, 2014

Noting that if $x, y > 0$, then $x + y \le \sqrt{2} \sqrt{x^2 + y^2}$, with equality iff $x = y$, then for all complex numbers $z$ satisfying $|z| = 1$
$$\begin{align*} |z - a| + |z + a| &\le \sqrt{2} \sqrt{|z + a|^2 + |z - a|^2} \\ &= \sqrt{2} \sqrt{(z + a)(\bar{z} + a) + (z - a)(\bar{z} - a)} \\ &= \sqrt{2} \sqrt{|z|^2 + za + a\bar{z} + a^2 + |z|^2 - za - a\bar{z} + a^2} \\ &= \sqrt{2} \sqrt{2 + 2a^2} = 2\sqrt{1 + a^2}, \end{align*}$$
with equality iff $|z - a| = |z + a|$, iff $2\Re(z) = z + \bar{z} = 0$, i.e., $z = i$.