Reverse Triangle Inequality

Theorem
Let $M = \left({X, d}\right)$ be a metric space.

Then:
 * $\forall x, y, z \in X: \left|{d \left({x, z}\right) - d \left({y, z}\right)}\right| \le d \left({x, y}\right)$

Proof
As $M = \left({X, d}\right)$ is a metric space, we have $\forall x, y, z \in X: d \left({x, y}\right) + d \left({y, z}\right) \ge d \left({x, z}\right)$.

So $d \left({x, y}\right) \ge d \left({x, z}\right) - d \left({y, z}\right)$ by subtracting $d \left({y, z}\right)$ from both sides.

If $d \left({x, z}\right) - d \left({y, z}\right) \ge 0$ we are done.

Otherwise, that means $d \left({x, z}\right) < d \left({y, z}\right)$, and instead we use $\forall x, y, z \in X: d \left({y, x}\right) + d \left({x, z}\right) \ge d \left({y, z}\right)$.

Hence $d \left({x, y}\right) \ge d \left({y, z}\right) - d \left({x, z}\right)$.

So combining them both together it follows that $\forall x, y, z \in X: \left|{d \left({x, z}\right) - d \left({y, z}\right)}\right| \le d \left({x, y}\right)$.