Serbian Mathematical Olympiad

Solution:

The Cauchy-Schwarz inequality for the triples $(x, \sqrt{y}, 1)$ and $(1, \sqrt{y}, z)$ gives $\frac{1}{x^{2} + y + 1} \leqslant \frac{1 + y + z^{2}}{(x + y + z)^{2}}$. Analogously, $\frac{1}{y^{2} + z + 1} \leqslant \frac{1 + z + x^{2}}{(x + y + z)^{2}}$ and $\frac{1}{z^{2} + x + 1} \leqslant \frac{1 + x + y^{2}}{(x + y + z)^{2}}$ hold. Adding these inequalities we obtain
$$\frac{1}{x^{2} + y + 1} + \frac{1}{y^{2} + z + 1} + \frac{1}{z^{2} + x + 1} \leqslant \frac{3 + x + y + z + x^{2} + y^{2} + z^{2}}{(x + y + z)^{2}} = S$$
It remains to prove that $S \leqslant 1$, and this is equivalent to $3 + x + y + z \leqslant 2(x y + y z + z x) = 2(x + y + z)$ by the condition of the problem, i.e. $x + y + z \geqslant 3$. This, however, follows from $x + y + z = x y + y z + z x \leqslant \frac{(x + y + z)^{2}}{3}$.
Equality holds only for $x = y = z = 1$.