2012/13 AHSMC Part 1 Problem 13

We have $$\begin{align*} \frac{1}{8} &= \frac{1}{a^2 + 4b + 4} + \frac{1}{b^2 + 4a + 4} \\ &= \frac{1}{(a - 2)^2 + 4a + 4b} + \frac{1}{(b - 2)^2 + 4b + 4a} \\ &\le \frac{1}{4(a + b)} + \frac{1}{4(a + b)} \\ &= \frac{1}{2(a + b)}. \end{align*}$$ Hence $a + b \le 4$. This maximum value is attained if and only if $a = b = 2$. The answer is (D).