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
 A: $$|1-\overline zw|^2=(1-\overline zw)(1-\overline wz)=1+|wz|^2-\overline z w-\overline wz,$$
$$|z-w|^2=(z-w)(\overline z-\overline w)=|z|^2+|w|^2-z\overline w-w\overline z.$$
The difference is
$$1+|wz|^2-|w|^2-|z|^2.$$
A: Use the fact that $|z|^2 = z \overline z$. Then,we have
$ |z - \overline \omega |^2 \le |1 - \overline z w|^2 \iff (z - \overline \omega) (\overline z - \omega) \le (1 - \overline z w) (1 - z \overline w) \iff z \overline z - z \overline \omega - \omega \overline \ + \omega \overline \omega \le 1 - z \overline \omega - \overline z \omega + z \overline z \omega \overline \omega $
After simplifying, the equivalence arrives at
$|z|^2 + |\omega|^2 <=  1 + |z|^2 |x|^2 \iff (|x| ^ 2 - 1) * (|y| ^ 2 - 1) \ge 0$, which is true since we assumed $|x| \le 1$ and $|y| \le 1$, so $|x| ^ 2 \le 1$ and $ |y| ^ 2 \le 1 $. So, we have proven the hint
Finally, since both terms are positive divide  to get the conclusion squared. And since, the absolute value is always positive, you can get rid of the square, to get the desired conclusion.
