Show that $d_b(x,y)=\frac{d(x,y)}{1+d(x,y)}$ is a metric. 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.
 A: 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, $d_{b}(x,z)\leq d_{b}(x,y)+d_{b}(y,z)$.
A: For any particular $x,y$ you know $d(x,y)\geq 0$. So define $a=d(x,y)$ and note that your new metric becomes
$$\frac a {1+a}$$
Since $d(x,y)$ is a metric by assumption, $a$ is not dependent on the ordering of $x$ and $y$ and so this is trivially symmetric. You then want to prove that $\frac{a}{1+a}=0\iff a=0$ and that $\forall a\ \frac{a}{1+a}\geq0$ (remember that $a=d(x,y)\geq0$ by assumption).
The triangle inequality is trickier. First show that $\frac a{1+a}$ is monotonically increasing. Denoting $a=d(x,z), b=d(x,y), c=d(y,z)$, you want to show that $a\leq b+c\implies \frac a{1+a}\leq \frac b{1+b}+\frac c{1+c}$. Since $\frac a{1+a}$ is monotonically increasing, we have that $a<b\implies \frac a{1+a}<\frac b{1+b}$, which you'll need.
A: If $(X, d)$ is any metric space, show that another metric on $X$ is defined by $$\overline d(x, y) = \frac{d(x, y)}{1 + d(x, y)}$$.
Proof:We must first show that $\overline d$ is a metric on $X$.


*

*Observe that $\displaystyle \overline d(x, y) = \frac{d(x, y)}{1 + d(x, y)}$ and $d(x, y) \geq 0$ since it is a metric of a metric space. It follows that $ \frac{d(x, y)}{1 + d(x, y)} \geq 0$ and thus Axiom 1 is satisfied for $\overline d$.

*Suppose $\overline d (x, y) = 0$. Then $\overline d(x, y) = \frac{d(x, y)}{1 + d(x, y)}$ which implies $d(x, y) = 0$. Since $d$ is a metric on $X$, by Axiom 2 it follows that $x = y$. Conversely suppose that $x = y$. Then $\overline d(x, y) = \frac{d(x, y)}{1 + d(x, y)} = \frac{0}{1 + 0} = \frac01 = 0$ and thus Axiom 2 is satisfied for $\overline d$.

*Observe that $\overline d(x, y) = \frac{d(x, y)}{1 + d(x, y)} = \frac{d(y, x)}{1 + d(y, x)} = \overline d(y, x)$ and thus Axiom 3 is satisfied for $\overline d$.

*Observe that given $d(z, w) \geq d(x, y)$ it follows that
            \begin{align*}
     d(x, y) &\leq d(z, w)\\
     d(x, y) + d(x, y)d(z, w) &\leq d(z, w) + d(z, w)d(x, y)\\
     d(x, y)(1 + d(z, w))&\leq d(z, w)(1 + d(x, y))\\
     \frac{d(x, y)}{1 + d(x, y)} &\leq \frac{d(z, w)}{1 + d(z, w)}
    \end{align*}
            Now observe that $d(x, y) \leq d(x, z) + d(z, y)$ since $d$ is a metric. It immediately follows that
            \begin{align*}
     \overline d(x, y)&= \frac{d(x, y)}{1 + d(x, y)}\\
     &\leq \frac{d(x, z) + d(z, y)}{1 + d(x, z) + d(z, y)}\\
     &= \frac{d(x, z)}{1 + d(x, z) + d(z, y)} + \frac{d(z, y)}{1 + d(x, y) + d(z, y)}\\
     &\leq \frac{d(x, z)}{1 + d(x, z)} + \frac{d(z, y)}{1 + d(y, z)}\\
     &= \overline d(x, z) + \overline d(z, y)\\
    \end{align*}
            Thus Axiom 4 is satisfied for $\overline d$.

