Proving pseudo-hyperbolic distance is distance The pseudo-hyperbolic distance on the unit disk is defined as:
$$\rho(z,w)=\left|\dfrac{z-w}{1-\bar wz}\right|.$$
I'd like to prove it's a distance. The real problem is, as always, the triangle inequality, because the other properties are mostly obvious. That is, I need to prove:
$$\rho(z,w)\leq\rho(z,t)+\rho(t,w),$$
for all $z,w,t\in\mathbb{D}$. I tried writing $z,t,w$ as real part plus $i$ times imaginary part, and ended up with a messy expression Wolfram can't handle. I tried polar coordinates, and the mess is even worse, and Wolfram's help is even less. I Googled first, but only found stuff about the Hyperbolic distance, and a pdf having this as an exercise, suggesting to also show that:
$$\rho(z,w)\leq\frac{\rho(z,t)+\rho(t,w)}{1+\rho(z,t)\rho(t,w)}.$$
But that didn't help. So here I am. How do I solve this?

It helps to know that $\rho$ is invariant under Möbius transformations. Indeed, $\rho(z,w)=|\phi(z)|$ where $\phi$ is any Möbius map such that $\phi(w)=0$ (they all agree up to rotation, which doesn't change the modulus). Since Möbius maps form a group, applying one of them to both $z$ and $w$ does not affect the above formula for $\rho$. 
In order to prove the triangle inequality, first map $t$ to $0$ by a suitable Möbius map. This reduces the task to
$$\left|\frac{z-w}{1-z\overline w}\right|\le |z|+|w| \tag{1}$$
Let $|z|=a$, $|w|=b$, and let $\theta $ be the polar angle between $z$ and $w$. Then $(1)$ becomes
$$ \frac{a^2+b^2-2ab\cos\theta }{1+a^2b^2-2ab\cos\theta } \le (a+b)^2 \tag{2}$$
To estimate the left side, rewrite it as 
$$
1 - \frac{(1-a^2)(1-b^2) }{1+a^2b^2-2ab\cos\theta }
\le 
1 - \frac{(1-a^2)(1-b^2) }{1+a^2b^2+2ab } =\frac{(a+b)^2}{(1+ab)^2} $$ and $(2)$ follows.
