# 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

Thanks

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Are $a,b,x,y$ somehow related? – Siméon Jan 19 '13 at 13:22
no, a, b just is 2 number satisfied $a,b<1$ – septimus Jan 19 '13 at 13:25
The question, as it stand, makes no sense: what's the relation between a,b and x,y...? – DonAntonio Jan 19 '13 at 13:54
So can you solve the other inequality. Thanks. This question almost driving me crazy. Take long time – septimus Jan 19 '13 at 14:15
a, b just to prove the inequality $a+b<1+ab$ when $a,b<1$ and apply for x, y. The aren't any relation between a,b and x,y. Like a lemma. – septimus Jan 19 '13 at 15:15

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.