Reverse Triangle Inequality

Theorem
Let $M = \struct {X, d}$ be a metric space.

Then:
 * $\forall x, y, z \in X: \size {\map d {x, z} - \map d {y, z} } \le \map d {x, y}$

Proof
Let $M = \struct {X, d}$ be a metric space.

By the triangle inequality, we have:
 * $\forall x, y, z \in X: \map d {x, y} + \map d {y, z} \ge \map d {x, z}$

By subtracting $\map d {y, z}$ from both sides:
 * $\map d {x, y} \ge \map d {x, z} - \map d {y, z}$

Now we consider 2 cases.


 * Case $1$: Suppose $\map d {x, z} - \map d {y, z} \ge 0$.

Then:
 * $\map d {x, z} - \map d {y, z} = \size {\map d {x, z} - \map d {y, z} }$

and so:
 * $\map d {x, y} \ge \size {\map d {x, z} - \map d {y, z} }$


 * Case 2: Suppose $\map d {x, z} - \map d {y, z} < 0$.

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

Hence:
 * $\map d {x, y} \ge \map d {y, z} - \map d {x, z}$

Since we assumed $\map d {x, z} - \map d {y, z} < 0$, we have that:
 * $\map d {y, z} - \map d {x, z} > 0$

and so:
 * $\map d {y, z} - \map d {x, z} = \size {\map d {y, z} - \map d {x, z} }$

Thus we obtain:
 * $\map d {x, y} \ge \size {\map d {x, z} - \map d {y, z} }$

Since these cases are exhaustive, we have shown that:
 * $\forall x, y, z \in X: \map d {x, y} \ge \size {\map d {x, z} - \map d {y, z} }$