Team Selection Test for IMO 2009

Letting $x=|A{B}_1|$, $y=|B{C}_1|$, $z=|C{A}_1|$, we have to show that $$\sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}}\le \frac{3}{\sqrt{2}}$$

or equivalently, $$\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}\le \frac{3}{\sqrt{2}}$$ for all positive real numbers $a$, $b$, $c$ satisfying $abc=1$. Assume without loss of generality that $ab\le 1$. Then $$\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}\le \sqrt{2}\left(\frac{1}{1+a^2}+\frac{1}{1+b^2}\right)$$ and $$\frac{1}{1+a^2}+\frac{1}{1+b^2}=1+\frac{1-a^2{b}^2}{(1+a^2)(1+b^2)}\le 1+\frac{1-a^2{b}^2}{(1+ab{)}^2}=\frac{2}{1+ab}$$ and $$\frac{1}{\sqrt{1+c^2}}\le \frac{\sqrt{2}}{1+c}$$ by the Cauchy-Schwarz inequality. Therefore, $$\begin{array}{rl}\sum _{cyc}\frac{1}{\sqrt{1+a^2}}& \le 2\sqrt{\frac{c}{1+c}}+\frac{\sqrt{2}}{1+c}=\frac{\sqrt{2}}{1+c}(\sqrt{2c(c+1)}+1)\\ & \le \frac{\sqrt{2}}{1+c}\left(\frac{2c+c+1}{2}+1\right)=\frac{3}{\sqrt{2}}\end{array}$$ by the AM-GM inequality.