# Inequality involving Mobius transformation

I have the following simple-looking inequality I have to show:

Let $z, w \in \mathbb D$, where $\mathbb D$ is the open unit disc in $\mathbb C$. Show that $$\left| \frac{z-w}{1-\overline{z}w} \right| \geq \left| \frac{|z|-|w|}{1-|z||w|} \right|.$$

It looks pretty straightforward, but I just can't seem to get it, and I think I might be missing something obvious. I've tried putting $z=|z|e^{i \alpha}$ and $w=|w|e^{i \beta}$ to get $$\left| \frac{z-w}{1-\overline{z}w} \right| = \left| \frac{|z|-|w|e^{i \theta}}{1-|z||w|e^{i \theta}} \right|$$ where $\theta = \beta - \alpha$, and can't get much out of this. I've tried squaring both sides etc., and a few other things. If anyone has any ideas, I'd be very grateful, thanks.

Recall that $\left| 1-\overline{z}w\right|^2 - \left|z-w\right|^2 = (1-|z|^2)(1-|w|^2)$. Similarly, we therefore get $\left| 1-|z|\cdot |w|\right|^2 - \left||z|-|w|\right|^2 = (1-|z|^2)(1-|w|^2)$. Hence we have $\alpha\in(0,1)$ such that $$\left|\frac{z-w}{1-\overline{z}w}\right|^2 = \frac{\left|z-w\right|^2}{\left|z-w\right|^2+\alpha};\qquad \left|\frac{|z|-|w|}{1-|z|\cdot |w|}\right|^2 = \frac{\left||z|-|w|\right|^2}{\left||z|-|w|\right|^2+\alpha}$$
It suffices to know $\left|z-w\right| \geq \left||z|-|w|\right|$ to deduce $$\frac{\left|z-w\right|^2}{\left|z-w\right|^2+\alpha} \geq \frac{\left||z|-|w|\right|^2}{\left||z|-|w|\right|^2+\alpha}$$ since $t\mapsto\frac{t}{t+\alpha}$ is an increasing mapping $[0,\infty)\to [0,1)$.