Nermin Hodzic, July 2015

Let x, y, z be nonnegative reals such that x2+y2+z2 = 3.Prove the inequalityx4y4 + y4z4 + z4x4 + (xyz)3 ≥ 4(xyz)2

Solution;

Let xyz = w3.Using AmGm inequality we have

3 = x2 + y2 + z2 ≥ 3 3√x2y2z2 = 3w2 ⇒ w ≤ 1

Using Schleifer inequality we have

x4y4 + y4z4 + z4x4 + 3 3√x8y8z8 ≥ 2x2y2z2(x2 + y2 + z2)⇔

x4y4 + y4z4 + z4x4 + 3 3√x8y8z8 ≥ 6x2y2z2

Hence it is su�cies to prove

6(xyz)2 − 3(xyz)83 ≥ 4(xyz)2 − (xyz)3 ⇔

(xyz)3 − 3(xyz)83 + 2(xyz)2 ≥ 0⇔

xyz − 3(xyz)23 + 2 ≥ 0⇔

w3 − 3w2 + 2 ≥ 0⇔ (w − 1)(w2 − 2w − 2) ≥ 0

Which is obvious since w − 1 ≤ 0 and w2 − 2w − 2 ≤ 0Equality holds if and only if w = 1 or w = 0,which yields (x, y, z) = (1, 1, 1) or (x, y, z) = (0, 0,

√3) and its cyclic permuta-

tions.

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