Romanian Mathematical Olympiad

We notice that $|z - 2i| = \left|\frac{2}{5}(z - 5i) + \frac{3}{5}z\right| \le \frac{2}{5}|z - 5i| + \frac{3}{5}|z|$.

In the same way, $|z - 3i| = \left|\frac{3}{5}(z - 5i) + \frac{2}{5}z\right| \le \frac{3}{5}|z - 5i| + \frac{2}{5}|z|$, whence $|z| + |z - 5i| \ge |z - 2i| + |z - 3i|$.

Equality takes place if there exists $\lambda \ge 0$ such that $z - 5i = \lambda z$ or if $z = 0$, i.e. $z = ai$, with $a \in (-\infty, 0] \cup [5, +\infty)$.