Let $f:\Bbb D \rightarrow \Bbb H$ be holomorphic with $f(0)=i$, show the following inequality holds:
$\frac{1-|z|}{1+|z|} \leq |f(z)| \leq \frac{1+|z|}{1-|z|}$
and that $|f'(0)|\leq 2$
This looks sort of like a Schwarz ineqality.
Let $g: \Bbb H \rightarrow \Bbb D$ be the conformal map
$g(z) = \frac{i-z}{i+z}$. Then $h=g\circ f$ is a map from the disc to itself with $f(0)=0$, so the Schwarz lemma gives
$|\frac{i-f(z)}{i+f(z)}| \leq |z|$
but I do not see how to proceed.
Let $h$ be the composed map as in your statement.
We know that $$|\dfrac{i-f(z)}{i+f(z)}| \leq |z|,$$ thus , $$|i-f(z)| \leq |z| |i+f(z)| \leq |z|+|z||f(z)|.$$ By the triangle inequality, we know that
$$|f(z)| \leq |z| +|z| |f(z)| +1.$$Rearrange, and we get that $$|f(z)|-|z||f(z)| \leq |z|+1,$$ and by dividing by $1-|z|$ we get $$|f(z)|\leq \dfrac{|z|+1}{|1-|z|}.$$
The other inequality is straightforward and follows by a similar strategy. Can you work this out or do you want me to write down the answer to that as well?