Inequality similar to the Schwarz-Pick theorem I am trying to prove that:
Let $f$ be analytic in the closed unit disc with $f(z)\leq 1$ for $|z|\leq1$.
For a given $0<r<1$ and $|z_1|,|z_2|\leq r$, $z_1\neq z_2$ then, 
$$\left|\frac{f(z_2)-f(z_1)}{z_2-z_1}\right|\leq \frac{1}{(1-r)^2}.$$
My attempt is:
part a  of the question asks to prove $$\left|\sum_k^\infty a_nz^n\right|\leq \frac{ |z^k|}{1-|z|}.$$ 
So,
\begin{eqnarray*}
 \left|\frac{f(z_2)-f(z_1)}{z_2-z_1}\right|&=& \frac{|\sum_1^\infty a_nz_2^n-\sum_1^\infty a_nz_1^n|}{|z_2-z_1|}\\ 
&\leq& \frac{\frac{ |z_1 |}{1-|z_1|}+\frac{ |z _2|}{1-|z_2|}}{|z_2-z_1|}\\ 
&\leq&\frac{1}{(1-r)^2} \frac{   |z_2| ({1-|z_1|})+  |z _1| (1-|z_2|) }{|z_2-z_1|}\\
\end{eqnarray*}
But could not prove that the last term is controlled by 1.
 A: Notice that
$$z_1^{n+1}-z_2^{n+1}=(z_1-z_2)\cdot\underbrace{\left(z_1^n+z_1^{n-1}z_2+z_1^{n-2}z_2^2+\dots+z_1^2z_2^{n-2}+z_1z_2^{n-1}+z_2^n\right)}_{\mathcal{Z}(n)}$$
Then:
\begin{align}
|f(z_2)-f(z_1)|=\left|\sum_{n\geq 1}a_n(z_2-z_1)\cdot\mathcal{Z}(n-1)\right|
&\leq |z_2-z_1|\cdot\sum_{n\geq1}\left|\mathcal{Z}(n-1)\right|\\
&=|z_2-z_1|\cdot\sum_{n\geq0}\left|\mathcal{Z}(n)\right|
\end{align}
Here, we have used that $|a_n|\leq 1$ for all $n$ (why?).
On the other hand, we have
\begin{align}
\left|\mathcal{Z}(n)\right|\leq\sum_{k=0}^n|z_1|^{n-k}|z_2|^k\leq\sum_{k=0}^nr^n=(n+1)r^n
\end{align}
It follows that
\begin{align}
\frac{|f(z_2)-f(z_1)|}{|z_2-z_1|}
&\leq \sum_{n\geq0}(n+1)r^n\\
&=\frac1r\sum_{n\geq1}nr^n
\end{align}
Do you think you can take it fom here?
A: It might be useful
\begin{align}
\left|\frac{f(z_2)-f(z_1)}{z_2-z_1}\right|
&= \left|\frac{1}{z_2-z_1}\int_{z_1}^{z_2}f'(u)du\right| \\
&= \left|\int_{0}^{1}f'(z_1+t(z_2-z_1))dt\right| \\
&\leq \int_{0}^{1}\dfrac{1-|f(z_1+t(z_2-z_1))|^2}{1-|z_1+t(z_2-z_1)|^2}dt \hspace {2cm} \text{Schwarz–Pick~theorem}\\
&\leq \dfrac{1}{1-r^2}\int_{0}^{1}dt \\
&\leq \dfrac{1}{(1-r)^2}
\end{align}
