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### Transcript of Octogon Mathematical Magazine, Vol. 23, No.1, April 2015 2016-05-29¢ ...

• 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 following inequality 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 Drăgan

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

√ n+ 1 +

√ n+ 2 +

√ n+ 3

� = �√

16n+ k � for each n ∈ N.

Marius Drăgan

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

Marius Drăgan

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

a2

�a2 � 1− 1

a2−b2−c2 � ≤ � 1 + 1

b2

�b2 � 1 + 1

c2

�c2 .

Marius Drăgan

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

� � b2 + 14

� � c2 + 14

� ≥ 26 (3a+ 2b+ c+ 1)2 .

Marius Drăgan

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

Marius Drăgan

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

22 n �

x8 + y8 �2n

= z2 + t2 + w2 has the integer solutions.

Marius Drăgan

PP24622. We consider the complex number Z such that |z| = 1. Re z ≥ 0, Im z > 0. If we denote x = π2 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 Drăgan

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

n� k=1

f �

1 (2k+1)

√ 2k−1

� ≤ nf

� 1 n

� .

Mihály 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 .

Mihály Bencze and György Szöllősy

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.

Mihály Bencze and Ferenc Olosz

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

n+ ��a1

a2

�n ≥� a1a2...an−1

an−1n + � an−11

a2a3...an .

Mihály Bencze

PP24627. If x0 ∈ R and xn+1 � 2x2n − xn + 3

� = 3x2n − xn + 2 for all n ∈ N,

then compute lim n→∞

n (1− xn) .

Mihály Bencze

PP24628. Compute F (m) = ∞� n=0

� n!m!(n+m)! (2n)!(2m)!

�2 .

Mihály Bencze

• 206 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015

PP24629. If a ≥ 1 and f, g : [0, a] → [0, 1] are two increasing functions and g is continuous, then compute lim

n→∞ n

� ag (0)−

a� a g (fn (x)) dx

� .

Mihály Bencze

PP24630. If a, b, c > 0, then ab4 + b3c2 + a5c+ 3a2b2c ≥ c

� a2b2 + b3c+ a3c

� +

3 √ a2b2c2

� a3 + abc+ b3

� .

Mihály Bencze

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

3 n�

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 ≥ 3n2 .

Mihály Bencze

PP24632. If x > 1 then π 2

6 > 13 36 +

∞� n=2

� 1− x−1

xn+1−1 − (n−1)x n+1

�2 .

Mihály Bencze

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

k=1

���zk + 1zk ��� 8 ≥ 16n

� 1 + 2n

n� k=1

Re � z2k ��2

.

Mihály Bencze

PP24634. Determine all x, y, z ∈ C for which 3An = 2n (A+ I3) + (−1)n (2I3 −A) for all n ∈ N∗, where

A =

 

0 x y y 0 z z x 0

  .

Mihály Bencze

PP24635. Prove that 14

� π2

6 − 1 � <

∞� n=2

� 1� 0

xndx x2+1

�2 < π

2

24 .

Mihály Bencze

• Proposed Problems 207

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

Mihály Bencze

PP24637. In all scalene triangle ABĈ 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

Mihály Bencze

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

a �

ama

�n + �

a �

ama

�p + �

a �

ama

�k ≥

≥ �

a amna+bm

p b+cm

k c +

a

ampa+bm k b+cm

n c +

a amka+bm

n b +cm

p c .

Mihály 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

z x+y 2

k − n�

k=1

w x+y 2

k

���� ≤ n�

k=1

��zxk − w y k

��+ n�

k=1

��zyk − wxk �� .

Mihály Bencze

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

λ = lim n→∞

xn. Compute lim n→∞

n

� λ− λ−xn

ln λ xn

� .

Mihály Bencze

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

(sinA)λ

(cos Aλ ) λ ≥

sinA �

cos A λ

.

Mihály Bencze

• 208 Octogon Mathematical Magazine, Vol. 23, No.1, April 2015

PP24642. Solve in R the equation n�

k=1

� x+k k+1

� = 2m, when [·] denote the

integer part.

Mihály Bencze

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

  n

n �

k=1

1 ak

 

n �

k=1

1 ak

≥ n�

k=1

a 1 ak k .

Mihály Bencze

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

k=1

� pk pk−1 �pn−k

.

Mihály 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−xn1+x1 = π − 4 (arctanx2)

2

.

Mihály 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.

Mihály Bencze

PP24647. Denote A (n) = Lpn−Ln

p when p is a prime and Ln denote the n th

Lucas numbers. Compute ∞� n=1

1 1+A2(n)

.

Mihály Bencze

PP24648. Solve in N the equation n�

k=1

(kr + 1) = pr, where r ∈ Z.

Mihály Bencze

• Proposed Problems 209

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

ln(mam2bmc) < 1.

Mihály Bencze

PP24650. If S (n, p) = n�

k=1

kp then

t� r=1

� 2t+1 2r

� S (n, 2r) = 12

� (n+ 1)2t+1 + n2t+1 − 2n− 1

� .

Mihály Bencze

PP24651. Compute lim n→∞

n

� lnΓ (x+ 1) + lnΓ (1− x)−

n� k=1

ζ (2k) x 2k

k

� ,

when x ∈ (−1, 1) .

Mihály 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∗.

Mihály Bencze

PP24653. In all triangle ABC holds 256sRr ≤ (a+ 2s) (b+ 2s) (c+ 2s) ≤ 4sR(R+2r)

r2 .

Mihály Bencze

PP24654. Compute lim n→∞

n

� ζ (λ)−

n� i=1

n� j=1

(i−1)!(j−1)! (i+j)!