Problem involving modulus of complex numbers

Question

Let $a, b \in \Bbb C$, such that either $|a| = 1$ or $|b| = 1$. Show that $|a-b|\le|a-\bar a b|$.

Answer 1

Counterexample: $$a=e^{i\pi/2}=i, \quad b=e^{i\pi}=-1,$$ one has $$|a-b|=\sqrt{2}\color{red}{>} |a-\bar{a}b|=0,$$ which makes your claim wrong.