Metric over 1 plus Metric forms Metric

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

Let $d_3: A^2 \to \R$ be the mapping defined as:
 * $\forall \tuple {x, y} \in A^2: \map {d_3} {x, y} = \dfrac {\map d {x, y} } {1 + \map d {x, y} }$

Then $d_3$ is a metric for $A$.

Proof
It is to be demonstrated that $d_3$ satisfies all the metric space axioms.

Proof of
So holds for $d_3$.

Proof of
Note that $\map f x = \dfrac x {1 + x}$ is an increasing function of $x$ for $x = 0$.

In order to simplify the algebra, let $a = \map d {x, y}$, $b = \map d {y, z}$ and $c = \map d {x, z}$.

Then:

But $a + b + 2 a b > c$.

Because $\map f x = \dfrac x {1 + x}$ is an increasing function:
 * $\dfrac {a + b + 2 a b} {1 + a + b + 2 a b} > \dfrac c {1 + c}$

and it follows by definition of $d_3$ that:
 * $\map {d_3} {x, y} + \map {d_3} {y, z} \ge \map {d_3} {x, z}$

So holds for $d_3$.

Proof of
So holds for $d_3$.

Proof of
So holds for $d_3$.

Thus $d_3$ satisfies all the metric space axioms and so is a metric.