TITLE: Show that $d_b(x,y)=\frac{d(x,y)}{1+d(x,y)}$ is a metric.
QUESTION [15 upvotes]: where $(X,d)$ is a metric and $x,y \in X$.
I know we need to show:

non-negativity: $d(x,y)\geq$ 0
$d(x,y)=0$ if and only if $x=y$
symmetry: $d(x,y)=d(y,x)$
$d(x,z)\leq d(x,y) + d(y,z)$

I think we need to consider the derivative of $f(t)=\frac{t}{1+t}$ however I am not sure what to do next.

REPLY [9 votes]: As you noted: the function $f(t)=\frac{t}{1+t}$ can be differentiated in $(-1,\infty)$, and you get $f '(t)=\frac{1}{(1+t)^{2}}>0$, which shows that $f$ is monotonically increasing. 
I think the only non-trivial axiom here is the triangle inequality, so I will demonstrate how it follows by this observation. Whenever I use the fact that $f$ is monotonically increasing, I denote it by $(*)$.
Take $x,y,z\in X$. If $d(x,z)\leq d(x,y)$, then
\begin{equation*}
d_{b}(x,z)=\frac{d(x,z)}{1+d(x,z)}\overset{(*)}{\leq}\frac{d(x,y)}{1+d(x,y)}=d_{b}(x,y)\leq d_{b}(x,y)+d_{b}(y,z).
\end{equation*}
Similarly, if $d(x,z)\leq d(y,z)$ then $d_{b}(x,z)\leq d_{b}(x,y)+d_{b}(y,z)$. 
So we may assume that $d(x,z)>d(x,y)$ and $d(x,z)>d(y,z)$. I denote this condition by $(**)$. Hence
\begin{align*}
d_{b}(x,z)&=\frac{d(x,z)}{1+d(x,z)}\leq\frac{d(x,y)+d(y,z)}{1+d(x,z)}=\frac{d(x,y)}{1+d(x,z)}+\frac{d(y,z)}{1+d(x,z)} \\
&\overset{(**)}{\leq}\frac{d(x,y)}{1+d(x,y)}+\frac{d(y,z)}{1+d(y,z)} =d_{b}(x,y)+d_{b}(y,z).
\end{align*}
So in any case, $d_{b}(x,z)\leq d_{b}(x,y)+d_{b}(y,z)$.