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:

$\forall x, y, z \in X: d \left({x, y}\right) + d \left({y, z}\right) \ge d \left({x, z}\right)$

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

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

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

Then we can use said assumption to re-write:

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

to:

$\left|d \left({x, z}\right) - d \left({y, z}\right)\right| \le d \left({x, y}\right)$

and the proof is finished.

Otherwise:

$d \left({x, z}\right) < d \left({y, z}\right)$

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:

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

Combining them both together it follows that:

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

$\blacksquare$