Reverse triangle inequality

For any metric space $(𝑀, 𝑑)$ , the reverse triangle inequality

| 𝑑 (𝑥, 𝑦) - 𝑑 (𝑦, 𝑧) | \leq 𝑑 (𝑥, 𝑧)

holds for any $𝑥, 𝑦, 𝑧 \in 𝑀$ . #m/thm/anal

Proof

Consider some $𝑥, 𝑦, 𝑧 \in 𝑀$ . By the triangle inequality, $𝑑 (𝑥, 𝑦) \leq 𝑑 (𝑥, 𝑧) + 𝑑 (𝑦, 𝑧)$ , and hence $𝑑 (𝑥, 𝑦) - 𝑑 (𝑦, 𝑧) \leq 𝑑 (𝑥, 𝑧)$ . Likewise $𝑑 (𝑦, 𝑧) \leq 𝑑 (𝑥, 𝑦) + 𝑑 (𝑥, 𝑧)$ , implying $𝑑 (𝑦, 𝑧) - 𝑑 (𝑥, 𝑦) \leq 𝑑 (𝑥, 𝑧)$ . Therefore $| 𝑑 (𝑥, 𝑦) - 𝑑 (𝑦, 𝑧) | \leq 𝑑 (𝑥, 𝑧)$ .

It follows immediately that the analogous reverse triangle inequality holds for any Normed vector space $𝑉$ :

| ‖ 𝑣 ‖ - ‖ 𝑢 ‖ | \leq ‖ 𝑣 - 𝑢 ‖

for any $𝑣, 𝑢 \in 𝑉$ .

#state/tidy | #lang/en | #SemBr