Prove that $|\frac{a-b}{1-\bar{a}b}|<1$ if $|a|<1$ and $|b|<1$.

Question

Prove that $$|\frac{a-b}{1-\bar{a}b}|<1$$ if $|a|<1$ and $|b|<1$.

After doing some calculation I ended up in a dead end

$$|\frac{a-b}{1-\bar{a}b}|=\sqrt{\frac{|a|^2+|b|^2-(a\bar{b}+b\bar{a})}{1+|a|^2|b|^2-(a\bar{b}+b\bar{a})}}$$

How do I proceed further ?

Is there any alternating ways like using cauchy's inequality ?

Answer 1

You can rewrite the inequality as follows:

$$(a-b)^2<(1-ab)^2 \Rightarrow a^2-2ab+b^2-1+2ab-(ab)^2<0$$

This transforms to: $(a^2-1)(b^2-1)>0$ or $(1-a^2)(1-b^2)>0$, which is clearly true.