�1 +�x+y+z

3

� 1λ

�λ

≥ 2��

1 +�x+y

2

� 1λ

�λ

.

Mihaly Bencze

PP25227. If ak > 1 (k = 1, 2, ..., n) (n ≥ 3) then solve in R the following

system:

n�k=1

axk =

�n�

k=1

ak − n

�y + n

n�k=1

ayk =

�n�

k=1

ak − n

�z + n

n�k=1

azk =

�n�

k=1

ak − n

�x+ n

.

Mihaly Bencze


PP25228. If a, b, c ∈ C are given, then solve the following system:(x1 + a)x2 + b = (x2 + a)x3 + b = ... = (xn + a)x1 + b = c.

Mihaly Bencze

PP25229. Let ABC be a triangle and MNK be the triangle for whichMN = ma, NK = mb, KM = mc. Let T ∈ Int (MNK) . Prove that(TM + TN + TK) (TM · TN + TN · TK + TK · TM) ≥≥ TM ·m2

a + TN ·m2b + TK ·m2

c .

Mihaly Bencze

PP25230. Let A1A2...An be a convex polygon, and M ∈ Int (A1A2...An) .

Prove that 12

� MA21+MA2

2−A1A22

MA1·MA2+ n

n−1 ≥ 0.

Mihaly Bencze

PP25231. Denote x1 < 0 < x2 the roots of equationx4 + a1x

3 + b1x2 + c1x+ d1 = 0. Determine all ak, bk, ck, dk ∈ C (k = 1, 2) for

which x1x2 is a root of the equation x6 + x4 + a2x3 + b2x

2 + c2x+ d2 = 0.

Mihaly Bencze

PP25232. Let ABC be a triangle. Solve in R the following system:

�x

(x2+1) cos A2

= y

(y2+1) cos B2

= z(z2+1) cos C

2

xy + yz + zx = 1

Mihaly Bencze

PP25233. Let ABC be a triangle with sides a, b, c. Solve in R the followingsystem:

�x

(x2+1)a= y

(y2+1)b= z

(z2+1)c

xy + yz + zx = 1.

Mihaly Bencze

PP25234. In all triangle ABC holds 8 (�

ma) (�

mamb) ≥ 27�

a2ma.

Mihaly Bencze


PP25235. If x0, y0, z0 > 0 and 2xn+1 = xn + 1yn; 2yn+1 = yn + 1

zn,

2zn+1 = zn + 1xn

for all n ≥ 1 then the sequences (xn)n≥1 , (yn)n≥1 and(zn)n≥1 are convergent and compute its limits.

Mihaly Bencze

PP25236. If ak, bk ∈ R (k = 1, 2, ..., n) and c ≥n�

k=1

(ak − bk)2 then

vs�n�

k=1

a2k

��n�

k=1

b2k − c

�+

�n�

k=1

b2k

��n�

k=1

a2k − c

�≤ 2

�n�

k=1

akbk

�2

.

Mihaly Bencze

PP25237. If ak ≥ −1 (k = 1, 2, ..., n) , then 9n�

k=1

a3k +3n�

k=1

a2k + n ≥ 5n�

k=1

ak.

Mihaly Bencze

PP25238. Determine all triangles ABC and all points M ∈ (BC) andK ∈ (BC) such that

�AB2 ·MC +AC2 ·MB

� �AB2 ·DK +AC2 ·DB2

�=

=�AB2 ·MC2 +AC2 ·MB2

� �AB2 ·KC +AC2 ·KB

�.

Mihaly Bencze

PP25239. If ak > 0 (k = 1, 2, ..., n) andn�

k=1

ak = 1 then

nn�

k=1ak

≤ �cyclic

11+a1+a2+...+an−1

≤ 1.

Mihaly Bencze

PP25240. If ak > 0 (k = 1, 2, ..., n) andn�

k=1

ak = n then

n�k=1

n−1−a3kak

≥ n (n− 2) .

Mihaly Bencze


PP25241. If ak > 0 (k = 1, 2, ..., n) then

1).n�

k=1

ak +�

cyclic

a1(a2+...+an)

2 ≥ 2nn−1

2).n�

k=1

a2k +�

cyclic

a21(a2+...+an)

4 ≥ 2n(n−1)2

Mihaly Bencze

PP25242. Prove that12

n�k=1

1k2

+ 12

n�k=1

�12 + 3

4 + ...+ kk+1

�2+

n�k=1

1k

�12 + 1

3 + ...+ 1k+1

�≥ n.

Mihaly Bencze

PP25243. If n, k ∈ N∗ then

12k

+ 13k

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

≤ n

�1− n

�(2k−1)(3k−1)...((n+1)k−1)

((n+1)!)k

�.

Mihaly Bencze


bc�1 + sin A

2

�2 ≤ s2 + 6r2 + 24Rr.

Mihaly Bencze


bc sin A2 ≤ 2r (4R+ r) .

Mihaly Bencze


(a− b)2 + 3�abc�1 + sin A

2

� �1 + sin B

2

� �1 + sin C

2

�� 23 ≤ 3s2.

Mihaly Bencze

PP25247. If ak ∈ N∗ (k = 1, 2, ..., n) such thatn�

k=1

a2k is a perfect square,

then exist p, q, r ∈ N∗ for which

n�

k=1ak

�

1≤i<j≤n

aiaj

2

= 1p + 2

q +1r .

Mihaly Bencze


PP25248. If ak ∈ N∗ (k = 1, 2, ..., n) such thatn�

k=1

a2k is a perfect square,

then exist br ∈ N∗ (r = 1, 2, ...,m+ 1) such that

n�

k=1ak

�

1≤i<j≤n

aiaj

m

=m+1�r=1

1br.

Mihaly Bencze

PP25249. If x0 = x1 = 1 and n (n+ 1)�xn+1

√xn−1 − xn

√xn�=

√xn−1xn

for all n ≥ 1, then compute limn→∞

n (4− xn) .

Mihaly Bencze


1).��

1 + sin A2

�2 ≤ s2

2Rr

2).��

1 + sin A2

�4 ≤ s2(s2−r2−4Rr)8R2r2

3).�

sin A2 ≤ s2+r2+4Rr

4Rr

4).��

1 + sin A2

�≤ s2

4Rr

Mihaly Bencze

PP25251. If a, b, c > 0 then 3�

a2�

ab ≥� 1a+b +

16

�� c(a−b)2

(a+c)(b+c)�

ab

�2

.

Mihaly Bencze

PP25252. In all triangle ABC holds� b2+c2

(2a2+2c2−b2)(2a2+2b2−c2)≤ s2−r2−4Rr

9s2r2.

Mihaly Bencze

PP25253. Determine all ak ∈ N∗ (k = 1, 2, ..., n) for whichn�

k=1

3k2+3k+1a3ka

3k+1

=n(n2+3n+3)

(n+1)3.

Mihaly Bencze


2[log2 x1] + x[log2 x3]2 = 2[log2 x2] + x

[log2 x4]3 = ... = 2[log2 xn] + x

[log1 x2]2 = 2 where


Mihaly Bencze


PP25255. Prove thatn−1�k=1

1cosx−cos kπ

n

= cosxsin2 x

− n cos(nx)sinx sinnx .

Mihaly Bencze

PP25256. Let a0, a1, ..., an−1 ∈ N∗ such that a0 + a1 + ...+ an−1 = n. Provethat k0a0 + k1a1 + ...+ kn−1an−1 =

kn−1k−1 if and only if

a0 = a1 = ... = an−1 = 1, when k ≥ 2.

Mihaly Bencze

PP25257. Let 1 = d1 < d2 < ... < dk = n the positive divisors of n. Provethat d1, d2, ..., dk are not in geometrical progression.

Mihaly Bencze

PP25258. In all triangle ABC holds r�s2 − (2R+ r)2

�≤ 4R3.

Mihaly Bencze

PP25259. Determine all functions f : N → N for which

f�x4 − x3f (y) + x2f2 (y)− xf3 (y) + f4 (y)

�=

x5+f(y5)x+y for all x, y ∈ N.

Mihaly Bencze


k=1

1

(k+2)((2k)!)1k≥ n(n+3)

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

Mihaly Bencze

PP25261. Determine all bijective functions f : [0,+∞) → [0,+∞) forwhich f (x+ y + z) =f (x)+ f (y)+ f (z)+2f−1 (f (x) f (y))+2f−1 (f (y) f (z))+2f−1 (f (z) f (x))for all x, y, z ≥ 0.

Mihaly Bencze

PP25262. Solve in Z the equation x3 + y3 = 7z + 2016.

Mihaly Bencze


PP25263. If ak ∈ N∗\ {1} (k = 1, 2, ..., n) then

logna1 (a2!) + logna2 (a3!) + ...+ lognan (a1!) ≥ nn�

k=1

(ak − 1) .

Mihaly Bencze

PP25264. Determine all xk ∈ N (k = 1, 2, ..., n) for whichn�

k=1

x2k +n�

k=1

xk =n�

k=1

xk.

Mihaly Bencze

PP25265. If xn+m + xn−m = x4n for all n,m ∈ N (n ≥ m) then determinex2016.

Mihaly Bencze


1).� a+b

b+c ≥ 134 − 3r(r+4R)

4s2

2).� a

b ≥ 4− 3r(4R+s)s2

3).� ra+rb

rb+rc≥ 4− 3s2

(4R+r)2

4).� sin2 A

2+sin2 B

2

sin2 B2+sin2 C

2

≥ 4− 3(s2+r2−8Rr)4(2R−r)2

5).� cos2 A

2+cos2 B

2

cos2 B2+cos2 C

2

≥ 4− 3(s2+(4R+r)2)4(4R+r)2

Mihaly Bencze


k=1

�2k (k + 1)

√6�> n

√6

3(n+1) where {·} denote the

fractional part.

Mihaly Bencze

PP25268. Solve the equationn�

k=1

�x

k(k+1)(k+2)

�= n, where [·] denote the

integer part.

Mihaly Bencze



a cosA b cosB c cosCa b c1 1 1

��= s

2Rr (a− b) (b− c) (c− a) .

Mihaly Bencze

PP25270. In all triangle ABC holds:1).�

a ma = 2�s2 − r2 − 4Rr

�

2).�

ma wa = s2

3).�

ha wa = 4s2r2��

s2+r2+4Rr4sRr

�2− 1

Rr

�

Mihaly Bencze


x21 + 8 = 6x2 +�1 +

√x3 − 2

x22 + 8 = 6x3 +�1 +

√x4 − 2

−−−−−−−−−−−−−x2n + 8 = 6x1 +

�1 +

√x2 − 2

Mihaly Bencze

PP25272. Determine all z ∈ C∗ for which��z + 1

z

�� =��z2n+1 + 1

z2n+1

�� = r ≥ 2

Mihaly Bencze

PP25273. Determine all ak > 0 (k = 1, 2, ..., n) for which holdsn�

k=1

akxakx ≥ n for all x > 0.

Mihaly Bencze

PP25274. Determine all n ∈ N (n ≥ 2) for whichn�

k=1

n

�8 + 1

k > n�

n√5 + n

√7− n

√4�.

Mihaly Bencze

PP25275. Determine all 0 < x1 ≤ x2 ≤ ... ≤ xn for whicharctgx1, arctgx2, ..., arctgxn are in arithmetical progression.

Mihaly Bencze



a3

�128R4s2r2 − a4 (b2 + c2 − a2)2 ≥ 3

√2srR .

Mihaly Bencze


1+sin2 A≤� 1

1+sinA sinB .

Mihaly Bencze

PP25278. In all acute triangle ABC holds� 1

1+cos2 A≤� 1

1+cosA cosB .

Mihaly Bencze

PP25279. Let be 1 = d1 < d2 < ... < dk = n the positive divisors of n.Determine all n for which k = 2016 and n = d21 + d22 + ...+ d22016.

Mihaly Bencze

PP25280. If ak ∈ R (k = 1, 2, ..., n) such that aiaj ≥ 1 (i, j ∈ {1, 2, ..., n})then

n�k=1

1a2k+1

≥ �cyclic

1a1a2+1 . If 0 < aiaj ≤ 1 (i, j ∈ {1, 2, ..., n}) then holds

the reverse inequality.

Mihaly Bencze

PP25281. Let ABCD be a convex quadrilateral, we construct in exteriorthe equilateral triangles ABE,BCF,CDG and DAH. Determine allquadrilaterals ABCD for which the lines AC,BD,EG,FH are concurents.


PP25282. If a, b > 1 then solve in R the following systemax1 + b−x2 = ax2 + b−x3 = ... = axn + b−x1 = a+ b



ma

�λ≥ 3�

2√3

�λfor all

λ ∈ (−∞, 0] ∪ [1,+∞) .

Mihaly Bencze


PP25284. If a > 1 and b ≥ −14 then solve in (0,+∞) the following system

a−bxloga x2

1 + log2a x3 = x4 + loga x5 + b

a−bxloga x3

2 + log2a x4 = x5 + loga x6 + b−−−−−−−−−−−−−−−−−a−bx

loga x1n + log2a x2 = x3 + loga x4 + b

.


PP25285. Let be P (x) = x4 + ax3 + bx2 + cx+ d when a, b, c, d ∈ R and forwhich P (x) = 1 have integer roots. Exist x0 ∈ Z for which P (x0) = 2016?

Mihaly Bencze


1).�

ama ≤√3�s2 − r2 − 4Rr

�

2).� ma

bc ≤√3(s2−r2−4Rr)

4sRr

3).�

a2ma ≤√3s�s2 − r2 − 4Rr

�

Mihaly Bencze

PP25287. If f : R → (0,+∞) is a continuous function, then

n�k=1

�

k+1�

k

xf(x)dx

�2

k+1�

k

f2(x)dx

≤ n(n2+3n+3)3 .

Mihaly Bencze

PP25288. If A,B ∈ M2 (R) and λ > 0 thendet�A2 − λB2

�+ λ det (AB −BA) ≤ (detA+ λ detB)2 ≤

≤ det�A2 + λB2

�+ λ det (AB +BA) .

Mihaly Bencze

PP25289. If f, g : [a, b] → R are continuous functions for which:b�xf (t) dt ≥ g (x) for all x ∈ [a, b] and

b�af (x) dx = α and

b�ag (x) dx = β then

1).�b3 − a3

� b�af2 (x) dx ≥ 1

3 (6aα+ 6β − 1)


2). if a ≥ 0 then�b2√b− a2

√a�2 b�

a|f (x)|3 dx ≥ 1

20 (15aα+ 15β − 2) .

Traian Ianculescu

PP25290. 1). If A,B ∈ M2 (R) then det�A2�−B2 + det (AB −BA) ≤

≤ (detA+ detB)2 ≤ det�A2 +B2

�+ det (AB +BA)

2). If A,B ∈ Mn (R), n ≥ 2 and AB = BA; a, b ∈ R, a2 < 4b thendet�b�A2 +B2

��+�a2 − 2b

�AB − a (A+B) + In ≥ 0

(In conection with problems L.906, L.927, L.1067, RMC)

Traian Ianculescu

PP25291. Solve in R the equation

loga

�a3 loga

�

x+a2

4x

�

12ax2−16x3

�− a

12x2

a2− 16x3

a3 −�x+ a2

4x

�where a > 1. (A generalization

of problem PP. 20281, Octogon Mathematical Magazine).

Traian Ianculescu

PP25292. If a, b, c, x, y > 0 then

1). a+ b+ c+ a(bx+cy)2

+ b(cx+ay)2

+ c(ax+by)2

≥ 6x+y

2). a2 + b2 + c2 +�

a(bx+cy)2

�2+�

b(cx+ay)2

�2+�

c(ax+by)2

�2≥ 3

x2+y2

(A generalization of problem 27070 GMB).

Traian Ianculescu

PP25293. Determine the number of increasing functionsf : {1, 2, ...,m} → {1, 2, ...,m} for which

��f2 (x)− f2 (y)�� ≤��x3 − y3

�� for allx, y ∈ {1, 2, ...,m} , where m ∈ N∗.

Mihaly Bencze

PP25294. If zk ∈ C (k = 1, 2, ..., n) , such thatn�

k=1

|zk| ≤ 1 then��

n�k=1

znk − nn�

k=1

zk

�� ≤ 1.

Mihaly Bencze


PP25295. Determine all n ∈ N for which exist ak ∈ N∗ (k = 1, 2, ..., n) such

thatn�

k=1

�a1+a2+...+am−1

am

�= n when [·] denote the integer part.

Mihaly Bencze

PP25296. In all triangle ABC holds 4 ≤ 6Rr + 9r(4R+r)

s2.

Mihaly Bencze

PP25297. If 0 < a < b then (2n+ 1) (2m+ 1)�bn+m+1 − an+m+1

�2 ≤≤ (n+m+ 1)2

�b2n+1 − a2n+1

� �b2m+1 − a2m+1

�for all n,m ∈ N.

Mihaly Bencze

PP25298. Solve the equation�

cyclic

(x− a1 − a2 − ...− an−1)n =

n�k=1

ank .

Mihaly Bencze

PP25299. If 0 < a < b thenn�

i=1(nki + 1)

�b1+

n�

i=1ki− a

1+n�

i=1ki

�n

≤�1 +

n�i=1

ki

�n n�i=1

�bnki+1 − anki+1

�,

for all ki ∈ N (i = 1, 2, ..., n).

Mihaly Bencze

PP25300. If en =�1 + 1

n

�nthen compute lim

n→∞nλ�eµn+1 − eµn

�where

λ, µ ∈ R.

Mihaly Bencze


�1 + 3

1x1

�(13− 7x2) = 16 (2 + 7x3)

�1 + 3

1x2

�(13− 7x3) = 16 (2 + 7x4)

−−−−−−−−−−−−−−−−�1 + 3

1xn

�(13− 7x1) = 16 (2 + 7x2)




−a+b+c

�λ≥ 3�R2r

�λfor all

λ ∈ (−∞, 0] ∪ [1,+∞) .

Mihaly Bencze

PP25303. In all triangle ABC holds� −a+b+c

−a+2b+2c ≥ 78 − 2(s2−2r2−8Rr)

5r(4R+r) .

Mihaly Bencze


k=1

logk+1 (k + 1) (k + 2) ≥�1 + n�ln (n+ 2)

�n.

Mihaly Bencze

PP25305. The triangle ABC is equilateral if and only if�sinA

�2 sin2C − sin2A = s2−r2−4Rr

2R2 .

Mihaly Bencze

PP25306. Solve in Z the equationn�

k=1

(1 + xk) = tn−1n�

k=1

xk.

Mihaly Bencze

PP25307. Determine all functions f : [1,+∞) → R for which

f

�n�

k=1

xk

�=�

cyclic

x1f (x2...xn) for all xk ≥ 1 (k = 1, 2, ..., n) .

Mihaly Bencze

PP25308. Let ABC be a triangle, denote RA the radius of interior circ

tangent in A and tangent to BC etc. Prove that�� 1

RA

�λ≤ 3�

23r

�λfor all

λ ∈ [0, 1] .

Mihaly Bencze

PP25309. In all triangle ABC holds� cos2 B

2+cos2 C

2−cos2 A

2�

(3n−1) cos2 C2+cos2 A

2

≥ 43n+1

�4R+r2R

for all n ∈ N∗.

Mihaly Bencze


PP25310. In all triangle ABC holds� ctgA

2

2tgA2+tgB

2

≥ 3.

Mihaly Bencze

PP25311. If ak > 0 (k = 1, 2, ..., n), λ ∈ R, b1, b2, ..., bn−1 > 0 thena λ1

aλ2 (b1a1+b2a3+b3a4+...+bn−1an)+

aλ2aλ3 (b1a2+b2a4+b3a5+...+bn−1a1)

+ ...

+ aλnaλ1 (b1an+b2a2+...+bn−1an−2)

≥ n2

(b1+...+bn−1)n�

k=1

ak

. A generalization of problem

3853 Crux Mathematicorum.

Mihaly Bencze


1).� cosA

ha≥ 2(

�

cos A2 )

2

4R+r

2).� cosA

ra≥ 4R(

�

cos A2 )

2

s2+r2+4Rr

Mihaly Bencze


�n�

k=1

arctgkλ+k2

− 1λ+1arctg

2n

�when λ > 0.

Mihaly Bencze

PP25314. Let ABC be a triangle. Determine all the numbers x, y ∈ R suchx+ y = 1 for which ax+cy

b + bx+azc + cx+bz

a ≤ Rr + 1.

Mihaly Bencze

PP25315. In all acute triangle ABC holds�� 1

ha+ 1

ra

�cosA

�λ≥ 3

(3r)λ

for all λ ∈ (−∞, 0] ∪ [1,+∞) .

Mihaly Bencze

PP25316. If 1n+1

√xn+1−1 − 1

n√xn−1 = 1 and x1 = 2 then

n�k=1

xk = (n+1)n

n! .

Mihaly Bencze

PP25317. If 0 ≤ a, b, c ≤ 1 and a+ b+ c = 2 then��

a2b� ��

ab2�≤ 1.

Mihaly Bencze


PP25318. If n, k ≥ 3 then exist p ∈ Z such that(k − 1)n−1 > kp+1−1

k−1 − n ≥ 0

Mihaly Bencze

PP25319. If f, g, h : (−1, 1] → R where f (a) =π�0

(1+cosx) cosxdx1+a2−2a cosx

,

g (a) =π�0

(1+cosx) cosxdx1+a2−2a cos 2x

, h (a) =π�0

(1+cosx) cosxdx1+a2−2a cos 4x

then

1). f (a) + f (−a) ≥ π2). f (a)h (a) = g2 (a) for all a ∈ (−1, 1) .

Gyorgy Szollosy

PP25320. If a, b > 0 then solve the equation ax3 − bx = b3

�b3a .

Gyorgy Szollosy

PP25321. Determine all a ∈ R for which all the roots of the equationx6 − 2 (a+ 1)x5 + 5ax4 − 5a2x2 + 2a2 (a+ 1)x− a3 = 0 are real numbers.

Gyorgy Szollosy

PP25322. Prove that 3−√3 <

3

�√85+92 − 3

�√85−92 <

√3.

Gyorgy Szollosy

PP25323. Solve in R the equation 3x + 35y = 2y − 2

3x = 5.

Gyorgy Szollosy

PP25324. If ε2 + ε+ 1 = 0 and a ∈ C then prove that the roots of theequation z2 − (a− ε) z + a2 + ε = 0 are rational functions of ε.

Gyorgy Szollosy

PP25325. Let be n ∈ N . Prove that the equation

22n �

x8 + y8�2n

= z2 + t2 + w2 have infinitely many integers solutions.

Marius Dragan


PP25326. Let be In =π�0

n�k=1

cos kxdx. If n ∈ {4k + 1, 4k + 2} then In = 0.

What happens if n ∈ {4k, 4k + 3}?

Gyorgy Szollosy

PP25327. If a, b, c > 0 and a = b+ c then�1 + 1

a2

�a2 �1− 1

a2−b2−c2

�≤�1 + 1

b2

�b2 �1 + 1

c2

�c2.

Marius Dragan

PP25328. Let be x, y, z ∈ N . Find all the rest dividing the numberx3yz + xy3z + xyz3 be 11.

Mihaly Bencze

PP25329. Find the best k ∈ Z such that�√n+

√n+ 1 +

√n+ 2 +

√n+ 3

�=�√

16n+ k�for each n ∈ N, where [·]


Marius Dragan

PP25330. If 0 < a ≤ b ≤ c then�a2 + 14

� �b2 + 14

� �c2 + 14

�≥ 26 (3a+ 2b+ c+ 1)2 .

Marius Dragan

PP25331. Solve in N the following equations1). 29x − 20x = y2 2). 29x − 21x = z2

Nicolae Papacu

PP25332. Determine all n ∈ N for which exist xi ∈ N (i = 1, 2, ..., n) such

that x1 < x2 < ... < xn and n+n�

i=1xi =

n�k=1

xi.

Mihaly Bencze

PP25333. Let f : [0, 1] → R a differentiable function for which1�0

f (x) dx = 1 then exist c ∈ (0, 1) for which1�0

f (x) dx = f (c) + cλ−1 (c− 1)

for all λ ∈ R.

Mihaly Bencze


PP25334. Solve in Z the equation x�y2 + 3z + 2

�= t2.

Mihaly Bencze

PP25335. If xk ∈�0, π2�(k = 1, 2, ..., n) then

2�

cyclic

sin (x1 + x2) ≥ 2n2

n�k=1

sinxk + 2n2

n�k=1

cosxk.

Mihaly Bencze

PP25336. If a > b > c > d > 0 then solve in R the following system:

(c+ d) (ax1 + bx2) = (a+ b) (cx3 + dx4)(c+ d) (ax2 + bx3) = (a+ b) (cx4 + dx5)−−−−−−−−−−−−−−−−−(c+ d) (axn + bx1) = (a+ b) (cx2 + dx3)

Mihaly Bencze

PP25337. If x1 = 2 and xn+1 =�exp�xn−1xn+1

��2for all n ≥ 1 then compute

limn→∞

n (1− xn) .

Mihaly Bencze


λ+sin2 A≥ 12

4λ+3 for all λ > 0.

Mihaly Bencze

PP25339. If a1 > a2 > ... > a2n > 0 then the function f : R → R when

f (x) =

n�

k=1axk

2n�

p=n+1axp

is injective.

Mihaly Bencze


�1 + 2 sin2 x1

� �1 + 2 cos2 x2

�= 4 sin 2x3�

1 + 2 sin2 x2� �

1 + 2 cos2 x3�= 4 sin 2x4

−−−−−−−−−−−−−−−−−�1 + 2 sin2 xn

� �1 + 2 cos2 x1

�= 4 sin 2x2

.

Mihaly Bencze


PP25341. In all triangle ABC holds� ln(ma)−ln(ha)

ln(mb)−ln(hb)rc ≥ 3s

√r.

Mihaly Bencze

PP25342. Let ABCD be a convex quadrilateral such thatmin {A∡, C∡} ≥ 60◦. Prove that 4BD2 ≥ 3max {AB,AD}max {CB,CD} .

Mihaly Bencze

PP25343. If xk ∈�0, π2�(k = 1, 2, ..., n) , then

n�k=1

1sin2 xk cos2 xk

≤ 3n+

�

1−n�

k=1sin2 xk

�2

n�

k=1sin2 xk

+

�

1−n�

k=1cos2 xk

�

n�

k=1cos2 xk

.

Mihaly Bencze

PP25344. Determine all λ > 0 for which��λ+ 1 + λ

√λ� 1

λ+1+�λ+ 1− λ

√λ� 1

λ+1

�= 2 where [·] denote the integer

part.

Mihaly Bencze

PP25345. Denote ak the greater odd divisor of k. Prove that

1).∞�n=1

12n−1a1a2...an

< e− 1

2).n�

k=1

k!a1a2...ak

≤ 2n − 1

Mihaly Bencze


w2a sinA cos2 B

2 ≤ sr(s2+(4R+r)2)8R2 .

Mihaly Bencze

PP25347. Solve in Z the equationx (x+ 2) (x+ 4) (x+ 6) (x+ 8) = y2 − 16.

Mihaly Bencze

PP25348. Determine all integers a, b, c ∈ Z such that� ab+c+1 ≤ 1 ≤� 1

b+c+1 .

Mihaly Bencze



w2a sinA ≤ sr(4R+r)

R .

Mihaly Bencze


w2a sinA cos2 A

2 ≤ sr((4R+r)2−s2)4R2 .

Mihaly Bencze

PP25351. In all triangle ABC holds� log2

�

21

sinB +21

sinC

�

sinA ≤ 4Rr .

Mihaly Bencze

PP25352. If xk ≥ 1 (k = 1, 2, ..., n) then�xn log2

�2

n−1

�2x1+x2+...+xn−2

�≤ 2

�1≤i<j≤n

xixj (A generalization of

problem 27061 GMB).

Mihaly Bencze

PP25353. If x0 =3√4 and xn+1 = xn + 1

x2nthen

n�k=1

(xk − 1)3 ≤ 3n(n+3)2 ≤

n�k=1

x3k.

Mihaly Bencze

PP25354. In all triangle ABC holds� (a3+2b3+c3+abc)(a3+c3+abc)

(a3+b3+abc)(b3+c3+abc)≤ 3(s2−3r2−6Rr)

2Rr .

Mihaly Bencze

PP25355. 1). If a > 0, a �= 1, b > 1 and n ∈ N∗ then compute

In (a, b) =b�1b

((1−n)x+ln a)xn−3a1x dx

xn−1a1x+1

2). Determine all continuous functions g :�12 , 2�→ R for which

2�12

�(2−x)ex

x(x2+ex)g (x) dx ≥ ln

4(4√e+1)

e2+4≥

2�12

g2 (x) dx.

Traian Ianculescu


PP25356. 1). If f : [a, b] → R is continuous and a > 0, α ∈ R, α �= 1, then

exist c ∈ (a, b) such that (α− 1) (ab)α−1b�af (x) dx =

�bα−1 − aα−1

�cαf (c)

2). If g : [a, b] → R is a continuous function and G (x) =x�0

g (t) dt, x ∈ [a, b]

and β < 0 then exists c1, c2 ∈ (0, b) , c1 �= c2 such that

1−βb1−β

b�0

G (x) dx = cβ+11 g (c2)

Traian Ianculescu

PP25357. If 0 < a < 1 < c < d then determine all differentiable functionwith continuous derivate for whichf (a · cx + b)− f (a · dx + b) = a (f (cx)− f (dx)) for all x ∈ R.

Mihaly Bencze

PP25358. Solve in Z the equation 2x

y+1 + 3y

2x+1 = 2.

Mihaly Bencze

PP25359. If a0 =13√4

and an+1

�1 + a2n

�= an for all n ≥ 1 the

a242 ∈�

110 ,

19

�and lim

n→∞3√nan = 1

3√3

Traian Ianculescu


k=1

�2n+2−k

k ≥ n�1 +

n√n!

n+1

�.

Traian Ianculescu

PP25361. If A ∈ M2 (R) and x ∈ R such that det�A2 − x2I2

�≥ 0 then

1). If detA+ x2 ≥ 0 thendet (An − xnI2) + det (An + xnI2) ≥ 22−n |xTr (A)|n for all n ∈ N

2). If A ∈ M2 (Q) and det�A2 − 2I2

�= 0 then A2 = 2I2.

Traian Ianculescu

PP25362. If ak, xk > 0 (k = 1, 2, 3, 4) then� x1a2a3

(x1+x2)(x2+x3)≤ (a1+a2+a3)

2

4(x1+x2+x3).

Traian Ianculescu


PP25363. If ai > 0 (i = 0, 1, ..., n) are in arithmetical progression with ratio

r > 0 thenn�

i=1

1ai

< nr

�1− n

�a0an

�. (A generalization of problem C.1107

GMB).

Traian Ianculescu

PP25364. If ak > 0 (k = 1, 2, ..., n) then�

cyclic

an2an1+an2+...+ann−1+a1a2...an

≥ 1 (A

generalization of problem C.3201 GMB).

Traian Ianculescu


1).�

aα�b2α + c2α − a2α

�≤ 3 (abc)α

2).� √

ara

≤ 3�

Rs when α ∈ [0, 1]

Traian Ianculescu

PP25366. If xi > 0 (i = 1, 2, ..., n) and f (λ) =�

(xλ1+xλ

2+...+xλk)

n�

i=1xλi

then

f (λ+ 1) ≥ f (λ) for all λ > 0, where k ∈ {1, 2, ..., n} .

Mihaly Bencze

PP25367. Solve in Z the equation 2x+1 + 3

y2+2+ 5

z3+4= 3.

Mihaly Bencze

PP25368. If x, y, z > 0 and x+ y + z = 1 then� x4

x2+y+z≥ 1

21 .

Mihaly Bencze

PP25369. If ak > 0 (k = 1, 2, ..., n), then�

cyclic

an1a2

≥� a1a2...an−1.

Mihaly Bencze

PP25370. If a, b, c > 0 and λ ∈ (0, 5) then� a3

b(a+λc) ≥5−λ9

�a.

Mihaly Bencze


PP25371. Let be aij > 0 (i = 1, 2, ..., n; j = 1, 2, ...,m) and xj > 0(j = 1, 2, ...,m) and f : R → (0,+∞) is a convex function. Prove thatm�j=1

fn+1(xij)a1ja2j ...anj

≥ mn+1

n�

i=1

�

m�

j=1aij

�fn+1

�1m

m�j=1

xj

�.

Mihaly Bencze

PP25372. If xk > 0 (k = 1, 2, ..., n) andn�

k=1

xk = 1 then

�cyclic

x31

x21+x2+...+xn

≥ 1n2−n+1

.

Mihaly Bencze

PP25373. In all triangle ABC holds� cos4 A

2

cos2 B2

≥ 4R+r2R

�2((4R+r)2−s2)(4R+r)2+s2

.

Mihaly Bencze

PP25374. If ak > 0 (k = 1, 2, ..., n) , then� a1a3+a22

(a1+a3)a2≥ n.

Mihaly Bencze

PP25375. Let f : R → (0,+∞) be a convex function and xk, ak > 0(k = 1, 2, ..., n). Determine all λ > 0 for whichn�

k=1

fλ(xk)ak

≥ n2

n�

k=1

ak

fλ

�1n

n�k=1

xk

�.

Mihaly Bencze

PP25376. If ak > 0 (k = 1, 2, ..., n) thenn�

k=1

ak +�

cyclic

(a2−a3)2

a1≥�

cyclic

|a1 − a2| .

Mihaly Bencze


tg2A2 ctgB2 ≥ 4R+r

s2

�(4R+ r)2 − 2s2.

Mihaly Bencze


PP25378. If ak > 0 (k = 1, 2, ..., n) then�

cyclic

a21a2

≥�

n�k=1

ak

��

n�

k=1a2k

�

cyclica1a2

.

Mihaly Bencze

PP25379. If a, b, c > 0 , a �= b and In =π�0

cosnxdxa2+b2−2ab cosx

for all n ∈ N, then�n≥0

In is convergent and compute its limit.

Gyorgy Szollosy

PP25380. If ak > 0 (k = 1, 2, ..., n) and f (u, v) =�

cyclic

au1av2

then

f (u+ 1, v + 1) ≥ f (u, v) for all u, v > 0.

Mihaly Bencze

PP25381. In all triangle ABC holds��−a+b+c

a > 2√2.

Mihaly Bencze

PP25382. Determine the rest of divisor of number 52016 with 2016.

Mihaly Bencze

PP25383. If a > 0, x0 > −1 and 3xn+1 (xn + 1)2 = a+ 2x3n + 3x2n − 1 thencompute lim

n→∞n (1− xn) .


PP25384. If xn = |zn + z−n| , when z ∈ C∗ and if x1 > 2 thenxn > xn−1 + x2 − x1 for all n ≥ 2.

Mihaly Bencze

PP25385. If A,B ∈ Mn (C) such that ABk = A−B for all n ∈ N∗.Determine all k ∈ N for which In +B is invertible. Determine all k ∈ N forwhich AB = BA.

Mihaly Bencze


PP25386. Solve in Z the following equation1

xy+z−1 + 1yz+x−1 + 1

zx+y−1 = 3n2−n+1

where n ∈ N∗ is given.

Mihaly Bencze


k=1

n(n−1)...(n−k+1)knk+1 = 1.

Mihaly Bencze


tg2A2 tg2B2 +�

ctg2A2 ctg2B2 ≥ 10.

Mihaly Bencze

PP25389. Prove that for all n, k ∈ N∗ exists m1,m2, ...,mk ∈ N∗ numbers

such that 1 + pk−1n(p−1) =

k�i=1

�1 + 1

mi

�for all p ∈ N, p ≥ 2.

Mihaly Bencze


√x1 − 2 +

�x22 − 4x3 + 3 =

�(x4 − 2)3

√x2 − 2 +

�x23 − 4x4 + 3 =

�(x5 − 2)3

−−−−−−−−−−−−−−−−−−√xn − 2 +

�x21 − 4x2 + 3 =

�(x3 − 2)3

Mihaly Bencze

PP25391. Let a, b, c be positive real numbers such that a2 + b2 + c2 = 3.

Prove that a�

bc2(a+b)+5c + b

�ca

2(b+c)+5a + c�

ab2(c+a)+5b <

�3332 .

Jose Luis Dıaz-Barrero

PP25392. Let a, b, c be three positive real numbers such thata2 + b2 + c2 = 9. Prove that b−1

(5√a+2

√b)2

+ c−1

(5√b+2

√c)2

+ a−1

(5√c+2

√a)2

≥ 149


PP25393. Let n ≥ 0 be an integer number. Calculate limx→+∞

xlnx

∞�n=0

12016n+x .



PP25394. Let n ≥ 0 be an integer number. Let n be a positive integer.

Show that 1n

n�k=1

�k2k

�nk

��2 ≥ 19

�32

�2n


PP25395. If a ∈�0, 14�then solve in R the following system:

x21 + 2ax2 +116 + a =

�a2 + x3 − 1

16

x22 + 2ax3 +116 + a =

�a2 + x4 − 1

16

−−−−−−−−−−−−−−−−x2n + 2ax1 +

116 + a =

�a2 + x2 − 1

16

.

Mihaly Bencze

PP25396. Solve in R the following system:a = x21 +

√a+ x2 = x22 +

√a+ x3 = ... = x2n +

√a+ x1 when a ∈ R is given.

Mihaly Bencze

PP25397. Solve in R the following system:�x1 + 3− 4

√x2 − 1 +

�x2 + 8− 6

√x3 − 1 =

�x2 + 3− 4

√x3 − 1 +�

x3 + 8− 6√x4 − 1 = ... =

�xn + 3− 4

√x1 − 1+

�x1 + 8− 6

√x2 − 1 = 1.

Mihaly Bencze


|x1 + 1|+ 3 |x2 − 1| = x3 + 2 + |x4|+ 2 |x5 − 2||x2 + 1|+ 3 |x3 − 1| = x4 + 2 + |x5|+ 2 |x6 − 2|−−−−−−−−−−−−−−−−−−−−−|xn + 1|+ 3 |x1 − 1| = x2 + 2 + |x3|+ 2 |x4 − 2|

.

Mihaly Bencze

PP25399. Determine all x ∈ R for whichn�

k=1

[kx]3 =�[nx]([nx]+[x])

2

�2for all

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

Mihaly Bencze


PP25400. If ak ∈ R∗ (k = 1, 2, ..., n) such thatn�

k=1

ak = 1 andn�

k=1

{ak} = 1.

Prove thatn�

k=1

{amk } = 1 for all m ∈ N, where {·} denote the fractional part.

Mihaly Bencze

PP25401. Determine all λ ∈ R for which the equation sinx = λx haveexactly 2016 solutions.

Mihaly Bencze

PP25402. Prove that x is integer number if and only ifn�

k=1

[kx]2 = n([nx]+[x])([2nx]+[x])6 for all n ∈ N∗, where [·] denote the integer

part.

Mihaly Bencze


� a(a+b)(a+c)2a(a+c)(a+b)+b(b+c)(b+a)+c(c+a)(c+b) ≤

34 .

Mihaly Bencze


log3 (1 + 2x1) = log2 (1 + x2)log3 (1 + 2x2) = log2 (1 + x3)−−−−−−−−−−−−log3 (1 + 2xn) = log2 (1 + x1)

.

Mihaly Bencze

PP25405. If ap > 1 (p = 1, 2, ..., n) and k ∈ {1, 2, ..., n} thenlog2a1+a2+...+ak

(a2+a3+...+ak+1)

a3+a4+...+ak+2+

log2a2+a3+...+ak+1(a3+a4+...+ak+2)

a4+a5+...+ak+3+ ...

+log2an+a1+...+ak−1

(a1+a2+...+ak)

a2+a3+...+ak+1≥ n2

kn�

p=1ap

.

Mihaly Bencze


PP25406. If (xn)n≥1 is a real sequence such that |xn+1 − xn| ≤ 1 andnyn = x1 + x2 + ...+ xn for all n ∈ N∗, then |yn+1 − yn|+ |yn+2 − yn+1| ≤ 1for all n ∈ N∗.

Mihaly Bencze

PP25407. If An =

��

cos2 nx cos2 ny cos2 nzsin 2nx sin 2ny sin 2nzcos 2nx cos 2ny cos 2nz

��where x, y, z ∈ R then

n�k=1

|Ak| < 2n.

Mihaly Bencze

PP25408. Solve in N the equation�2n

n

�= 2k, when [·] denote the integer

part.

Mihaly Bencze

PP25409. Solve in N the equation�3n

n

�= 3k, when [·] denote the integer

part.

Mihaly Bencze

PP25410. Solve in N the equation�2n+3m

nm

�= 6k, when [·] denote the

integer part.

Mihaly Bencze

PP25411. Let a1, a2, ..., an2 be an arithmetical progression, and let be the

following tabel:

��

a1 a2 ... anan+1 an+2 ... a2n... ... ... ...a(n−1)2+1 a(n−1)2+2 ... an2

��. Prove that if we elimine

n elements from this tabel in the rest n (n− 1) elements exist n numbers inarithmetical progression.

Mihaly Bencze


PP25412. If xk ∈ R (k = 1, 2, ..., n) such thatn�

k=1

sinxk = 0, then

n�k=1

|cosx− sinxk| ≤ n for all x ∈ R.

Mihaly Bencze

PP25413. Denote An the set of every integers of form ±1± 2± ...± n. Byexample A2 = {−3,−1, 1, 3} and A3 = {−6,−4,−2, 0, 2, 4, 6} . Denote |An|the number of elements of the set An. Compute

∞�n=1

1|An| .

Mihaly Bencze

PP25414. If a, b ∈ C then (|1 + ab|+ |a+ b|)2 ≥��(1 + ab)2 − (a+ b)2

�� .

Mihaly Bencze

PP25415. If A ∈ M2 (R) then83 det

�A2 +A+ I2

�≥ (1− detA)2 + (1 + TrA)2 .

Mihaly Bencze

PP25416. If A ∈ Mn (R) is an antisimmetric matrice(aij + aji = 0 for all i, j ∈ {1, 2, ..., n}) then for all xk ≥ 0 (k = 1, 2, ...,m)

holdsm�k=1

det (A+ xkIn) ≥�det

�A+ m

�m�k=1

xkIn

��m

.

Mihaly Bencze

PP25417. In all triangle ABC holds�� ab

(b+c)(c+a) ≤�

6(s2−r2−Rr)s2+r2+2Rr

.

Mihaly Bencze

PP25418. In all triangle ABC holds s2+(4R+r)2

4sR − 3s2(4R+r) ≥

√3.

Mihaly Bencze

PP25419. Solve in Z the equation 1x + 1

y+2 + 1z+3 = 3.

Mihaly Bencze


PP25420. If x1 = 2 and x2n+1 = xn + 1n for all n ≥ 1 then compute

limn→∞

n (1− xn) .

Mihaly Bencze

PP25421. In all triangle ABC holds� a(a+b)3(a+c)

(b+c)(b(a+b)3+(a+c)c3+7(a+c)(a+b)3)≤ 1

3 .

Mihaly Bencze

PP25422. Prove that

n�

k=1(−1)k−1 1

k2(nk)

Hn

Hn

≥n�

k=1

(Hk)1k when Hn =

n�k=1

1k .

Mihaly Bencze

PP25423. Let G = (0,+∞) and (log3 (x ◦ y))n = (log3 x)n + (log3 y)

n − 3n

where n = 2k + 1, k ∈ N∗ and for all x, y ∈ G

1). Prove that (G, ◦) is Abelian group

2). Prove that (G, ◦) ∼= (R,+)

Mihaly Bencze

PP25424. If Hn = 1 + 12 + ...+ 1

n then

n�k=1

�1 + 1

k

�nk

k+1�0

xn−1 ln�

kk+1 − x

�dx = − (Hn + ln (n+ 1)) .

Mihaly Bencze

PP25425. If ak > 0 (k = 1, 2, ..., n), then

�cyclic

a1

�1aλ2

+ 1aλ3

+ ...+ 1aλn

�≥

n�

k=1ak

(n−1)λ−1

n�

k=1

1ak

− nn�

k=1

ak

λ

for all

λ ∈ (−∞, 0] ∪ [1,+∞) .

Mihaly Bencze

PP25426. If a > 1, b ≥ −14 then solve in (0,+∞) the equation

a−bxloga x + log2a x = x+ logb x+ b.

Gyorgy Szollosy


PP25427. If Hn = 1 + 12 + ...+ 1

n thenn�

k=0

Hkk ≥ 1

2 (Hn)2 + ln2 n

2n .

Mihaly Bencze

PP25428. Solve in R the equation 9x2 − 9x− 2 = 12 sinπx.

Gyorgy Szollosy

PP25429. Solve in R the equation�65

� 1x + 2

1+( 95)

x−1 = 3.

Gyorgy Szollosy

PP25430. If n,m ∈ N (n ≥ 2) , α, ak > 0 (k = 1, 2, ..., n) such thatn�

k=1

aαk ≤ n then 1n + m

n�

k=1ak

≥ 1+mnn�

k=1ak

.

Gyorgy Szollosy

PP25431. Solve in R the equation 3x + 35y = 2y − 2

3x = 5.

Gyorgy Szollosy

PP25432. In all scalene triangle ABC holds�� b−a

ma−mb− 2

3

�2+

32(s2+r2−2Rr)s2+r2+2Rr

< 64�

s2−r2−Rrs2+r2+2Rr

�2.

Mihaly Bencze


ma−mb− 2

3

��c−b

mb−mc− 2

3

�≤ 16(s2+r2−2Rr)

s2+r2+2Rr.

Mihaly Bencze

PP25434. In all scalene triangle ABC holds�b−a

ma−mb− 2

3

��c−b

mb−mc− 2

3

��a−c

mc−ma− 2

3

�≤ 128Rr

s2+r2+2Rr.

Mihaly Bencze


ma−mb

�2≥ 16

27

�� ma+mba+b

�2.

Mihaly Bencze


PP25436. In all scalene triangle ABC holds� b−a

ma−mb<

2(7s2−r2+2Rr)3(s2+r2+2Rr)

.

Mihaly Bencze

PP25437. Denote F the area of the triangle with sides cos A2 , cos

B2 , cos

C2

when ABC is a given triangle. Prove that

16F 2 ≤ cos A2 cos B

2 cos C2

�cos A

2 + cos B2 + cos C

2

�.

Mihaly Bencze

PP25438. In all triangle ABC holds3�s2 − r2 − 4Rr

�2 − 12s2r2 ≥�

mamb (a− b)2 .

Mihaly Bencze

PP25439. In all triangle ABC holds� (a+b)2m2

b+(a+c)m2c

2a+b+c ≥�mamb.

Mihaly Bencze

PP25440. In all triangle ABC holds7(s2−r2−4Rr)

4 ≥� ama ≥ s(s2+r2−2Rr)2R .

Mihaly Bencze

PP25441. In all triangle ABC holds 3�

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

916

�(b+ c) (b+ c− a) (b− c)2

�a2 + b2 + c2 + 3a (b+ c)

�≥

≥��

(a+ b)m2b +�

(a+ c)m2c

�2.

Mihaly Bencze


6�

(b− c) (bmc − cmb) (bmc + cmb) ≥��

(b+ c) (a2 + b2 + c2) |b− c|�2

.

Mihaly Bencze

PP25443. If Sn =n�

k=1

2k−1�

k2

k+1

�when [·] denote the integer part, then

�2−n−1Sn

�= n

2 − 1 for all n = 2k.

Mihaly Bencze


PP25444. Let A1A2...An be a convex polygon. Prove that

n+�

cyclic

�a1

a2+...+an

�1 +�

a2+...+ana2+...+an−a1

��=�

cyclic

�a2+...+an

a2+...+an−a1

�when [·]


Mihaly Bencze

PP25445. Let ABCD be a convex quadrilateral such that A∡ ≥ 90◦,C∡ ≥ 90◦. Denote E respective F the proiection of A and C to BD. Provethat 1

AE·CF ≥ 1AB·DC + 1

AD·BC .

Mihaly Bencze

PP25446. Prove thatm�

n=1

1n�

k=1[√k2−k+1+

√k2+k+1]

= mm+1 .

Mihaly Bencze

PP25447. Let a, b, c > 0 then determine all λ ∈ R for which� ab ≥ λ

� abbλ+ca

.

Mihaly Bencze

PP25448. In all triangle ABC holds� a+b

c

��b+c−a

a +�

c+a−bb

�≤�s2+r2−2Rr

2Rr

�2− 2(s2−r2−Rr)

Rr .

Mihaly Bencze

PP25449. If a1 = 1 and ak > 0 (k = 1, 2, ..., n) such that

1a1+a2

+ 1a2+a3

+ ...+ 1an−1+an

= an − 1 for all n ≥ 1 thenn�

k=1

1a2ka

2k+1

= nn+1 .

Mihaly Bencze


1).��

a2 + b2

3 + c2�≥ 8s2Rr

�s2 + r2 + 2Rr

�

2).��

r2a +r2b3 + r2c

�≥ 4s4Rr2

3).��

sin A2

�4+ 1

3

�sin B

2

�4+�sin C

2

�4� ≥ r2((2R−r)(s2+r2−8Rr)−2Rr2)512R5

Mihaly Bencze


PP25451. Let A1A2...An be a convex polygon. Prove that�

cyclic

√an(a1+a2+...+an−1−an)

a1+a2+...+an−1≤ n

2 .

Mihaly Bencze


λr2br2c+((4R+r)2−2s2)

2 ≥ 3λ+3 for all

λ > 0.

Mihaly Bencze

PP25453. If ak > 0 (k = 1, 2, ..., n) and λ > 0 then

�cyclic

1

λa1+2n�

k=1ak

≤ 1(n+1)2

�1λ

n�k=1

+n�

cyclic

1a1+a2

�.

Mihaly Bencze

PP25454. In all triangle ABC holds� rarb

5ra+2(ra+rc)≤ 4R+r

12 + s2+r(4R+r)64R .

Mihaly Bencze

PP25455. In all triangle ABC holds� (s−a)(s−b)

5(a+c)−b ≤ s24 +

r(s2+(4R+r)2)8sR .

Mihaly Bencze

PP25456. If ak, bk ∈ R (k = 1, 2, ..., n) , then�

n�k=1

a2k

��n�

k=1

b2k

�≥�

n�k=1

akbk

�2

+ 2n(n−1)

��

1≤i<j≤n|aibj − ajbi|

�2

.

Mihaly Bencze

PP25457. If ak ∈ [0, 1] (k = 1, 2, ..., n) then�

cyclic

a1a31+2(a2+...+an)+n+1

≤ 13 .

Mihaly Bencze

PP25458. If x, ak > 1 (k = 1, 2, ..., n) , then loga1a2...an x ≤ n2n�

k=1

logak x.

Mihaly Bencze


PP25459. Determine an (n ≥ 1) if a1 = 1 and1

a1a3a5+ 1

a3a5a7+ ...+ 1

a2n−1a2n+1a2n+3= n(n+2)

3a2n+1a2n+3.

Mihaly Bencze

PP25460. Determine an (n ≥ 1) if a1 = 1 and

14

a1a3+ 24

a3a5+ ...+ n4

a2n−1a2n+1=

n(n+1)(n2+n+1)6a2n+1

.

Mihaly Bencze

PP25461. Determine all function f : N → N for which(f(n3)+1)(f3(n)−1)

(f(n2)−n+1)(f2(n)+n+1)= f (n− 1) (f (n) + 1) for all n ∈ N∗.

Mihaly Bencze

PP25462. In all triangle ABC holds1).� a2

a4+b2c2≤ 1

4R

2).� r2a

r4a+r2br2c≤ 4R+r

2s2r

3).� (sin A

2 )4

(sin A2 )

8+(sin B

2sin C

2 )4 ≤ 4R(2R−r)

r2

4).� (cos A

2 )4

(cos A2 )

8+(cos B

2cos C

2 )4 ≤ 4R(4R+r)

s2

Mihaly Bencze

PP25463. If ak > 0 (k = 1, 2, ..., n) , then

�cyclic

(n−1)an−11 +a2a3...an

an−22

≥ n2 n

�n�

k=1

ak.

Mihaly Bencze


[x1] +�x2 +

13

�+�x3 +

23

�= [3x4]

[x2] +�x3 +

13

�+�x4 +

23

�= [3x5]

−−−−−−−−−−−−−−−[xn] +

�x1 +

13

�+�x2 +

23

�= [3x3]

, when [·] denote the integer part.

Mihaly Bencze


PP25465. Solve in R the equation�x2�+�x+ 1

2

�= [2

√x] , when [·] denote

the integer part.

Mihaly Bencze

PP25466. If a1 = 0 and (an+1 − an) (an+1 − an − 2) = 4an for all n ≥ 1

thenn�

k=2

1ak

= n−1n .

Mihaly Bencze


�x21 − x2 − 2

�= [x3]�

x22 − x3 − 2�= [x4]

−−−−−−−−−�x2n − x1 − 2

�= [x2]

, when


Mihaly Bencze

PP25468. Solve in R the following equation

n {x}+n�

k=1

�x+ 1

k

�= 1 +

n�k=1

{kx} , when {·} denote the fractional part.

Mihaly Bencze

PP25469. If Sn =n�

k=1

�k3

�when [·] denote the integer part, then compute

∞�n=3

1S2n.

Mihaly Bencze

PP25470. Determine all a, b, c, d ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} for which��ab�2n −

�cd�2n��

cd�2n

+�cd�2n+1

+ 1�is divisible by abcd.

Mihaly Bencze

PP25471. If a1 =16 and (n+ 3) an+1 = (n+ 1)

�an + 1

2

�for all n ≥ 1, then

computen�

k=1

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

Mihaly Bencze



�

tgA2tgB

2

1+ 4�

9tgB2tgC

2

≥ 16 .

Mihaly Bencze

PP25473. Determine all ak ∈ R (k = 1, 2, ..., n) for which a1 = 1 andn�

k=1

a3k =

�n�

k=1

ak

��n�

k=1

a2kk

�

Mihaly Bencze

PP25474. If xk ∈ R (k = 1, 2, ..., n) ,λk > 0 (k = 1, 2, ..., n) such thatn�

k=1

λk = 1, then |λ1x1 + λ2x2 + ...+ λnxn|+ |λ1x2 + λ2x3 + ...+ λnx1|+

...+ |λ1xn + λ2x1 + ...+ λnxn−1| ≤n�

k=1

|xk| .

Mihaly Bencze


�1

10−x1

�= 1

10−[x2]�1

10−x2

�= 1

10−[x3]

−−−−−−−−�1

10−xn

�= 1

10−[x1]

when [·]


Mihaly Bencze

PP25476. Solve in Z the equation 1x2+1

+ 2y3+3

+ 3z4+5

= 3.

Mihaly Bencze

PP25477. In all triangle ABC hold� (r2a+s2)(r2b+s2)

r2a+r2b+2s2≥ 2.

Mihaly Bencze

PP25478. Determine all function f : N∗ → N∗ for whichn�

k=1

k4

f(k) =n4f2(n+1)4f2(n)

for all n ∈ N∗.

Mihaly Bencze



�4x1+44x2−5

�= 2x3−1

3�4x2+44x3−5

�= 2x4−1

3

−−−−−−−−�4xn+44x1−5

�= 2x2−1

3

, when [·]


Mihaly Bencze

PP25480. Solve in R the following system:�x21�− 4 [x2] =

�x22�− 4 [x3] = ... =

�x2n�− 4 [x1] = −3, when [·] denote the

integer part.

Mihaly Bencze

PP25481. Compute S (p, r, n) =n�

k=1

pk�

kr

k+1

�, when [·] denote the integer

part. We have S (2, 2, n) = n · 2n − 2n+1 + 2.

Mihaly Bencze


1[x1]

+ 1{x2} = 2016x3

1[x2]

+ 1{x3} = 2016x4

−−−−−−−−−−1

[xn]+ 1

{x1} = 2016x2

, when

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

Mihaly Bencze

PP25483. Determine (an)n≥1 such that a1 = 1 andn�

k=1

ka2k = 14n

2a2n+1 for

all n ≥ 1.

Mihaly Bencze

PP25484. If xk > 0 (k = 1, 2, ..., n) and m ∈ N, then�

cyclic

xm+21 +xm+2

2

xm+11 +xm+1

2

+�

cyclic

xm1 +xm

2

xm−11 +xm−1

2

≤ 2�

cyclic

xm+11 +xm+1

2xm1 +xm

2.

Mihaly Bencze


PP25485. In all triangle ABC holds� tg2Atg3A+tgA

+� ctg2A

ctg3B+ctgA≥� sin2 2A

1+cos2 2A.

Mihaly Bencze

PP25486. In all triangle ABC holds�(tgA+ ctgA)

�tg2A

tg3A+ctgA+ ctg2A

ctg3A+tgA

�≤ 6.

Mihaly Bencze

PP25487. In all triangle ABC holds 27s2(9t2−1)

≤� 1(st)2−(rarb)

2 ≤ 3t2−2(st)2(t2−1)

for all t > 1.

Mihaly Bencze

PP25488. If x ∈�0, π2�then

�sin2 x

�1 +�

1cos2 x

��+�cos2 x

�1 +�

1sin2 x

��+ 2 =

�1

sin2 x

�+�

1cos2 x

�, when


Mihaly Bencze


3 +�

cyclic

�a

b+c

�1 +�

b+cb+c−a

��=�

cyclic

�b+c

b+c−a

�, when [·] denote the integer

part.

Mihaly Bencze

PP25490. If a1 = 3 then determine (an)n≥1 such thatn�

k=1

(2k−1)(2k+1)a22k−1a

22k+1

= n−13a2n+1

for all n ≥ 2.

Mihaly Bencze

PP25491. If (an)n≥1 is an arithmetical progression with ratio r and a1 ≥ 12 ,

then compute the integer part of the expression�

cyclic

3

�1 + r2

a1a2a3.

Mihaly Bencze


PP25492. Determine all n, k, r, s ∈ N and all prime p and q such thatn4 + (4k + 1)n2 + 4k2 + 4k + 3 = pr + qs.

Mihaly Bencze


x1

�n�

k=1

xk

�= 13

x2

�n�

k=1

xk

�= 23

−−−−−−−−xn

�n�

k=1

xk

�= n3

and prove

thatn�

k=1

3

�x2k =

3

�n2(n+1)2(2n+1)3

108 .

Mihaly Bencze

PP25494. In all triangle ABC holds� tgA

2tgB

2

1+λtgC2

�

tgA2tgB

2

≥ 3λ+3 for all λ > 0.

Mihaly Bencze


rarbs2

�2+�rbrcs2

�2+�

s2

rcra

�2≥ 3

√3.

Mihaly Bencze

PP25496. Denote f (n) the number of triples (a, b, c) with properties1 ≤ a ≤ b ≤ c ≤ n when a, b, c are in geometrical progression with ratio a

natural number. Prove that∞�k=1

1f2(k)

> π2

24 .

Mihaly Bencze

PP25497. Solve in N the equation 2015x(x+1) +

2016y(y+1) +

2017z(z+1) = 1.

Mihaly Bencze

PP25498. Compute limx→∞

x

�1

ln 2016 − xlnx

∞�n=0

12016n+x

�



PP25499. If a, b, c > 0 and a+ b+ c = 1, then 111 +

� a2

5a+2 ≥� (a+b)2

5(a+b)+4


PP25500. If ak > 0 (k = 1, 2, ..., n) andn�

k=1

ak = 1 thenn�

k=1

a2k5ak+2 ≥ 1

2n+5 .


PP25501. In all triangle ABC holds� ab

5b+2(a+b) ≤s6 +

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

32s(s2+r2+2Rr).

Mihaly Bencze


ra+ 1

ha

3

�

1+ r2

�

1rb

+ 1hb

�

− r2

�

1rd

+ 1hd

�

≥ 2r .

Mihaly Bencze

PP25503. In all triangle ABC holds 2s2

3s2−r2−4Rr+� w2

abc ≤ 3.

Mihaly Bencze

PP25504. In all triangle ABC holds� (ha)

32

(ha)32√ha+2hb

+� (rb)

32

(ra)32√ra+2rb

≥ 2√r.

Mihaly Bencze

PP25505. In all tetrahedron ABCD holds1).� 3√hbhd

ha3√

hbhd+r(hd−hb)≥ 1

r

2).� 3

√rbrd

ra3√

rbrd+r(rd−rb)≥ 1

r

Mihaly Bencze


((4R+r)2+9r2a)2 ≥ (4R+r)2

972 .

Mihaly Bencze

PP25507. In all triangle ABC holds� (s−a)6

(a2+9(s−a)2)2 ≥ s2

972 .

Mihaly Bencze



1).� cos A

2cos B

2

cos A2+cos B

2−cos C

2

≥� cos A2

2).� mamb

ma+mb−mc≥�ma

Mihaly Bencze

PP25509. In all triangle ABC holds� c2−(a−b)2

3c−a−b ≥ 2s.

Mihaly Bencze

PP25510. In all triangle ABC holds� (cos A

2+cos B

2 )2

(cos A2+cos B

2 )2−cos2 C

2

≥ 4.

Mihaly Bencze

PP25511. In all triangle ABC holds� a2+b2+3c2+8mamb

4c2−a2−b2+8mamb≥ 4.

Mihaly Bencze

PP25512. In all triangle ABC holds� (a+b)λ

(a+b)λ−cλ≥ 3·2λ

2λ−1for all λ ≥ 1.

Mihaly Bencze

PP25513. Let A1A2...An be a convex polygon. Prove that� (a1+a2+...+an−1)λ

(a1+a2+...+an−1)λ−aλn

≥ n(n−1)λ

(n−1)λ−1for all λ ≥ 1.

Mihaly Bencze

PP25514. If ak, bk, xk > 0 (k = 1, 2, ..., n) and x =n�

k=1

xk then�

n−2n−1

n�k=1

a2k +n�

k=1

a2kxk

x−xk

��n−2n−1

n�k=1

b2k +n�

k=1

b2kxkx−xk

�≥

≥ 4(n−1)2

��

1≤i<j≤n

�aibiajbj

�2

.

Mihaly Bencze

PP25515. If ak > 0, λk > 0 (k = 1, 2, ..., n) andn�

k=1

λk = 1 then

n

�n�

k=1

ak ≤ 1n

�aλ11 aλ2

2 ...aλnn + aλ2

1 aλ32 ...aλ1

n + ...+ aλn1 aλ1

2 ...aλn−1n

�≤ 1

n

n�k=1

ak.


(A generalization of Heinz-mean).

Mihaly Bencze

PP25516. In all scalene triangle ABC holds� a(b−a)

(b+c)(ma−mb)<

4(2s2−3Rr)3(s2+r2+2Rr)

.

Mihaly Bencze

PP25517. In all scalene triangle ABC holds8Rr(5s2+r2+4Rr)

3(s2+r2+2Rr)+� b3−a3

ma−mb<

2(9s2+r2+4Rr)3 .

Mihaly Bencze

PP25518. In all scalene triangle ABC holds� b2−a2

ma−mb< 16s

3 .

Mihaly Bencze

PP25519. Compute� � (x−sin y cos y−sin2 x)(y−sinx cosx−cos2 y)dxdy

(1−sin 2x)(1−sin 2y) .

Mihaly Bencze

PP25520. In all scalene triangle ABC holds94

� (b−a)mamb

m3a−m3

b≤� ma+mb

a+b ≤ 98

� (b−a)(m2a+m2

b)m3

a−m3b

.

Mihaly Bencze

PP25521. In all acute triangle ABC holds�

(1 + sinA) (1 + cosA) ≥ (3+2√2)sr

R2 .

Mihaly Bencze

PP25522. In all acute triangle ABC holds� 2+3 sinA+2 cos 2A+sinA cos 2A

1+cos 2A ≥ (17+12√2)s

R .

Mihaly Bencze

PP25523. In all acute triangle ABC holds��1 + 1

sinA

� �1 + 1

cosB

�≥ 3(3 + 2

√2).

Mihaly Bencze


PP25524. Compute� �

�

x+√

y2+y�

(y+√x2+x)dxdy

(x+√x2+x+

√x+

√x+1+1)

�

y+√

y2+y+√y+1+1

� .

Mihaly Bencze

PP25525. Let (an)n≥1 be an arithmetical progression formed with naturalnumbers. Prove that if an thermen of the given arithmetical progression is2025 then the given progression have infinitely many therms formed byperfect square.

Mihaly Bencze

PP25526. If xn+1 · xn = x2n + 1 for all n ≥ 1 and x1 > 1 then compute

limn→∞

n�√

e−�xn+1

xn

�n�.

Mihaly Bencze

PP25527. If xk > 1 (k = 1, 2, ..., n) , then

�cyclic

logx1

�x20142 + x20082 − x20072 + 1

�≥ 2011n.

Mihaly Bencze

PP25528. If xk > 1 (k = 1, 2, ..., n) , then

�cyclic

logx1

�x20142 + x20112 + 2x20082 − x20072 + 1

�≥ 2008n.

Mihaly Bencze

PP25529. If Ak ∈ Mn (C) (k = 1, 2, ..., n) such thatAn−1

1 = −A2A3...An;An−12 = −A3A4...AnA1, ..., A

n−1n = − (A1A2...An−1)

then An1 = An

2 = ... = Ann.

Mihaly Bencze

PP25530. Solve in R the equation

2�√

2x − 1 +√3x − 2x + ...+

�(n+ 1)x − nx

�= (n+ 1)x + n− 1 when

n ∈ N∗.

Mihaly Bencze


PP25531. Let (an)n≥1 be an arithmetical progression formed by naturalnumbers. Prove that if an term of the given arithmetical progression is 2197the given progression have infinitely many therms formed by perfect cubes.

Mihaly Bencze

PP25532. If ak > 1 (k = 1, 2, ..., n) and

t =a21

(n−1)a2+a3+

a22(n−1)a3+a4

+ ...+ a2n(n−1)a1+a2

thenn�

k=1

logak t ≥ n.

Mihaly Bencze

PP25533. If xk > 1 (k = 1, 2, ..., n) , then�

cyclic

logx1

�2x32 − 1

�≥ n.

Mihaly Bencze

PP25534. If xk > 1 (k = 1, 2, ..., n) , then�cyclic

logx1

�x122 + x42 − x2 + 1

�≥ 9n.

Mihaly Bencze

PP25535. If xk > 1 (k =, 2, ..., n) , then�cyclic

logx1

�x122 − x92 + 2x42 − x2 + 1

�≥ 4n.

Mihaly Bencze

PP25536. If xk > 1 (k = 1, 2, ..., n) then�

cyclic

log2x31+x2

1

�2x42 + 1

�≥ n.

Mihaly Bencze

PP25537. If xk > 1 (k = 1, 2, ..., n) , then�cyclic

logx1

�2x42 − x32 − x22 + 1

�≥ 3n.

Mihaly Bencze

PP25538. If xk > 1 (k = 1, 2, ..., n) then�

cyclic

logx1

�2x42 − 2x32 + 1

�≥ 2n.

Mihaly Bencze


PP25539. If xk > 0 (k = 1, 2, ..., n) then�cyclic

logx1+1

�x42 + 9x32 + 31x22 + 49x2 + 31

�≥ 4n.

Mihaly Bencze

PP25540. If xk > 1 (k = 1, 2, ..., n) and m ∈ N∗

then�

cyclic

logx1

�x2m2 − (2m− 1)x2 + 2m− 1

�≥ n.

Mihaly Bencze

PP25541. If ak > 1 (k = 1, 2, ..., n) and λ > 0 then�cyclic

loga1�a3λ2 − 2a2λ2 + 4

�≥ 2nλ.

Mihaly Bencze

PP25542. If ak > 1 (k = 1, 2, ..., n) then�

cyclic

loga31+a1

�a42 + a22 + 1

�≥ n.

Mihaly Bencze

PP25543. In all triangle ABC holds� a(a+b)(a+c)

(b+c)(b2+c2+ab+ac)≤ 3

2 .

Mihaly Bencze

PP25544. In all triangle ABC holds� a(a+b)(a+c)

(b+c)(bc+(a+c)(a+b)) ≤ 2.

Mihaly Bencze

PP25545. In all triangle ABC holds� (b+c)√c+a−b

a√b

+� (c+a)

√b+c−a

b√a

≤ s2−r2−RrRr .

Mihaly Bencze

PP25546. In all triangle ABC holds� √

c(a+b−c)

a+b ≤ 32 .

Mihaly Bencze

PP25547. In all triangle ABC holds17s2+9r2+18Rr4(s2+r2+2Rr)

− 12

�s2−r2−Rrs2+r2+2Rr

�2≤

√2s� 1√

a+b≤ 4s2+2r2+5Rr

s2+r2+2Rr.

Mihaly Bencze


PP25548. In all triangle ABC holds (b+c)√c+a−b

a√b

+ (c+a)√b+c−a

b√a

≤ (b+c)(c+a)ab

and his permutations.

Mihaly Bencze

PP25549. In all scalene triangle ABC holds��ctgA−B2 ctgC

2

��+�� sin2 A+sin2 B

sin(A−B) sinC

�� > 8.

Mihaly Bencze

PP25550. In all triangle ABC holds�s2 + r2 + 4Rr

� �s2 + r2 + 2Rr

�≥ 24Rr

�s2 − r2 −Rr

�.

Mihaly Bencze


≤ 2 + 3 3

�2Rr

s2+r2+2Rr.

Mihaly Bencze


4�


�2≤ 2 + 2

�s2+r2−Rrs2+r2+2Rr

�+ 3 3

��2Rr

s2+r2+2Rr

�2.

Mihaly Bencze

PP25553. In all triangle ABC holds��2

3

� ra3√

s2r ≤ 2.

Mihaly Bencze

PP25554. In all triangle ABC holds (4R+ r)3 + 15s2r ≥ 12s2R+ 6s23√s2r

Mihaly Bencze

PP25555. In all triangle ABC holds� (b+c)2(c+a)(a+b)

a2(c+a)(a+b)+bc(b+c)2≥ 2�s2+r2+2Rrs2−r2−Rr

�2.

Mihaly Bencze


≥ 136 − s2+r2+4Rr

2(s2−r2−4Rr).

Mihaly Bencze


PP25557. In all triangle ABC holds 3� 1

3a+b ≤ s2+r2+4Rr16sRr + 5s2+r2+4Rr

2s(s2+r2+2Rr).

Mihaly Bencze


+ s2+r2+4Rr2(s2−r2−4Rr)

≤ 52 .

Mihaly Bencze


4

�a2

b ≥ 3 4

�2s3 .

Mihaly Bencze

PP25560. If x, y, z > 0 then�� x4

x3+y3

�� y3

x(x3+y3)

�≥ 9

4 .

Mihaly Bencze

PP25561. In all triangle ABC holds� a

b2+c2≥ 8(s2−r2−Rr)

5(s2+r2+2Rr).

Mihaly Bencze

PP25562. In scalene triangle ABC holds� 1

|a2−b2| ≥3s2−7r2−28Rrs2−r2−4Rr

.

Mihaly Bencze


≥ 92 + 7

8

�s2+r2−12Rrs2+r2+4Rr

�.

Mihaly Bencze

PP25564. In all triangle ABC holds� (b+c)(c+a)

(b+c)(c+a)+ab ≤ 5(s2+r2+2Rr)2(s2−r2−Rr)

.

Mihaly Bencze

PP25565. In all triangle ABC holds λ�

3�

a2b2s(s2+r2+4Rr)

− 1�≥ s2−7r2−10Rr

2(s2+r2+2Rr)

for all λ ∈�14 ,

12

�.

Mihaly Bencze

PP25566. In all triangle ABC holds 4�


�≥�2√2− 1

� �

a2b�

ab2+ 3.

Mihaly Bencze


PP25567. In all triangle ABC holds2�s2 − r2 −Rr

�2 ≤�s2 + r2 + 2Rr

� �2s2 + 2r2 +Rr

�.

Mihaly Bencze

PP25568. In all triangle ABC holds��(c+ a)2 (b+ c)− ab (c+ a) + b2 (b+ c)

�≥ 4s2

�s2 + r2 + 2Rr

�2.

Mihaly Bencze


6�


�2−3

�

s2+r2−2Rrs2+r2+2Rr

�

2

�

(s2−r2−Rr)2+3R2r2

(s2+r2+2Rr)2− s2+r2−2Rr

s2+r2+2Rr

� ≥� (b+c)2

a2+(b+c)2.

Mihaly Bencze

PP25570. In all triangle ABC holds� a(a+c)(a+b)

(b+c)(bc+(c+a)(a+b)) ≤ 2.

Mihaly Bencze

PP25571. In all triangle ABC holds9(s2+r2+2Rr)5s2+r2+4Rr

≤ s2+r2+2Rr4s2

≤ 3 3√s2+r2+2Rr3√2Rr+ 3√s2+r2+2Rr

.

Mihaly Bencze

PP25572. In all convex quadrilateral ABCD holds��(b+ c)2 (c+ d)− a2b

�≥�

�(b+ c)3 − a3

�.

Mihaly Bencze

PP25573. In all triangle ABC holds1).� ma

2ma+mb+mc> 2

3

2).� cos A

2

2 cos A2+cos B

2+cos C

2

> 23

Mihaly Bencze

PP25574. In all triangle ABC holds�1 + 3

�2Rr

s2+r2+2Rr

�3≤ 4s2

s2+r2+2Rr≤�

5s2+r2+4Rr3(s2+r2+2Rr)

�3.

Mihaly Bencze


PP25575. If x, y, z, t ∈ R then1

sin2 x+sin2 y+sin2 z+sin2 t+ 1

cos2 x+cos2 y+cos2 z+cos2 t≥

≥ 12 + 64

27

�sin2 x sin2 y sin2 z sin2 t+ cos2 x cos2 y cos2 z cos2 t

�.

Mihaly Bencze

PP25576. In all tetrahedron ABCD holds

288r2 + 16 (ha + hb + hc + hc)

2 + 16 (ra + rb + rc + rd)

2 ≥

≥ 70

�1

1hahb

+ 1hahc

+ 1hahd

+ 1hbhc

+ 1hbhd

+ 1hchc

+ 11

rarb+ 1

rarc+ 1

rard+ 1

rbrc+ 1

rbrd+ 1

rcrc

�.

Mihaly Bencze

PP25577. Solve in Z the equationx21−x2

x3x4−x25+

x22−x3

x4x5−x26+ ...+ x2

n−x1

x2x3−x4= x2

x1+ x3

x2+ ...+ x1

xn.

Mihaly Bencze


sin A2 sin B

2 ≤ 3 + r2R −

�6 + r

2R .

Mihaly Bencze


1).� (a+b)2

a+b−c ≥ 8s

2).� (ma+mb)

2

ma+mb−mc≥ 4 (ma +mb +mc)

3).� (cos A

2+cos B

2 )2

cos A2+cos B

2−cos C

2

≥� cos A2

4).� a2

3a−b−c ≥ 2s

Mihaly Bencze

PP25580. In all triangle ABC holds� w2

a(b+c−a)bc +

8ss2+r2+2Rr

=(5s2+r2+4Rr)

2

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

Mihaly Bencze

PP25581. In all triangle ABC holds� bc

w2a≥ 4s.

Mihaly Bencze


PP25582. In all triangle ABC holds(s2+r2+4Rr)

2

2(3s2−r2−4Rr)+� bw2

ac ≤ 2

�s2 − r2 − 4Rr

�.

Mihaly Bencze

PP25583. In all triangle ABC holds� (b+c−a)w2

abc ≥ 4s3

3s2−r2−4Rr.

Mihaly Bencze

PP25584. In all triangle ABC holds� (b+c−a)3w2

abc ≥ 4s(s2−2r2−8Rr)

2

3s2−r2−4Rr.

Mihaly Bencze

PP25585. In all triangle ABC holds� (b+c)w2

abc ≤ 3s.

Mihaly Bencze

PP25586. In all triangle ABC holds2(s2−r2−4Rr)

2

3s2−r2−4Rr+� a2w2

abc ≤ 2

�s2 − r2 − 4Rr

�.

Mihaly Bencze

PP25587. In all triangle ABC holds 2s2(2R−r)2

3s2−r2−4Rr+� w2

ar2a

bc ≤ (4R+ r)2−2s2.

Mihaly Bencze

PP25588. In all triangle ABC holds 43

�s2−r2−Rrs2+r2+2Rr

�2+� w2

abc ≤ 3.

Mihaly Bencze


4�


�2+� w2

abc = 3 + 4

�s2+r2−2Rrs2+r2+2Rr

�.

Mihaly Bencze

PP25590. In all triangle ABC holds1).� b+c

a2≥ s2+r2+4Rr

2sRr

2).� a

(s−a)2≥ 2(4R+r)

sr

3).� hb+hc

h2a

≥ 2r

4).� rb+rc

r2a≥ 2

r

5).� sin2 B

2+sin2 C

2

sin4 A2

≥ 2(s2+r2−8Rr)r2


6).� cos2 B

2+cos2 C

2

cos4 A2

≥ 2 + 2�4R+r

s

�2

Mihaly Bencze

PP25591. If xk > 0 (k = 1, 2, ..., n) thenn�

k=1

1xk(1+x2

k)≥ 8n5

�

n+n�

k=1

xk

�4 .

Mihaly Bencze

PP25592. Solve in Z the equation 2016x2 − 2015y2 = 1.

Mihaly Bencze


k=1

xxkk = yy.

Mihaly Bencze

PP25594. If x, y, z ∈ (0, 1) and x+ y + z = 1 then� xλ+1

1−xλ + 13λ−1

≥� (x+y)λ+1

2λ−(x+y)λfor all λ ≥ 1.

Mihaly Bencze

PP25595. If xk ∈ (0, 1) (k = 1, 2, ..., n) , then

n�k=1

kxλ+1k

1−xλ+1k

≥

�

n�

k=1xk

�λ+1

�

n(n+1)2

�λ−�

n�

k=1xk

�λ for all λ ≥ 1.

Mihaly Bencze

PP25596. Let ABC be a triangle and M ∈ Int (ABC). Prove that

1).�

MA2 · ra ≥ 14R+r

�a2rbrc

2).�

MA2 sin2 A2 ≥ 2R

2R−r

�a2 sin2 B

2 sin2 C2

3).�

MB2 cos2 A2 ≥ 2R

4R+r

�a2 cos2 B

2 cos2 C2

Mihaly Bencze

PP25597. Solve in Z the equation 5xy+z + 7

zt+u = 2.

Mihaly Bencze


PP25598. If a > 1 and xk > 0 (k = 1, 2, ..., n) , then

n�k=1

(axk − 1) ≤�a

1n

n�

k=1xk

− 1

�n

.

Mihaly Bencze

PP25599. Let ABC be a triangle and M ∈ Int (ABC) . If α = MAB∡,

β = MBC∡, γ = MCA∡ then ctgα+ ctgβ + ctgγ ≥ s2−r(4R+r)2sr + 3 3

�2R2

sr .

Mihaly Bencze

PP25600. In all triangle ABC holdsu2λ(ab)2λ−1

(ha−r)(hb−r) +r2λ(bc)2λ−1

(hb−r)(hc−r) +t2λ(ca)2λ−1

(hc−r)(ha−r) ≥(uab+rbc+tca)2λ

32λ−2r2(5s2+r2+4Rr).

Mihaly Bencze

PP25601. If uk, vk > 0 (k = 1, 2, 3) then

u1

�v2v3v1

�lg v2v3 + u2

�v3v1v2

�lg v3v1 + u3

�v1v2v3

�lg v1v2 ≥ 3 3

√u1u2u3.

Mihaly Bencze

PP25602. In all triangle ABC holdsu2λa2λ−1

ha−r + v2λb2λ−1

hb−r + t2λc2λ−1

hc−r ≥ (ua+vb+tc)2λ

4·32λ−2sr.

Mihaly Bencze

PP25603. If a, b, c > 0, then determine all x, y ∈ R for which a, b, c are in

geometrical progression if and only if

��

ax (ab)y bx

bx (bc)y cx

cx (ca)y ax

��= 0.

Mihaly Bencze


ra�bca

�lg bc ≥ 3

3√s2r.

Mihaly Bencze

PP25605. Let A1A2...An be a convex polygon. Prove that�n�

k=1

AksinAk

�sin

�1

(n−2)π

n�k=1

A2k

�≥ (n− 2)π.

Mihaly Bencze


PP25606. If xk > 0 (k = 1, 2, ..., n) thenn�

k=1

xk�0

e−t2dt ≤ narctg

�1n

n�k=1

xk

�.

Mihaly Bencze

PP25607. If zk ∈ C (k = 1, 2, ..., n) , thenn�

k=1

z2k ≥ Im (�

z1z2) .

Mihaly Bencze

PP25608. In all triangle ABC holds�� A

sinB

�sin�1π

�AB�≥ π.

Mihaly Bencze

PP25609. In all triangle ABC holds�� 1

ra sinA

�sin�r� A

ra

�≥ 1

r .

Mihaly Bencze


1).� 1

ha−r =2(s2−r2−Rr)r(s2+r2+2Rr)

2).� 1

(ha−r)(hb−r) =s2+r2−2Rr

r2(s2+r2+2Rr)

Mihaly Bencze

PP25611. In all triangle ABC holds� a2λ−1

ha−r ≥�23

�2λ−2 s2λ−1

r for allλ ∈ (−∞, 0] ∪ [1,+∞) .

Mihaly Bencze


log ab+c

4bb+4(c+a) ≥

32 .

Mihaly Bencze

PP25613. Determine all function f : R → R such thatn�

k=1

xk ≤n�

k=1

5xkf (xk) ≤�

n�k=1

xk

�5

n�

k=1xk

for all xk ∈ R (k = 1, 2, ..., n) .

Mihaly Bencze

PP25614. Determine all functions N∗ → N∗ such thatn�

k=1

kf2 (k) = n2f2(n+1)4 for all n ∈ N∗.

Mihaly Bencze



�2−

√2 + x1 = x2�

2−√2 + x2 = x3

−−−−−−−−−�2−

√2 + xn = x1

..

Mihaly Bencze

PP25616. If xk ∈ R (k = 1, 2, ..., n) such that

x1√x2x3...xn = 1

2x2

√x3x4...xnx1 =

23

−−−−−−−−−−xn

√x1x2...xn−1 =

nn+1

thenn�

k=1

√xk

k2= n

(n+1)34.

Mihaly Bencze

PP25617. If a1 = 1 and an+1 = an + a[n+12 ] for all n ≥ 1 then compute

∞�n=1

1a2n.

Mihaly Bencze

PP25618. If xk ∈�ea, eb

�(k = 1, 2, ..., n) then�

n�k=1

1lnxk

�ln

�n�

k=1

xk

�≤ n2(a+b)2

4ab .

Mihaly Bencze


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

.

Mihaly Bencze

PP25620. Let f : N∗ → N∗ for which f (1) = 1, f (p) = 1 + f (p− 1) for all

p prime and f (p1p2...pn) =n�

k=1

f (pk) for all prime p1, p2, ..., pn. Prove that

n�k=1

2f(k) ≤�n(n+1)

2

�2≤

n�k=1

3f(k).

Mihaly Bencze



�x21 + 4

�52x2−10x3+7 = x4�

x22 + 4�52x3−10x4+7 = x5

−−−−−−−−−−−�x2n + 4

�52x1−10x2+7 = x3

.

Mihaly Bencze

PP25622. If ak ∈ (0, 1) (k = 1, 2, ..., n) and λ > 0 then�

cyclic

loga12λa2a2+λ2 ≥ n

2 .

Mihaly Bencze


5x1 = 3x2+{x3}

5x2 = 3x3+{x4}

−−−−−−5xn = 3x1+{x2}

, when {·}

denote the fractional part.

Mihaly Bencze

PP25624. Prove that 2n+1 + 3n+1−12·3n + 4n+1−1

3·4n + 6n+1−15 ≥ 4n+ 1 for all

n ∈ N.

Mihaly Bencze

PP25625. If xk ≥ 2 (k = 1, 2, ..., n) , then�

cyclic

(log2 a1)2

1+2 log2 a2≥ n

3 .

Mihaly Bencze

PP25626. Solve in C the dollowing system:

x1x2 + x3 + x4 = 2x5 + x26x2x3 + x4 + x5 = 2x6 + x27−−−−−−−−−−−xnx1 + x2 + x3 = 2x4 + x25

.

Mihaly Bencze

PP25627. If ak > 0 (k = 1, 2, ..., n) , thena1a2

+ 2�

a2a3

+ 3 3

�a3a4

+ ...+ n n

�ana1

≥ n(n+1)2 .

Mihaly Bencze


PP25628. Determine all functions f : R → R such thatf (f (x) y) + f (xf (y)) = xy + f (xy) for all x, y ∈ R.

Mihaly Bencze

PP25629. If S =n�

k=1

zk when zk ∈ C (k = 1, 2, ..., n) and t ∈ N then

(n− 2t) |s|2 + t2n�

k=1

|zk|2 =n�

k=1

|s− tzk|2 .

Mihaly Bencze


2016x1 + 2017x2 = 2019x3 + 20142016x2 + 2017x3 = 2019x4 + 2014−−−−−−−−−−−−−−2016xn + 2017x1 = 2019x2 + 2014

.

Mihaly Bencze

PP25631. If xk > 0 (k = 1, 2, ..., n) then�

cyclic

�3�x21 + x22

�+ 10x1x2 ≤ 4

n�k=1

xk.

Mihaly Bencze


2sin 3x1 = sin3 x2 + 8sinx3

2sin 3x2 = sin3 x3 + 8sinx4

−−−−−−−−−−−2sin 3xn = sin3 x1 + 8sinx2

.

Mihaly Bencze


22x1−1

= 1 + log2 (1 + x2)

22x2−1

= 1 + log2 (1 + x3)−−−−−−−−−−−−22

xn−1= 1 + log2 (1 + x1)

.

Mihaly Bencze


PP25634. If ak > 0 (k = 1, 2, ..., n) , then

2nn�

k=1

n√ak ≤ 4

n�k=1

√ak + 2n (n− 2) .

Mihaly Bencze

PP25635. Determine all increasing function f : R → R for whichf (xf (y) + yf (z) + zf (x)) = f (x) y + f (y) z + f (z)x for all x, y, z ∈ R.

Mihaly Bencze

PP25636. If zk ∈ C (k = 1, 2, ..., n) such that Re (zk) ≥ 0 (k = 1, 2, ..., n) ,

then 2n�

k=1

|zk|2 +�

(z1z2 + z1z2) ≥ 0.

Mihaly Bencze

PP25637. Determine all function f : (0,+∞) → R for which�n�

k=1

xk

�lg

�n�

k=1

xk

�≤�x2x3...xnf (x1) ≤ f

�n�

k=1

xk

�for all xk > 0

(k = 1, 2, ..., n) .

Mihaly Bencze

PP25638. If ε = −1+i√3

2 then solve on C the following system

(z1 − 1)2 + (z2 − ε)2 + (z3 − ε)2 = 3z24(z2 − 1)2 + (z3 − ε)2 + (z4 − ε)2 = 3z25−−−−−−−−−−−−−−−−(zn − 1)2 + (z1 − ε)2 + (z2 − ε)2 = 3z23

.

Mihaly Bencze


1 + log5 x1 = log2�4 +

√5x2�

1 + log5 x2 = log2�4 +

√5x3�

−−−−−−−−−−−−1 + log5 xn = log2

�4 +

√5x1�

.

Mihaly Bencze


PP25640. If xk > 0 (k = 1, 2, ..., n) then

1).� (1+x3)

2

1+33√

1+(x1x2)2≥ n

2).� (1+x2)(1+x3)

1+33√

1+(x1x2)2≥ n

Mihaly Bencze

PP25641. If xk > 0 (k = 1, 2, ..., n) such that�

cyclic

n2x1

x2+x3 = n2 then

x1 = x2 = ... = xn.

Mihaly Bencze

PP25642. If A (a) , B (b) , C (c) then determine all x, y, z ∈ R for whichABC is equilateral if and only if ax (b− c) + by (c− a) + cz (a− b) = 0.

Mihaly Bencze

PP25643. Let be bk ≥ 1 (k = 1, 2, ..., n) be a geometrical progression with

ratio q. Prove that n lg b�lg b1 +

n−12 lg q

�≤

n�k=1

�lg Sk

k

�2≤

n (lg b1)2 + n(n−1)

2 lg b1 lg q +n(n−1)(2n−1)

12 (lg q)2 when Sn = b1 + b2 + ...+ bn.

Mihaly Bencze

PP25644. Solve in R the folowing system:

1 +�3 + 2

√2�x1 ≤ 6

�√2 + 1

�x2

1 +�3 + 2

√2�x2 ≤ 6

�√2 + 1

�x3

−−−−−−−−−−−−−1 +�3 + 2

√2�xn ≤ 6

�√2 + 1

�xn

.

Mihaly Bencze

PP25645. If ÷a1, a2, ..., an are an arithmetical progression with positive

numbers and Sn =n�

k=1

ak then

a1Sn ≤n�

k=1

�Skk

�2≤ na21 +

12a1r (n− 1) + 1

12n (n− 1) (2n− 1) r2 when r

denote the ratio.

Mihaly Bencze


PP25646. If ak > 0 (k = 1, 2, ..., n) such that�

cyclic

na1+a2+...+an−1

an = nn then

a1 = a2 = ... = an.

Mihaly Bencze

PP25647. If a, b ∈ R such that (a+ 1)2 + b2 ≥ 4 then�a3 − 3ab2

�2+�3a2b− b3

�2 ≥ 1.

Mihaly Bencze

PP25648. If ak ∈ (0, 1) (k = 1, 2, ..., n) , then�

cyclic

log a1+a2+...+an−1n−1

an ≥ n.

Mihaly Bencze


k=1

1(k+1)

m√k≤ m

�1− 1

m√n+1

�for all n,m ∈ N∗.

Mihaly Bencze


log9�1 + x21 + x32

�= 2 log4 x3

log9�1 + x22 + x33

�= 2 log4 x4

−−−−−−−−−−−−−log9�1 + x2n + x31

�= 2 log4 x2

.

Mihaly Bencze

PP25651. Determine all functions f : R → R for which2 (f ◦ f) (x+ y + z+) = x+ y + z + f (1− x) + f (1− y) + f (1− z) for allx, y, z ∈ R.

Mihaly Bencze


x21 +3�x62 + 7x33 =

�x44 + 8x5

x22 +3�x63 + 7x34 =

�x45 + 8x6

−−−−−−−−−−−−−x21 +

3�x62 + 7x33 =

�x44 + 8x5

Mihaly Bencze


PP25653. If a, b > 0 then solve in R the following system:

a+√x1 − a2 = b+

√x2 − b2

a+√x2 − a2 = b+

√x3 − b2

−−−−−−−−−−−−−a+

√xn − a2 = b+

√x1 − b2

.

Mihaly Bencze


x1�3 +

√5�lg x2

+ 20 = 10�5 +

√5�lg x3

x2�3 +

√5�lg x3

+ 20 = 10�5 +

√5�lg x4

−−−−−−−−−−−−−−−−−xn�3 +

√5�lg x1

+ 20 = 10�5 +

√5�lg x2

.

Mihaly Bencze

PP25655. If 1 < a1 < a2 < ... < an are natural numbers, thenn�

k=1

log2

�1− 1

a2k

�≥ lg2

n+22(n+1) .

Mihaly Bencze

PP25656. Determine all z ∈ C such that

�|z − (2016 + i)| =

√2

|z − (1 + 2016i)| ≤√2

.

Mihaly Bencze

PP25657. Prove that for all n ∈ N the equation z4n+2 − z4n+1 + z + 1 = 0have a root z1 ∈ C\R for which |z1| = 1.

Mihaly Bencze

PP25658. Determine the integer part oflog2 2011 + log3 2012 + log4 2013 + log5 2014 + log6 2015 + log7 2016.

Mihaly Bencze

PP25659. Determine all ak > 0 (k = 1, 2, ..., n) for whichn�

k=1

k3(ak)!a2k

= (n+ 1)!− 1.

Mihaly Bencze

Octogon Mathematical Magazine, Vol. 23, No.1, April 2015 … · 2016-05-29 · x,y∈ R for which 2...

Documents

Transcript of Octogon Mathematical Magazine, Vol. 23, No.1, April 2015 … · 2016-05-29 · x,y∈ R for which 2...