1. Complex Inequalities

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

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

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

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

Fernando Revilla

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

Is there any easier method to this problem?