Let x = ak1, y = ak2, z = ak3 for k ≥ 0, then (a31)k+ (a32)k+ (a33)k+ 3(a1a2a3)k ≥ (a21a2)k+ (a1a22)k+ (a22a3)k+ (a2a23)k+ (a23a1)k+ (a3a21)k

Problem 11692

(American Mathematical Monthly, Vol.120, February 2013) Proposed by Cezar Lupu (USA) and Stefan Spataru (Romania).

Let a1, a2, a3, a4 be real numbers in (0, 1) with a4= a1. Show that 3

1 − a1a2a3

+

3

X

k=1

1 1 − a³_k ≥

3

X

k=1

 1

1 − a²_kak+1

+ 1

1 − aka²_k+1

 .

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

By Schur’s inequality, for any x, y, z ≥ 0,

x³+ y³+ z³+ 3xyz ≥ x²y + xy²+ x²z + xz²+ y²z + yz². Let x = a^k₁, y = a^k₂, z = a^k₃ for k ≥ 0, then

(a³₁)^k+ (a³₂)^k+ (a³₃)^k+ 3(a1a2a3)^k ≥ (a²₁a2)^k+ (a1a²₂)^k+ (a²₂a3)^k+ (a2a²₃)^k+ (a²₃a1)^k+ (a3a²₁)^k. Finally, since a₁, a₂, a₃∈ (0, 1), by summing for k ≥ 0, the geometric series converge and easily we

obtain the required inequality.