EDIT: I give a new proof, that is better than the old one (see the old one below).
I am going to prove something stronger: $|a+b|+ |a-b| \ge 2\max \{ |a|,|b| \}$. Assume WLOG that $|a| \ge |b|$. Square both sides to get the following equivalent inequality (using $|a\pm b|^2 = |a|^2 + |b|^2 \pm 2(a\overline{b} + b\overline{a})$:
$2(|a|^2+|b|^2) + 2|a^2-b^2| \ge 4|a|^2$ Which is the same as: $|a^2-b^2| \ge |a^2|-|b^2|$ Which in turn is a direct application of the triangle inequality.
My old proof:
Since both sides are non-negative, if I take the square of both sides I get an equivalent inequality. By using $|a\pm b|^2 = |a|^2 + |b|^2 \pm 2(a\overline{b} + b\overline{a})$, we get: $|a|^2 + |b|^2 + 2|a^2-b^2| \ge 2|ab|$ Since $(|a|-|b|)^2 \ge 0$, we have $|a|^2+|b|^2 \ge 2|ab|$, and because $|a^2-b^2|\ge 0$, the inequality follows.
Equality occurs when $|a^2-b^2|=0$ and $|a|=|b|$, i.e. $a=\pm b$.
(I must admit I prefer ashley's proof, although after my edit my proof works for complex numbers too)