Schwarz lemma problem 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.
 A: Here is a proof that uses only Schwarz' lemma and the triangle inequality.
Set $a=f(0)$ and 
$$\phi_a(z)=\frac{z-a}{1-\overline{a}z}.$$
Then $\phi_a\circ f$ maps the unit disc to the unit disc and fixes zero. Thus, by Schwarz' lemma,
$$|\phi_a(f(z))|=\left|\frac{f(z)-a}{1-\overline{a}f(z)}\right|\leq|z|,$$
and so
\begin{equation}|f(z)-a|\leq|z||1-\overline{a}f(z)|\leq|z|+|z||a||f(z)|.\tag{1}\end{equation}
Applying the triangle inequality, we arrive at
$$|f(z)|\leq|z|+|z||a||f(z)|+|a|.$$
Then
$$|f(z)|-|z||a||f(z)|\leq|z|+|a|$$
$$|f(z)|\leq\frac{|z|+|a|}{1-|a||z|}=\frac{|f(0)|+|z|}{1-|f(a)||z|},$$
Where the final equality uses the fact that $f(0)=a$. This is the second desire inequality.
To obtain the first inequality, we begin with
$$|a|=|a-f(z)+f(z)|\leq |a-f(z)|+|f(z)|\leq |z|+|z||a||f(z)|+|f(z)|,$$
where the last inequality follows from (1). Then
$$|a|-|z|\leq |z||a||f(z)|+|f(z)|$$
$$\frac{|a|-|z|}{|z||a|+1}=\frac{|f(0)|-|z|}{1+|f(0)||z|}\leq |f(z)|.$$
A: Expanding on the answer of mike.  Set
$$
\frac{f(z) - f(0)}{1-\overline{f(0)}f(z)} = w.
$$
Then $|w| \leq |z|$.  Solve for $f(z)$:
$$
f(z) = \frac{w + f(0)}{1 + \overline{f(0)}w}.
$$
The right hand side is an automorphism of the unit disc as a function in $w$.  It maps circles $|w| = r$ to (non concentric) circles around $f(0)$.  The following inequalities hold for such a Möbius transformation:
$$
\frac{|f(0)| - |w|}{1 - |f(0)| |w|} \leq \left| \frac{w + f(0)}{1 + \overline{f(0)}w} \right| \leq \frac{|f(0)|+|w|}{1 + |f(0)| |w|}.
$$
The left hand side decreases in $|w|$ and the right hand side increases in $|w|$.  Since $|w| \leq |z|$, the inequalities for $|f(z)|$ follow.
A: set $ h = (f(0) - f(z))/ (1- \bar f(0) f(z))$ and solve explicity for f, though I would prefer that the signs in the denominator were reversed, but maybe that's the hard part
A: 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 
$$
\frac{|f(0)|-|z|}{1-|f(0)||z|}\leq|f(z)|\leq\frac{|f(0)|+|z|}{1+|f(0)||z|}
$$
