Let $D$ be a simply-connected set, and let $\Delta=\{z:|z|<1\}$.
$f \colon \Delta \to D$ is bijective and holomorphic. Denote by $\partial D$ the boundary of $D$.
The exercise is to prove that $$\forall z_0 \in \Delta \ : \ \text{dist}(f(z_0),\partial D) \leq (1-|z_0|^2)|f'(z_0)|\leq 4 \ \text{dist}(f(z_0),\partial D)$$
I managed to prove the right inequality:
let $g(z) = \frac{z_0 - z}{1-\bar{z_0}z}$ be the univalent Möbius transformation taking $0$ to $z_0$. Then, applying Koebe theorem to $f \circ g\ \ $yields
$\{z : |z - f(z_0)| < \frac{|(f \circ g) ' (0)|}{4}\} \subseteq D$, and
$|(f \circ g) ' (0)|=|(f'(g(0)) \cdot g'(0)|=|f'(z_0) \cdot (\frac{z_0 - z}{1-\bar{z_0}z})'(0)|=|f'(z_0) \cdot (\frac{-1 \cdot (1-\bar{z_0}z)+\bar{z_0}(z_0-z)}{(1-\bar{z_0}z)^2})(0)|=|f'(z_0) \cdot (-1+|z_0|^2)|=|f'(z_0) \cdot (1-|z_0|^2)|$.
But how to prove the left inequality?
Thank you!
To prove the left inequality, let $r = \operatorname{dist}(f(z_0),\partial D)$ and $G = D_r(f(z_0)) = \{ w : \lvert w - f(z_0)\rvert < r\}$. Then $G\subset f(\Delta) = D$, and we can apply the Schwarz lemma to $h \colon \Delta \to \Delta$ defined by
$$h\colon z \mapsto \frac{f^{-1}\bigl(f(z_0) + rz\bigr) - z_0}{1 - \overline{z_0} f^{-1}\bigl(f(z_0) + rz\bigr)}\,.$$
That is, we obtain
\begin{align} 1 &\geqslant |h'(0)| \\ &= \frac{1 - \lvert z_0\rvert^2}{(1 - \lvert z_0\rvert^2)^2}\cdot \frac{1}{\lvert f'(z_0)\rvert}\cdot r \end{align}
(by using the formula $(f^{-1})'(w)=\frac{1}{f'(f^{-1}(w))}$),
which clearly is equivalent to
$$(1-\lvert z_0\rvert^2)\lvert f'(z_0)\rvert \geqslant r$$
as desired.