Inequality with complex variables

Question

I was trying to prove that $$ (w - r)(\overline{w} - r) \leq (1-wr)(1-\overline{w}r) $$ where $w$ is a complex number with absolute value smaller than 1 and $r \in \mathbb{R}$, $r <1$. I saw some solution online, which says we substitute $w = me^{ix}$ so that the LHS becomes $(m-r)^2$. How is that true?

We have $$ (w - r)(\overline{w} - r) = (me^{ix}-r)(me^{-ix}-r) = e^{ix}e^{-ix}(m-re^{-ix})(m-re^{ix}) =(m-re^{-ix})(m-re^{ix}) $$ right?

Answer 1

The statement from the online solution looks wrong.

LHS=$|w|^2-(w+\bar w)r+r^2$.

RHS=$1-(w+\bar w)r+|w|^2r^2$.

Now compare $|w|^2+r^2$ with $1+|w|^2r^2$.

Proof of original assertion.

$1 \gt r^2$

$1-|w|^2 \gt r^2(1-|w|^2)=r^2-r^2|w|^2$

$1+r^2|w|^2\gt r^2+|w|^2$

Answer 2

LHS is not $(m-r)^2$, a trivial example would be $w = i$ and $r=1$! In that case, $(m-r)^2=0$ which is apparently not true.

What you can say though is that LHS: $$(\omega-r)(\bar{\omega}-r)=|\omega-r|^2=(m\cos(x)-r)^2+(m\sin(x))^2$$