, 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
$$\begin{aligned} & (a+1)(b+2) + (b+1)(c+2) + (c+1)(a+2) \\ &= ab + bc + ca + 3(a+b+c) + 6 \\ &= \frac{1}{2}((a+b+c+3)^2 - (a^2 + b^2 + c^2 - 3)) \\ &\le \frac{1}{2}(a+b+c+3)^2. \end{aligned}$$