Problem

[99]

Let \(x, y, z\) be three positive reals such that \(xyz \ge 1\). Prove that $$ \frac{x^5\;-\;x^2}{x^5\;+\;y^2\;+\;z^2}\;+\;\frac{y^5\;-\;y^2}{x^2\;+\;y^5\;+\;z^2}\;+\;\frac{z^5\;-\;z^2}{x^2\;+\;y^2\;+\;z^5}\;\geq\;0. $$

Solution

Attributes Олімпіадна
Source International Mathematical Olympiad
Year 2005
Number 3
Difficulty 10.0
Themes