Saudi Arabia Mathematical Competitions 2012

We have $x^2+1\ge 2x$, $y^2+4\ge 4y$, $z^2+9\ge 6z$, hence $$\frac{1}{x^2+1}+\frac{1}{y^2+4}+\frac{1}{z^2+9}\le \frac{1}{2x}+\frac{1}{4y}+\frac{1}{6z}.(1)$$ 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 $$\frac{1}{2x}+\frac{1}{4y}+\frac{1}{6z}\le \frac{7}{12}\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}.(2)$$ 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 $x=1$, $y=2$, $z=3$.