Analytic function on unit disk, Schwartz lemma type inequality Let $f:\Bbb{D}\to\Bbb{C}$ be an analytic function that satisfies the inequality $|f(z)|\le\dfrac{1}{1-|z|}.$
How can I show that $|f'(0)|\le 4.$   

Also, I am looking for the equality case. Is this a sharp upper bound?

 A: We can show that $|f'(0)|\le 4$ using Cauchy's Integral Formula
$$f'(0)=\frac{1}{2\pi i}\oint_{|z|= r<1}\frac{f(z)}{z^2}\,dz \tag1$$
Taking the magnitude of both sides of $(1)$ reveals
$$\begin{align}
|f'(0)|&=\left|\frac{1}{2\pi i}\oint_{|z|=r<1}\frac{f(z)}{z^2}\,dz\right|\\\\
&\le \frac{1}{2\pi }\oint_{|z|= r<1}\frac{|f(z)|}{|z^2|}\,|dz|\\\\
&=\frac1{2\pi}\frac{(2\pi r )\left(\frac{1}{1-r}\right)}{r^2}\\\\
&=\frac{1}{r(1-r)}\tag 2
\end{align}$$
The minimum of $\frac{1}{r(1-r)}$ occurs when $r=1/2$ in which case we have
$$|f'(0)|\le 4$$
as was to be shown!
A: With the Schwarz lemma I get : 
If $r \in (0,1)$ then for $|z| < 1$ : $|f(rz)| \le \frac{1}{1-r}$ and $|f(0)| \le 1$ so that $$g(z)=\frac{f(rz)-f(0)}{1+\frac{1}{1-r}}$$
is analytic on $\mathbb{D} \to \mathbb{D}$ and $g(0) = 0$. Thus by Schwarz's lemma $$1 \ge  |g'(0)| = \left|\frac{r f'(0)}{1+\frac{1}{1-r}}\right|=|f'(0)|\frac{r(1-r)}{2-r}$$
whose optimum is at $r \approx 0.6$ with $\frac{r(1-r)}{2-r} \approx 0.18$
I wonder if there is a way to improve this to $|f'(0)| \le 4$ as Mark Viola did with the Cauchy's integral formula.
