Follow up to a question, why does proof $\rho(x,y) = \frac{d(x,y)}{1+d(x,y)}$ work

Question

This is a follow up to a well known question

Showing $\rho (x,y)=\frac{d(x,y)}{1+d(x,y)}$ is a metric

A general proof is as follows: Let $x,y,z \in (X, \rho)$

\begin{align} \rho(x,z) &= \dfrac{d(x,z)}{1+d(x,z)}\\ & \leq \dfrac{d(x,y)+d(y,z)}{1+d(x,y)+d(y,z)}\\ &= \dfrac{d(x,y)}{1+d(x,y)+d(y,z)} + \dfrac{d(y,z)}{1+d(x,y)+d(y,z)} \\&\leq \rho(x,y) + \rho(y,z) \end{align}

This might be very simple but how do you rigorously argue that

$$\dfrac{d(x,z)}{1+d(x,z)}\leq \dfrac{d(x,y)+d(y,z)}{1+d(x,y)+d(y,z)}$$

If only the numerator guy got bigger due to triangle inequality, then I would have no problem. However, the thing in the denominator also got bigger. How do we know that this inequality holds?

Answer 1

$0 \le a < b \implies 0 \le \frac{a}{b} \le\frac{a+n}{b+n}< 1$ where $n\ge0$ with $\lim_{n\to\infty}\frac{a+n}{b+n}=1$

We can use this since, of course, the difference between the numerators is the same as the difference between the denominators.

see this question if you want a proof of the inequality.

Answer 2

One general reason, as user1952009 notes in the comments, is that $f(x) = \frac{x}{1 + x}$ is increasing and concave on $[0, \infty)$.