Show that $d(z,a)=|\frac{z-a}{1-\overline{a}z}|$ is a metric on the unit disk 
For any $z, a$ in the open unit disk, we define$$d(z,a)=\left \vert \frac{z-a}{1-\overline{a}z} \right \vert$$  Show that $d$ is a metric on the open unit disk.  

It is easy to show $d(z,a)=d(a,z)$ and $d(z,a)=0$ $\rm iff$ $z=a$. However, I am having trouble showing that it satisfies triangle inequality. Any help would be appreciated!
 A: We can write every automorphism of the unit disk in the form
$$T \colon z \mapsto \frac{az+b}{\overline{b}z + \overline{a}}, \qquad \lvert a\rvert^2 - \lvert b\rvert^2 = 1.$$
Using that representation, an elementary but tedious computation shows
$$\frac{Tz - Tw}{1 - \overline{Tw}Tz} = \frac{(\lvert a\rvert^2 - \lvert b\rvert^2)(z-w)}{(\lvert a\rvert^2 - \lvert b\rvert^2)(1 - \overline{w}z)}\cdot \frac{b\overline{w}+a}{\overline{b}w+\overline{a}},$$
from which we deduce that for all $z, w \in \mathbb{D}$ and $T \in \operatorname{Aut}(\mathbb{D})$, we have
$$d(Tz,Tw) = d(z,w).$$
That allows us to reduce checking the triangle inequality $d(x,z) \leqslant d(x,y) + d(y,z)$ to checking the cases $x = 0$ and $z > 0$ (the case $x = z$ is trivial), so let us check that for all $0 < r < 1$ and $y \in \mathbb{D}$, we have
$$d(0,r) \leqslant d(0,y) + d(y,r).$$
The equality $d(0,z) = \lvert z\rvert$ is easily read off the definition of $d$, so we need verify that
$$d(y,r) \geqslant r - \lvert y\rvert$$
for $0 < r < 1$ and $y \in \mathbb{D}$. For $\lvert y\rvert \geqslant r$ or $y = 0$, that is clear, so let us consider the case $0 < \rho = \lvert y\rvert < r$. It is visually clear that on the circle $\lvert y\rvert = \rho$, the expression is minimised for $y = \rho$, so let us, before proving that, verify
$$r - \rho \leqslant d(r,\rho) = \frac{r-\rho}{1 - \rho r}.$$
Since $0 < \rho r < 1$, that is the case.
Now, let us find the minimum of
$$\delta(\varphi) = \left\lvert \frac{\rho e^{i\varphi} - r}{1 - r\rho e^{i\varphi}}\right\rvert^2 = \frac{\rho^2 + r^2 - 2r\rho\cos\varphi}{1+r^2\rho^2 - 2r\rho\cos\varphi} = 1 - \frac{(1-\rho^2)(1-r^2)}{1+r^2\rho^2 - 2r\rho\cos\varphi}.$$
Since the numerator is positive (and the denominator too, for all $\varphi$), $\delta(\varphi)$ is minimised when the denominator is minimised, and that is evidently the case for $\varphi = 0$.
