Ultrametric space

Remarks

The strong triangle inequality implies the usual one, since $\max(a, b) \le a + b$ for $a, b \ge 0$.

Example: fix a prime $p$ and let $v_p$ be the p-adic valuation on $\Q$. Put $d(x, y) = p^{-v_p(x - y)}$ for $x \ne y$ and $d(x, x) = 0$. Since $v_p(a + b) \ge \min(v_p(a), v_p(b))$, applying it to $a = x - y$ and $b = y - z$ gives the strong triangle inequality.

Non-example: the usual metric $d(x, y) = \abs{x - y}$ on $\R$. Here $d(0, 2) = 2$, while $\max(d(0, 1), d(1, 2)) = 1$.

In an ultrametric space, if $d(x, y) \ne d(y, z)$, then $d(x, z) = \max(d(x, y), d(y, z))$. Every open ball $B(a, r)$ equals $B(b, r)$ for each $b \in B(a, r)$.