1. ## Complex Inequalities

Suppose that $z$ and $w$ are in the closed unit disk.

Prove the inequality $1- \vert w \vert^2 \leq 2 \vert 1-z \bar w \vert$.

I've tried to square both sides and expand the equation. However, I'm stuck there. Is there any other way that I can approach this question?

2. Hint :

Use

$|1-z\bar{w}|\geq |\;1-|z\bar{w}|\;|=1-|z||\bar{w}|\geq \ldots$

Fernando Revilla

3. $2 \vert 1- z \bar w \vert
\geq 2(\vert 1- \vert z \bar w \vert \vert )
\geq 2(1- \vert zw \vert )
\geq 2(1- \vert w \vert )$
(Since $\vert z \vert \leq 1$)
Now we claim that $2 \vert w \vert - \vert w \vert^{2} \leq 1$.
We let $y=2 \vert w \vert - \vert w \vert^{2}$ and substitute $\vert w \vert$ with $x$, where $0 < x \leq 1$. Then we differentiate the equation $y$ to get the maximum point. From here we can get $y \leq 1$. Hence $2 \vert w \vert - \vert w \vert^{2} \leq 1$.
Therefore, $2 \vert 1- z \bar w \vert
\geq 2(1- \vert w \vert )
\geq 2(1- \frac{1+ \vert w \vert^2}{2})
=1- \vert w \vert^2$
.

Is there any easier method to this problem?