, 2008

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

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