How to prove this complex inequality elegantly? Question Let $z_{1,2}\in U(0,1)\subset \Bbb C$, prove that 
$$\frac{|z_1|-|z_2|}{1-|z_1||z_2|}\le\left|\frac{z_1+z_2}{1+\overline{z_1}z_2}\right|\le\frac{|z_1|+|z_2|}{1+|z_1||z_2|}$$
Actually I haven't come up with any reasonably good proof so far. All I could do was simply use brute force, i.e., relations like $|z|^2=z\overline z$. When I finished my brute-force proof and rewound it, I found it could be simplified into the following form, which looks not as horrible:
Let $w=2|z_1z_2|-z_1\bar {z_2}-z_2\bar{z_1}\ge 0$. Square the inequality, and denote the middle one as $\frac AB < 1$. Then 
$$\frac{A-w}{B-w}\le\frac AB\le \frac{A+w}{B+w}$$
which is the desired result.
Seems good. But indeed doesn't. Because it comes in hindsight: it's only after I had brute-forced and rewound that I formulated this short one. 
So apart from this one, is there any other more elegant or advanced proof? Incidentally, the structure $\displaystyle\frac{|x|\pm|y|}{1\pm|x||y|}$ frequently occurs, is it of any significance?
 A: The following is not elegant or advanced but without any brute force, I think.
We may assume that $0<|z_2|\le |z_1|$. Then the inequality we want to prove is$$
\frac{|z_1|-|z_2|}{\frac{1}{|z_1|}-|z_2|}\le\left|\frac{z_1+z_2}{\frac{1}{\overline{z_1}}+z_2}\right|\le\frac{|z_1|+|z_2|}{\frac{1}{|z_1|}+|z_2|}.
$$
Let $\theta =\arg z_1$ and replace $z_2$ by $z_2e^{i\theta }$, then the middle term becomes 
$$\left|\frac{z_1+z_2e^{i\theta }}{\frac{1}{\overline{z_1}}+z_2e^{i\theta }}\right|
=\left|\frac{z_1e^{-i\theta }+z_2}{\frac{1}{\overline{z_1}}e^{-i\theta }+z_2}\right|
=\left|\frac{r+z_2}{\frac{1}{r}+z_2}\right|
,$$
where $r=|z_1|$ is real positive.
Therefore it is sufficient to prove $$
\frac{r-|z_2|}{\frac{1}{r}-|z_2|}
\le \left|\frac{r+z_2}{\frac{1}{r}+z_2}\right|\le 
\frac{r+|z_2|}{\frac{1}{r}+|z_2|}$$ for $z_2$ with $|z_2|\le r$. Let $z_2=\rho e^{i\varphi }$ ( $\rho \le r,$ $0\le \varphi <2\pi$). Then we have
$$
 \left|\frac{r+z_2}{\frac{1}{r}+z_2}\right|^2=r^2\frac{r^2+\rho ^2+2r\rho \cos \varphi }{1+r^2\rho ^2+2r\rho \cos \varphi }.
$$
Fix $\rho $ and use the inequality you mentioned (note that $1+r^2\rho ^2>r^2+\rho ^2$), then we have
\begin{align}
r^2\frac{r^2+\rho ^2-2r\rho}{1+r^2\rho ^2-2r\rho}&\le
 \left|\frac{r+z_2}{\frac{1}{r}+z_2}\right|^2\le r^2\frac{r^2+\rho ^2+2r\rho}{1+r^2\rho ^2+2r\rho},
\\
\frac{(r-\rho)^2}{(\frac{1}{r}-\rho)^2}&\le
 \left|\frac{r+z_2}{\frac{1}{r}+z_2}\right|^2\le \frac{(r+\rho)^2}{(\frac{1}{r}+\rho)^2}
.\end{align}
This completes the proof.
