Show $g:X^2\to \mathbb{R}$ is a metric.

Question

$d$ is a metric on a set $X$.

Let $g:X^2\to\mathbb{R}$ such that $g(x,y)=\dfrac{d(x+y)}{1+d(x,y)}$

I have to prove that $g$ is a metric. I'm struggling to show triangle inequaility i.e. $g(x,y)\leq g(x,z)+g(y,z)$.

So far I tried $g(x,y)=\dfrac{d(x,y)}{1+d(x,y)}\leq \dfrac{d(x,z)+d(y,z)}{1+d(x,y)}\leq \dfrac{d(x,z)}{1+d(x,y)}+\dfrac{d(y,z)}{1+d(x,y)}\leq\dfrac{d(x,z)}{d(x,y)}+\dfrac{d(y,z)}{d(x,y)}$

and i'm stuck here. Starting from $g(x,z)+g(y,z)$ had a similar problem.

Another thing I noticed is $\dfrac{d(x,y)}{1+d(x,y)}\leq d(x,y)\leq d(x,z)+d(y,z)$.

Any hint?

Answer 1

In this answer, let $r,s,t$ range over $\mathbb{R}^{\geq0}$.

$f(t)=t/(1+t)$ is an increasing concave function.

Try to prove the following inequality:

$r/(1+r)+s/(1+s)\geq (r+s)/(1+r+s)$.

If you just want to read the proof, look at the first two paragraphs of the accepted answer in the following link. The example of $f$ in that answer is different, but you can use the same sort of reasoning for this problem.

Prove variant of triangle inequality containing p-th power for 0 < p < 1

Now, $g(x,y)+g(y,z)$

$=f\bigg(d(x,y)\bigg)+f\bigg(d(y,z)\bigg)$

$\geq f\bigg(d(x,y)+d(y,z)\bigg)$

$\geq f(d(x,z))$ [[using the fact that $f$ is an increasing function]]

$=g(x,z)$.