So I've been trying to solve this one for a few hours and am now out of ideas on how to approach this problem.
Here are the inequalities:
$$\text{show that if}$$
$$z,w \in \Bbb C$$ $$|z| < 1 \text{ and } |w| < 1 $$
$$\text{then}$$
$$\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\left|\frac{z-w}{1-\overline{z}w}\right|< 1$$
My thoughts have been to first
$$\left|z-w\right|< \left| 1-\overline{z}w \right|$$
and then
$$\left|z-w\right|^2< \left| 1-\overline{z}w \right|^2 \ \ => \ \ (z-w)(\overline{z}-\overline{w})< (1-\overline{z}w)(1-z\overline{w})$$
which would give me
$$|z|^2 +|w|^2<1+|z|^2|w|^2 \ \ \ => \ \ (1-|z|^2)(1-|w|^2)>0$$
But after that I don't know where to go next, or even if I'm on the right track..
I would really appreciate some help on how to approach and solve this problem!
Fix $w$ and consider the function $f(z)=\frac{z-w}{1-\bar{z}w}$. Now show that $f(f(z))=z$. What kind of functions satisfy the functional equation $f(f(x))=x$?