Reverse Triangle Inequality

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Theorem

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


Then:

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


Normed Division Ring

Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring.

Then:

$\forall x, y \in R: \norm {x - y} \ge \bigsize {\norm x - \norm y}$


Normed Vector Space

Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.

Then:

$\forall x, y \in X: \norm {x - y} \ge \size {\norm x - \norm y}$


Real and Complex Numbers

Let $x$ and $y$ be elements of either the real numbers $\R$ or the complex numbers $\C$.

Then:

$\cmod {x - y} \ge \size {\cmod x - \cmod y}$


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:

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

$\blacksquare$