I have to show, that if $|f(z)| \leq \frac{1}{\sqrt{1-|z|^2}}$ on the open unit disk $\mathbb{D}$ then $|f'(0)| \leq 2$.
I thought I could use Cauchy's estimates theorem ,where $\left|f(z)\right| \le M$ and $$ \left|f^{(n)}(z_0)\right| \le \frac{n! M}{R^n} \quad n \in \Bbb N, n \ge 1 $$ Here, $z_0 = 0$ for all $z \in \mathbb{D}$ , the open unit disk:
$$ f'(0) \le \frac{1}{\sqrt{1-|z|^2}} $$
But obviously that isn't enough. So I suppose by Cauchys Integral formula, $$ \left\lvert\, f'(0) \right|=\left\lvert\,\frac{1}{2\pi i}\int_{|z|=1}\!\frac{f(z)\,dz}{(z-0)^2}\right|\le max _{|z|=1} \left\lvert\, \frac{1}{\sqrt{1-|z|^2}} \frac{1}{z^2} \right| $$ I don't know how to go on. I would appreciate any hint.
The relation shows that on the disc of radius $0<r<1$, $|f(z)| \le \frac{1}{\sqrt{1-r^2}}$, so Cauchy shows that $|f'(0)| \le \frac{1}{r\sqrt{1-r^2}}$, where $r$ is arbitrary as above. Hence the hope is to show that for such an $r$ we get what we need.
But now it is obvious that the denominator is highest when $r^2=\frac{1}{2}$ and that for $r=\frac{1}{\sqrt 2}$ the inequality becomes precisely $|f'(0)| \le 2$, so we are done