Team Selection Test for IMO 2011

Since $4(x+2{)}^2-3(x+1)(x+5)=(x-1{)}^2\ge 0$, we have $\frac{x+2}{(x+1)(x+5)}\ge \frac{3}{4(x+2)}$. Therefore it suffices to show that $$\frac{a+1}{b+2}+\frac{b+1}{c+2}+\frac{c+1}{a+2}\ge 2$$ for all positive real numbers satisfying $a^2+b^2+c^2\ge 3$.

The Cauchy-Schwarz Inequality gives $$((a+1)(b+2)+(b+1)(c+2)+(c+1)(a+2))\left(\frac{a+1}{b+2}+\frac{b+1}{c+2}+\frac{c+1}{a+2}\right)\ge (a+b+c+3{)}^2.$$ We finish by observing that $$\begin{array}{rl}& (a+1)(b+2)+(b+1)(c+2)+(c+1)(a+2)\\ & =ab+bc+ca+3(a+b+c)+6\\ & =\frac{1}{2}((a+b+c+3{)}^2-(a^2+b^2+c^2-3))\\ & \le \frac{1}{2}(a+b+c+3{)}^2.\end{array}$$