Reverse Triangle Inequality

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Theorem

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

Then:

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

Normed Division Ring

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

Then:

$\forall x, y \in R: \norm {x - y} \ge \big\lvert {\norm x - \norm y} \big\rvert$

Normed Vector Space

Let $\left({X, \left\lVert{\, \cdot \,}\right\rVert}\right)$ be a normed vector space.

Then:

$\forall x, y \in X: \left\lVert{x - y}\right\rVert \ge \big\lvert{\left\lVert{x}\right\rVert - \left\lVert{y}\right\rVert}\big\rvert$

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

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:

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

$\blacksquare$