Prove that for any positve real Prove that for any positive real numbers $x,y,z$ such that $xyz \geq 1$
$$\frac{x^5-x^2}{x^5+y^2+z^2} + \frac{y^5-y^2}{y^5+z^2+x^2} +\frac{z^5-z^2}{z^5+x^2+y^2} \geq 0.$$
This problem is from the $2005$ IMO competition.

Since,$$\frac{x^5-x^2}{x^5+y^2+z^2} - \frac{x^5-x^2}{x^3(x^2+y^2+z^2)} = \frac{x^2(x^3-1)^2(y^2+z^2)}{x^3(x^2+y^2+z^2)(x^5+y^2+z^2)} \ge 0$$
Hence, it suffices to prove $\displaystyle \sum\limits_{cyc} \frac{x^5-x^2}{x^3(x^2+y^2+z^2)} \ge 0$
$$\sum\limits_{cyc} \frac{x^5-x^2}{x^3(x^2+y^2+z^2)} = \frac{\sum\limits_{cyc} \left(x^2 - \frac{1}{x}\right)}{x^2+y^2+z^2} \ge \frac{\sum\limits_{cyc} \left(x^2 - yz\right)}{x^2+y^2+z^2} = \frac{\sum\limits_{cyc} \left(x-y\right)^2}{2(x^2+y^2+z^2)} \ge 0$$
Since, $xyz \ge 1$.
Here are a few more proofs of the inequality, possibly neater.
