Show that if $z,w\in\mathbf{C}$, $|z|<1$ and $|w|<1$, then $\left|\frac{z-w}{1-\overline{z}w}\right|$<1? Every way I try to approach this turn it into proving the inequality $|z-w|<|1-\overline{z}w|$. Not sure at all how to approach it at the moment.
 A: Since $|z|<1,\ |w|<1$ we can square both sides to find $|z|^{2}<1,\ |w|^{2}<1$. Now subtract $1$ from each quantity to get $$(|z|^{2}-1)<0,\ (|w|^{2}-1)<0$$ We now have two quantities that are less than zero (negative) so if we multiply the two we will have a positive number. Thus,$$0<(|z|^{2}-1)(|w|^{2}-1)$$ Expanding the quantity yields $$0<1-|z|^{2}-|w|^{2}+|z|^{2}|w|^{2}$$ which can be rearranged as $$|z|^{2}+|w|^{2}<1+|z|^{2}|w|^{2}$$ Now add the quantity $-z \bar w- \bar zw$ to both sides. We will have $$|z|^{2}+|w|^{2}-z \bar w- \bar zw<1+|z|^{2}|w|^{2}-z \bar w- \bar zw$$ Convince yourself that we can unfoil each side into two quantities as follows: $$(z-w)(\bar z -\bar w)<(1-\bar zw)(1-z \bar w)$$ and using properties of complex conjugates, this is equivalent to  $$(z-w)\overline{(z-w)}<(1-\bar zw)\overline{(1-\bar zw)}$$ which is further equivalent to $$|z-w|^{2}<|1-\bar zw|^{2}$$ From here you can take the square root of each side, and divide by the quantity on the RHS. You should be left with $$\frac{|z-w|}{|1-\bar zw|}<1$$ which is equivalent to what you are trying to show. Hopefully that helps! In my experience, the trick with these kind of problems is starting by squaring each side. 
A: I don't see this viewpoint, I imagine this is what Pedro means.
Make a Möbius transformation, with a fixed $ \beta =z,$ as
$$ f(w) = \frac{w - \beta}{\bar{\beta} w - 1}  $$
Carefully calculating $f(1),f(-1), f(i)$ shows that all have magnitude $1.$ That is, $f$ maps the standard unit circle to itself. This uses the fact that, given some complex number $\gamma,$ the eight numbers $\gamma, - \gamma, i \gamma, -i \gamma, \bar{\gamma}, - \bar{\gamma}, i \bar{\gamma}, -i \bar{\gamma}$
all have the same magnitude.
The remaining question is this: does $f$ map the interior of the circle to the exterior or to itself? Well, $f(0) = \beta,$ and $|\beta| < 1.$
So $f$ maps the interior of the circle to itself.
A: Let me sketch how I approached this when I saw it.  The proof comes down to geometry, but there are some preliminary steps.  Note that this argument is longer
than the ones which involve clever algebraic manipulations, but perhaps seeing
a different approach, with the motivations explained, will help.

When you see $z - w$ and $1 - \bar{z} w$, it makes sense to consider multiplying
the first expression by $\bar{z}$ to make it look more like the second.  The problem
then becomes to prove that
$$ \mid \, |z|^2 - \bar{z} w \! \mid < |z| \,| 1 - \bar{z} w|.$$
Now $z$ is just some number of norm $r < 1$, say.
And what is $\bar{z} w$?  Well, $w$ is any complex number of norm $< 1$,
while $\bar{z}$ is a complex number of norm $r$, so $\bar{z} w$ can be any
complex number of norm $ < r$.
So we have the following problem: show that if $v$ is any complex number with $| v| < r$, then
$$| \,r^2 - v | < r \, | 1 - v |.$$
At this point, complex numbers don't really play a role: you can just think of $v$
as any vector lying within the interior of the disk of radius $r$, and try to
prove this.  Drawing a picture will help.
Note that since the only input is that $r < 1$ and that $|v| < r$,
the argument can't be too complicated; you should expect it to come down
to some fairly simple geometry.  I'll leave the details to you.
A: Boy this site is cruel to new users.
$$|z-w|^2=|z|^2+|w|^2-z\overline{w}-w\overline{z}$$
$$|1-w\overline{z}|=1-z\overline{w}-w\overline{z} +|z|^2|w|^2$$
sp you need 
$$|z|^2+|w|^2 \leq 1+|z|^2|w|^2$$
And this reduces to 
$$0 \leq (|z|^2-1)
(|w|^2-1)$$
