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  • 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)!