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

First show that $\frac{a}{1+a}\le \frac{b}{1+b},$ if $0\le a\le b$. Secondly, try to prove that $\frac{a+b}{1+c+d}\le\frac{a}{1+c}+\frac{b}{1+d}$, if $0\le a,b,c,d$. And thirdly, use that your function $d(x,y)$ is a metric.

This result doesn't depend on $d(x,y)$ being the Euclidean metric - if $\rho(x,y) = \frac{d(x,y)}{1+d(x,y)}$ where $d(x,y)$ is a metric on $X$, $\rho$ is a metric which induces the same topology as $d$. This metric is also bounded.

Here is a hint to showing this: You've already got non-negativity (with zero iff equality) and symmetry. Consider $f(x)=\frac{x}{1+x}$ on $x>0$. Use the mean value theorem to show that $f(a+b) - f(b) \leq f(a)$. Now, compare this with the definition of $\rho$.

Side note: Another useful metric from $d$ is $\bar{d}(x,y) = \min\{d(x,y),1\}$, which is known as the standard bounded metric corresponding to $d$. This also gives the same topology as $d$ does on $X$.

I just wanted to add an answer that will show you a more general view on the question.

Suppose you have a metric $d$, the question is when for some function $f$, $f(d)$ is also a metric.

First, you obviously need

a) $f$ is defined on $\mathbb{R}_+$, $f(0)=0$ and $f(x)>0$ for $x>0$.

You also need $f$ to satisfy $f(d(x,z))+f(d(y,z))\ge f(d(x,y))$.

Now, if $f$ is also

b) increasing, and

c) concave,

then $f(d(x,z))+f(d(y,z))\ge f(d(x,z)+d(y,z))$ (because for a concave function $f(a+b)-f(a)\le f(b)-f(0)=f(b)$, or put it differently $f(a)+f(b)\ge f(a+b)$),

and $f(d(x,z)+d(y,z))\ge f(d(x,y))$ (because $f$ is increasing and $d(x,z)+d(y,z)\ge d(x,y)$).

So, given any function $f$ that starts at $0$, and is positive, increasing and concave for positive reals ($\frac{x}{1+x}$, $\arctan x$, $\sqrt{x}$ etc.) for any metric $d$ you have another metric $f(d)$. In fact, I think, if $f$ is continuous at $0$, metrics $f(d)$ and $d$ are equivalent, but I am not sure about this last statement.