Showing a complex inequality Let $z,w \in \mathbb C$ s.t $\bar z w \neq 1$ and $|z|\leq 1, |w| \leq 1$ then $ |\dfrac{z-w}{1-\bar z w}| \leq 1$
The hint is to show that $ |z-w|^2\leq |1-\bar z w|^2$ but i cant relate those 2

$$|1-\overline zw|^2=(1-\overline zw)(1-\overline wz)=1+|wz|^2-\overline z w-\overline wz,$$
$$|z-w|^2=(z-w)(\overline z-\overline w)=|z|^2+|w|^2-z\overline w-w\overline z.$$
The difference is
$$1+|wz|^2-|w|^2-|z|^2.$$
