Let $f$ be a holomorphic map of the open unit disk to itself. Prove that $|f'(z)| \leq \frac{1}{1-|z|}$ for all $z$ in the disk. [duplicate]

Consider a holomorphic functionf $f$ from $\mathbb{D}$ to $\mathbb{D}$ where $\mathbb{D}$ represents the unit disc centred at the origin.

Prove that $|f'(z)| \leq \frac{1}{1-|z|}$

I think the Cauchy inequalities should be useful, but I couldn't get anything significant out of them.

The stronger inequality $$|f'(z)|\leq \frac{1}{1-|z|^2}$$ holds. Here are two proofs. Proof 1. For any $w \in \mathbb{D}$ the function $$z\mapsto \frac{z-w}{1-\overline{w}\,z}$$ maps $\mathbb{D}$ onto $\mathbb{D}$. In particular $$\frac{f(z)-f(0)}{1-\overline{f(0)}\,f(z)}$$ maps $\mathbb{D}$ into $\mathbb{D}$ and since it vanishes at $z=0$ we can use the Schwarz lemma to bound its derivative at $0$. This results in the inequality

$$\frac{|f'(0)|}{1-|f(0)|^2}\leq 1.$$

So $|f'(0)|\leq 1-|f(0)|^2\leq 1$ for any function $\mathbb{D}\to \mathbb{D}$. Now let $w\in \mathbb{D}$ and apply this result to

$$z\mapsto f\left(\frac{z+w}{1+\overline{w}z}\right).$$

Proof 2. Let $w\in \mathbb{D}$ and $r$ be a real number such that $|w|<r<1$. By the Cauchy integral theorem we have

$$ f'(w)=\frac{1}{2\pi \textrm{i}} \oint_{|z|=r}\frac{f(z)\,dz}{(z-w)^2}$$

and so

$$ |f'(w)|\leq\frac{1}{2\pi} \oint_{|z|=r}\frac{|dz|}{|z-w|^2}.$$

Now for $|z|=r$ we have $|z-w|^2=(z-w)(\overline{z}-\overline{w})=(z-w)(r^2z^{-1}-\overline{w})$ and $|dz|=-\textrm{i}rz^{-1}dz$ . Plug this in to get

$$\frac{1}{2\pi} \oint_{|z|=r}\frac{|dz|}{|z-w|^2} = \frac{1}{2\pi \textrm{i}} \oint_{|z|=r}\frac{r\, dz}{(z-w)(r^2-\overline{w}z)} = \frac{r}{r^2-|w|^2}.$$

(The last equality is simply Cauchy's integral formula for the holomorphic function $r/(r^2-\overline{w}z)$.) Finally take $r \uparrow 1$.