Given a metric space $(X,d)$, I am asked to show that the function $\hat{d}$ is also a metric. It's defined like this:
$$\begin{align*} \hat{d}\colon X \times X & \longrightarrow\mathbb{R} \\[-1ex] (x,y) & \longmapsto \frac{d(x,y)}{1+d(x,y)} \end{align*}$$
What I am struggeling with is the triangle inequality. The other points are easy to see (when keeping in mind that $d$ is a metric):
$x=y \Longleftrightarrow d(x,y)=0$ $\Longrightarrow \hat{d}(x,y)=\frac{0}{1+0}=0$.
$d(x,y)\geq 0 \Longrightarrow \frac{d(x,y)}{1+d(x,y)}=\frac{\text{<positive number or zero>}}{1+\text{<positive number or zero>}}\geq 0 \Longleftrightarrow \hat{d}(x,y)\geq 0$
$d(x,y)=d(y,x) \Longrightarrow \hat{d}(x,y)=\frac{d(x,y)}{1+d(x,y)}=\frac{d(y,x)}{1+d(y,x)}=\hat{d}(y,x)$
As for the triangle inequality, this is what I have so far:
We know that $d(x,y)+d(y,z)\geq d(x,z)$ (since $d$ is a metric). We need to somehow use this fact later. I'll shorten the notation a bit by defining $d_1:=d(x,y)$, $d_2=d(y,z)$ and $d_3=d(x,z)$ (and similarly for $\hat{d}$).
$$\hat{d}(x,y)+\hat{d}(y,z)=\frac{d_1}{1+d_1}+\frac{d_2}{1+d_2}=\frac{d_1(1+d_2)+d_2(1+d_1)}{(1+d_1)(1+d_2)}=\frac{d_1+d_2+2\cdot d_1d_2}{1+d_1+d_2+d_1d_2}\geq \frac{d_3+2\cdot d_1d_2}{1+d_3+d_1d_2}$$
The problem I have is the cross terms $d_1d_2$. How do I get $\frac{d_3}{1+d_3}$ at the end? I to me it seems like:
$$\frac{d_3+2\cdot d_1d_2}{1+d_3+d_1d_2} > \frac{d_3}{1+d_3}$$
But let's say that $d_1=1, d_2=2$ and $d_3=3$ so that $d_1+d_2=d_3$. Then I would expect that $\hat{d}_1+\hat{d}_2=\hat{d}_3$, but:
$$ \frac{1}{1+1}+\frac{2}{1+2}=\frac{1}{2}+\frac{2}{3}=\frac{7}{6} \overset{?}{=} \hat{d}_3$$
$$ \hat{d}_3=\frac{3}{3+1}=\frac{3}{4} \neq \hat{d}_1 + \hat{d}_2 $$
But is this unreasonable? If a point $y$ lies between $x$ and $z$ in the metric $d$, then it appears it doesn't lie between $x$ and $z$ in the metric $\hat{d}$?
Let's consider $\hat{d}_1=0.2, \hat{d}_2=0.4, \hat{d}_3=0.6 \Longrightarrow \hat{d}_1+\hat{d}_2=\hat{d_3}$. We have that if $\frac{A}{1+A}=B$, then $A=\frac{B}{1-B}$. We get that:
$$\begin{align*} d_1=&\frac{\hat{d}_1}{1-\hat{d}_1}=\frac{0.2}{1-0.2}=\frac{0.2}{0.8}=\frac{1}{4}\\[-1ex] d_2=& \frac{\hat{d}_2}{1-\hat{d}_2}=\frac{0.4}{1-0.4}=\frac{0.4}{0.6}=\frac{2}{3}\\[-1ex] d_3=& \frac{\hat{d}_3}{1-\hat{d}_3}=\frac{0.6}{1-0.6}=\frac{0.6}{0.4}=\frac{3}{2}\\[-1ex] \end{align*}$$
If we try to add $d_1$ and $d_2$ we get that:
$$\frac{1}{4}+\frac{2}{3}=\frac{11}{12}\neq d_3=\frac{3}{2}=\frac{18}{12}$$
So clearly $d_1+d_2=d_3\;\not\!\!\!\implies \hat{d}_1+\hat{d}_2=\hat{d}_3$. However, it is still possible to have $\hat{d}_1+\hat{d}_2=\hat{d}_3$ and it seems like $\hat{d}_1+\hat{d}_2 \geq \hat{d_3}$
But how do I show that the triangle equality generally holds for $\hat{d}$? Can I use the properties of $d$ to prove this, or do I need another approach?
