A bound on a Möbius transform Suppose $z,w \in \mathbb{C}$ s.t. $|z|,|w| < 1$. How would you show that $\frac{|z-w|}{|1-\bar{w}z|} \leq |z| + |w|$?
I calculated that 
\begin{align*}
&\frac{|z-w|}{|1-\bar{w}z|} \leq |z| + |w| \\
&\Rightarrow \frac{|z|^2-2Re(\bar{w}z)-|w|^2}{1-2Re(\bar{w}z) + |w|^2|z|^2} \leq (|z|+|w|)^2 \\
&\Rightarrow -2Re(\bar{w}z) \leq 2|z||w| + (|w|^2|z|^2 - 2 Re(\bar{w}z))(|z| +|w|)^2\
\end{align*}
If $2Re(\bar{w}z) < 0$, then $2|z||w| \geq 2Re(\bar{w}z)$ and $(|w|^2|z|^2 - 2 Re(\bar{w}z)) \geq 0$ imply that the inequality is true.
I haven't been able to figure out what happens in the case that $2Re(\bar{w}z) >0$ though.
On a separate note, whenever I come across these types of problems, I always try to multiply everything out and see if I can get enough terms to cancel so that the inequality becomes obvious. 
I'm not quite sure if that's the smartest way to go about things, though and if there is some intuition that I should be using that I don't know about.
 A: A hint: Think in terms of hyperbolic distance. If $d(z,w)$ is the hyperbolic distance between $z$ and $w\in D$ then $|z-w|/|1-\bar w z|= \tanh d(z,w)$, and in particular $|z|=\tanh d(0,z)$.
Another hint: You may assume $w=\rho>0$ and $z=r e^{i\alpha}$. Now compute $|z-w|^2$ and $|z-{1\over \bar  w}|^2$ by means of the cosine theorem and show that the quotient of the two is largest when $t:=\cos\alpha=-1$.
A: As Christian Blatter mentioned, this has a geometric interpretation in terms of hyperbolic metrics, but you can also do it directly: First note that
$${|z - w|^2 \over |1 - \bar{w}z|^2} = {|z|^2 - 2Re(\bar{w}z) + |w|^2 \over 1 - 2Re(\bar{w}z) + |w|^2|z|^2}$$
(You made a small error in your calculation). Next, observe that $|z|^2 + |w|^2 \leq 1 + |w|^2|z|^2$, since this is equivalent to $(1 - |w|^2)(1 - |z|^2) \geq 0$. Thus the above expression is of the form ${\displaystyle {A + c \over B + c}}$, where $0 \leq A \leq B$ and both $A + c$ and $B + c$ are nonnegative. Such an expression increases as $c$ increases. Since $- 2Re(\bar{w}z)$ is at most $2|w||z|$, we thus have
$$ {|z|^2 - 2Re(\bar{w}z) + |w|^2 \over 1 - 2Re(\bar{w}z) + |w|^2|z|^2}
\leq {|z|^2 + 2|w||z| + |w|^2 \over 1 + 2|w||z| + |w|^2|z|^2}$$
$$ = {(|z| + |w|)^2 \over (|w||z| + 1)^2}$$
Showing this is at most $(|z| + |w|)^2$ is equivalent to showing $(|w||z| + 1)^2 \geq 1$, which is clearly true. 
