Prove that $x-\mid y\mid\le\left|\frac{x-y}{1-xy}\right|<1$ Let a be a real number, b is a complex number, $a \in (0,1)$ and $|b|<1$
Prove that $$x-\mid y\mid\le\left|\frac{x-y}{1-xy}\right|<1$$
I have solved the left side:  $$\left|\frac{x-y}{1-xy}\right|\le1$$
We have: 
$$\left|\frac{x-y}{1-xy}\right|^2=\frac{(x-y)(x-\overline{y})}{(1-xy)(1-x\overline{y})}=\frac{x^2-(xy+x\overline{y})+|y|^2}{1-(xy+x\overline{y})+x^2|y|^2}$$
If $a,b<1$ then $0<(a-1)(b-1)$ so $a+b<1+ab$
Using this inequality with $a=x^2<1$ and $b=|y|^2$, we get
$|x|^2−(xy+x\overline{y})+|y|^2 <1−(xy+x\overline{y})+x^2|y|^2$
since $1−(xy+x\overline{y})+x^2|y|^2>0$ we done
What about the other side? Please help me with this.
Thanks
 A: More generally, if $z,w$ are complex numbers and $|z|,|w|<1$, then
$$|z|-|w|\le \left|\frac{z-w}{1-z\overline w}\right| <1 \tag1$$
The upper estimate is standard: 
$$|z-w|^2-|1-z\overline w|^2=|z|^2+|w|^2-1-|zw|^2=(1-|z|^2)(|w|^2-1)<0$$
The left side of (1) is trivial when $|z|\le |w|$ or $w=0$. Suppose $|z|>|w|>0$. Let $r=|z|$. The image of the circle of radius $r$ under the map $\phi(z)=\frac{z-w}{1-z\overline w}$ is again a circle, denoted $C$. The nearest point of $C$ to $0$ is the endpoint of line segment that begins at $0$, meets $C$ orthogonally, and does not pass through the center of $C$. Applying $\phi^{-1}$, we find that the point  that achieves $\min_{|z|=r}|\phi(z)|$ lies on hyperbolic geodesic that begins at $w$, meets $|z|=r$ orthogonally, and does not pass through $0$. This geodesic is the line segment from $w$ to $rw/|w|$. Therefore, it suffices to prove $|\phi(z)|\ge |z|-|w|$  when $z$ and $w$ lie on the same radius of the unit disk. But in this case 
$$|z|-|w|=|z-w|\le \left|\frac{z-w}{1-z\overline w}\right|$$
because $z\overline w$ is positive.
