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$\sqrt{x} + \sqrt{y} + \sqrt{z} = 1$.

Source: APMO 2007

March 31, 2007
inequalities

Problem Statement

Let x;yx; y and zz be positive real numbers such that x+y+z=1\sqrt{x}+\sqrt{y}+\sqrt{z}= 1. Prove that x2+yz2x2(y+z)+y2+zx2y2(z+x)+z2+xy2z2(x+y)1.\frac{x^{2}+yz}{\sqrt{2x^{2}(y+z)}}+\frac{y^{2}+zx}{\sqrt{2y^{2}(z+x)}}+\frac{z^{2}+xy}{\sqrt{2z^{2}(x+y)}}\geq 1.