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Yet another classical inequality

Source: Romanian District Olympiad 2006, Grade 9, Problem 1

March 11, 2006
inequalitiesfunctionalgebra

Problem Statement

Let x,y,zx,y,z be positive real numbers. Prove the following inequality: 1x2+yz+1y2+zx+1z2+xy≤12(1xy+1yz+1zx). \frac 1{x^2+yz} + \frac 1{y^2+zx } + \frac 1{z^2+xy} \leq \frac 12 \left( \frac 1{xy} + \frac 1{yz} + \frac 1{zx} \right).
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