TITLE: Schwarz lemma problem
QUESTION [10 upvotes]: This is taken from an old complex analysis qualifying exam.
Problem
Let $\Delta$ denote the unit disc $\{z\in\mathbb{C}:|z|<1\}$.
Suppose $f:\Delta\rightarrow\Delta$ is holomorphic. Show that
$$\frac{|f(0)|-|z|}{1-|f(0)||z|}\leq|f(z)|\leq\frac{|f(0)|+|z|}{1+|f(0)||z|}$$
for all $|z|<1$.
Attempt
Since $f(0)\in\Delta$, define $\phi\in\operatorname{Aut}(\Delta)$ as
$$\phi(z)=\frac{f(0)-z}{1-\overline{f(0)}z}.$$
Then $\phi\circ f$ maps the unit disc into the unit disc and fixes zero. Thus, by Schwarz' lemma, we have
$$\left|\frac{f(0)-f(z)}{1-\overline{f(0)}f(z)}\right|\leq|z|.$$
But I don't see how this will lead to either of the desired inequalities. Any help would be greatly appreciated.

REPLY [3 votes]: First we define
$$
\rho(z,w)=\left|\dfrac{z-w}{1-\overline w z}\right|
$$
Then is easy prove that
$$
1-\rho(z,w)^2=\dfrac{(1-|z|^2)(1-|w|^2)}{|1 - \overline w z|^2}
$$
Now you can check that 
$$
z \overline w+ \overline z  w\le 2|z||w|
$$
and this inequality help to obtained
$$
(1-|w||z|)^2\le |1-\overline w z|^2
$$
So:
\begin{align*}
   \rho(z,w)^2=\left|\dfrac{z-w}{1-\overline w z}\right|^2
=&1-  \dfrac{(1-|z|^2)(1-|w|^2)}{|1 - \overline w z|^2}\\
\ge& 1- \dfrac{(1-|z|^2)(1-|w|^2)}{(1 - |\overline w| |z|)^2}\\
=&\dfrac{(1 - |\overline w| |z|)^2-  (1-|z|^2)(1-|w|^2)}{(1 - |\overline w||z|)^2}\\
=&\dfrac{ |z|^2-2|z||w|+|w|^2}{(1 - |\overline w||z|)^2}\\
=&\dfrac{ (|z|-|w|)^2}{(1 - |\overline w||z|)^2}\\
\end{align*}
Now chaning $z$ by $f(z)$, and $w$ by $f(0)$ we have
$$
\left|\dfrac{ (|f(z)|-|f(0)|)}{(1 - |f(0)||f(z)|)}\right|\le\left|\dfrac{f(z)-f(0)}{1-\overline{f(0)}f(z)}\right|\le|z|
$$
the last inequality by Schwarz-Pick. Then
$$
\left|\dfrac{ (|f(z)|-|f(0)|)}{(1 - |f(0)||f(z)|)}\right|\le|z|
$$
we obtained 
$$
\frac{|f(0)|-|z|}{1-|f(0)||z|}\leq|f(z)|\leq\frac{|f(0)|+|z|}{1+|f(0)||z|}
$$