The ultrametric (strong triangle) inequality

Write $x = p^{v_p(x)} u$, $y = p^{v_p(y)} w$ with $u, w$ units in $\mathbb{Z}_p$ (i.e. $|u|_p = |w|_p = 1$), and suppose without loss of generality $v_p(x) \le v_p(y)$, so $|x|_p \ge |y|_p$. Factor out the smaller power: $x + y = p^{v_p(x)}(u + p^{v_p(y)-v_p(x)} w)$. The term in parentheses is a genuine element of $\mathbb{Z}_p$ (a sum of elements of $\mathbb{Z}_p$), so its $p$-adic absolute value is $\le 1$; hence $|x+y|_p \le p^{-v_p(x)} = |x|_p = \max(|x|_p, |y|_p)$, proving the inequality. If moreover $v_p(x) < v_p(y)$ strictly (i.e. $|x|_p \ne |y|_p$), then $p^{v_p(y)-v_p(x)} w \equiv 0 \pmod p$ while $u$ is a unit, so $u + p^{v_p(y)-v_p(x)} w \equiv u \not\equiv 0 \pmod p$ is again a unit; thus $|x+y|_p = p^{-v_p(x)}$ exactly, giving equality.