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.