3
$\begingroup$

I have the following simple-looking inequality I have to show:

Let $z, w \in \mathbb D$, where $\mathbb D$ is the open unit disc in $\mathbb C$. Show that $$\left| \frac{z-w}{1-\overline{z}w} \right| \geq \left| \frac{|z|-|w|}{1-|z||w|} \right|.$$

It looks pretty straightforward, but I just can't seem to get it, and I think I might be missing something obvious. I've tried putting $z=|z|e^{i \alpha}$ and $w=|w|e^{i \beta}$ to get $$\left| \frac{z-w}{1-\overline{z}w} \right| = \left| \frac{|z|-|w|e^{i \theta}}{1-|z||w|e^{i \theta}} \right|$$ where $\theta = \beta - \alpha$, and can't get much out of this. I've tried squaring both sides etc., and a few other things. If anyone has any ideas, I'd be very grateful, thanks.

$\endgroup$
1
$\begingroup$

Recall that $\left| 1-\overline{z}w\right|^2 - \left|z-w\right|^2 = (1-|z|^2)(1-|w|^2)$. Similarly, we therefore get $\left| 1-|z|\cdot |w|\right|^2 - \left||z|-|w|\right|^2 = (1-|z|^2)(1-|w|^2)$. Hence we have $\alpha\in(0,1)$ such that $$\left|\frac{z-w}{1-\overline{z}w}\right|^2 = \frac{\left|z-w\right|^2}{\left|z-w\right|^2+\alpha};\qquad \left|\frac{|z|-|w|}{1-|z|\cdot |w|}\right|^2 = \frac{\left||z|-|w|\right|^2}{\left||z|-|w|\right|^2+\alpha}$$

It suffices to know $\left|z-w\right| \geq \left||z|-|w|\right|$ to deduce $$\frac{\left|z-w\right|^2}{\left|z-w\right|^2+\alpha} \geq \frac{\left||z|-|w|\right|^2}{\left||z|-|w|\right|^2+\alpha}$$ since $t\mapsto\frac{t}{t+\alpha}$ is an increasing mapping $[0,\infty)\to [0,1)$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.