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
Let $M = \left({X, d}\right)$ be a metric space. By the Triangle Inequality we have:
 * $\forall x, y, z \in X: d \left({x, y}\right) + d \left({y, z}\right) \ge d \left({x, z}\right)$

By subtracting $d \left({y, z}\right)$ from both sides:
 * $d \left({x, y}\right) \ge d \left({x, z}\right) - d \left({y, z}\right)$

Now we consider 2 cases.

Case 1: Suppose $d \left({x, z}\right) - d \left({y, z}\right) \ge 0$.

Therefore:
 * $d \left({x, z}\right) - d \left({y, z}\right) = \left|d \left({x, z}\right) - d \left({y, z}\right)\right|$

and so
 * $ d \left({x, y}\right) \geq \left|d \left({x, z}\right) - d \left({y, z}\right)\right|$

Case 2: Suppose $d \left({x, z}\right) - d \left({y, z}\right) < 0$.

Applying the triangle inequality again, we have:
 * $\forall x, y, z \in X: d \left({y, x}\right) + d \left({x, z}\right) \ge d \left({y, z}\right)$

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

Since we assumed $d \left({x, z}\right) - d \left({y, z}\right) < 0$, we have that $d \left({y, z}\right) - d \left({x, z}\right)>0$ and so
 * $d \left({y, z}\right) - d \left({x, z}\right) = \left|d \left({y, z}\right) - d \left({x, z}\right) \right|$

Thus we obtain
 * $ d \left({x, y}\right) \geq \left|d \left({x, z}\right) - d \left({y, z}\right)\right|$

Since these cases are exhaustive, we have shown that:
 * $\forall x, y, z \in X: d \left({x, y}\right) \geq \left|d \left({x, z}\right) - d \left({y, z}\right)\right|$