For the open unit disk $\mathbb{D}$ and $A \subset \mathbb{D}$ a subset I want to look at a conformal equivalence (a holomorphic bijection with $f'(z) \neq 0$ on all of $\mathbb{D}$)
$$
f: \mathbb{D} \rightarrow A
$$
with $f(0) = 0$.
I want to show that
$$
\frac{|f(u)-f(v)|^2}{(1-|f(u)|^2)(1-|f(v)|^2)} \leq \frac{|u-v|^2}{(1-|u|^2)(1-|v|^2)}.
$$
I found it to be sufficient if $f$ was lipschitz continuous with Lipschitz constant $1$.
Using the Schwarz Lemma $|f(z)| \leq |z|$ and changing the equation to $\frac{1}{1-|f(z)|^2} \leq \frac{1}{1-|z|^2}$ it holds that
\begin{align*}
\frac{|f(u)-f(v)|^2}{|u-v|^2} \leq 1 = \frac{(1-|u|^2)(1-|v|^2)}{(1-|u|^2)(1-|v|^2)} \leq \frac{(1-|u|^2)(1-|v|^2)}{(1-|f(u)|^2)(1-|f(v)|^2)} \\
\Leftrightarrow \frac{|f(u)-f(v)|^2}{(1-|f(u)|^2)(1-|f(v)|^2)} \leq \frac{|u-v|^2}{(1-|u|^2)(1-|v|^2)}.
\end{align*}
But is $f$ lipschitz-continuous? with $L=1$? Does it follow from the Schwarz-Lemma? Or is there a better way to prove this?
One approach was the Schwarz-Pick-Lemma, which states that $$ \left| \frac{f(u)-f(v)}{1-\overline{f(u)}f(v)} \right| \leq \left| \frac{u-v}{1-\overline{u}v} \right| $$ but the inequality is so symmetric, that I don't know how to solve it.