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Unusual inequality, x+y+z=0

Source: Balkan Mathematical Olympiad 2011. Problem 2.

May 6, 2011
inequalitiesalgebrathree variable inequality

Problem Statement

Given real numbers x,y,zx,y,z such that x+y+z=0x+y+z=0, show that x(x+2)2x2+1+y(y+2)2y2+1+z(z+2)2z2+1≥0\dfrac{x(x+2)}{2x^2+1}+\dfrac{y(y+2)}{2y^2+1}+\dfrac{z(z+2)}{2z^2+1}\ge 0 When does equality hold?
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