Complex numbers inequality

Question

Let $z$ , $w$ $ \ \in \mathbb{C}$ with $|z|$ , $|w| < 1$. Then prove that $|\frac{z-w}{1-z\bar{w}}| < 1$.

What I have noticed : $|z-w| = |w||\frac{z\bar{w}}{|w|} -1| $ but I don't know how to proceed.

Answer 1

$$|\frac{z-w}{1-z\bar{w}}| < 1<=>|z-w|<|1-z\overline w|<=>(z-w)(\overline z-\overline w)<(1-z\overline w)(1-\overline zw)<=>|z|^2+|w|^2<1+|z|^2|w|^2<=>|z|^2(1-|w|^2)<1-|w|^2<=>|z|^2<1$$ which is true.