Ukrajina 2008

Cauchy-Bunyakowsky inequality implies that you just have to prove the following inequality: $$\begin{gathered} \left(\frac{x}{x^2+y+z}+\frac{y}{y^2+z+x}+\frac{z}{z^2+x+y}\right)(x+y+z)\le 3.\text{ And since} \\ (x+y+z{)}^2\le 3(x^2+y^2+z^2)=9,\text{ it’s enough to prove that} \end{gathered}$$ $$\frac{x}{x^2+y+z}+\frac{y}{y^2+z+x}+\frac{z}{z^2+x+y}\le 1.$$

We obtain $(a^2+b+c)(1+b+c)\ge (a+b+c{)}^2$ for non-negative $a,b,c$. Therefore, $\frac{a}{a^2+b+c}\le \frac{a(1+b+c)}{(a+b+c{)}^2}$. If we add all these inequalities, we obtain that it's enough to prove that: $\frac{x(1+y+z)+y(1+z+x)+z(1+x+y)}{(x+y+z{)}^2}\le 1$. This inequality is equivalent to $x+y+z\le x^2+y^2+z^2=3$, which has already been proved.