Saudi Arabian Mathematical Competitions, 2012

We have $x^2+1 \ge 2x$, $y^2+4 \ge 4y$, $z^2+9 \ge 6z$, hence
$$\frac{1}{x^2+1} + \frac{1}{y^2+4} + \frac{1}{z^2+9} \le \frac{1}{2x} + \frac{1}{4y} + \frac{1}{6z}. \quad (1)$$
Using Cauchy-Schwarz inequality it follows
$$\left(\frac{1}{2x} + \frac{1}{4y} + \frac{1}{6z}\right)^2 \le \left(\frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2}\right) \left(\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2}\right) = \frac{1}{4} \cdot \frac{49}{36} \left(\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2}\right),$$
hence
$$\frac{1}{2x} + \frac{1}{4y} + \frac{1}{6z} \le \frac{7}{12} \sqrt{\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2}}. \quad (2)$$
From (1) and (2) we get
$$\frac{1}{x^2+1} + \frac{1}{y^2+4} + \frac{1}{z^2+9} \le \frac{7}{12} \sqrt{\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2}}.$$
with equality if and only if $x = 1$, $y = 2$, $z = 3$.