An inequality about a bounded holomorphic function on unit disk Suppose $f$ is a bounded holomorphic function on the open unit disk $D$. Show that $$\left(1-\left|z\right|\right)\left|f'\left(z\right)\right|\leq\sup_{w\in D}\left|f\left(w\right)\right|$$
for all $ z\in D$.
The first thing that occurs to me is the maximal principle. But it didn't work. I don't know how to find connection between right side and left side of the inequality.
 A: I tkink that your $f^{\prime}(t)$ is $f^{\prime}(z)$. If true: 
Let $z\in D$, and $0<r<1-|z|$. Let $\gamma$ the circle centered at $z$, with radius $r$. Then $\gamma$ is included in $D$. We have
 $$f^{\prime}(z)=\frac{1}{2i\pi}\int_{\gamma}\frac{f(t)}{(t-z)^2}dt$$
 Hence with $M=\sup_{|z|< 1}|f(z)|$, we have $$ |f^{\prime}(z)|\leq \frac{1}{2\pi}\frac{M2\pi r}{r^2}=\frac{M}{r}$$
 Now let $r\to 1-|z|$. 
A: Let's use a consequence of Cauchy's integral formula. The following holds.
$U \subset \mathbb{C}$ open$, f: U \to \mathbb{C}$ holomorphic, then $f$ can be differentiated any number of times. Furthermore, we have:
$$f^{(n)}(z_0) = \frac{n!}{2\pi i} \int\limits_{\partial B} \frac{f(\xi)}{(\xi - z_0)^{n + 1}} d\xi$$ where $B \subset U$ is a disk and $\partial B$ is the boundary.
Okay. You are interested in $|f'(z)|$, so let's use the formula:
$$f'(z) = \frac{1!}{2\pi i} \int\limits_{\partial B} \frac{f(\xi)}{(\xi - z)^2} d\xi$$
We must now guess a good disk so that we obtain what is needed. Let's use $D_r(z)$ which is a disk centered at $z$ with radius $r = 1 - |z|$. For a given $z$ the radius is the distance of $z$ to $1$. You see that our disk is a subset of the unit disk $D$. You also already see, at the boundary of this disk $r = 1 - |z|$ is pretty close to the denominator $\xi - z$ and matches the leftmost factor of your result $1 - |z|$.
We now estimate the size of the integral by the length multiplied with the biggest element. The length is $2\pi (1 - |z|)$.
$$f'(z) \le \frac{1!}{2\pi i} (2\pi \cdot (1 - |z|)) \sup_{\xi \in D_r(z)} \frac{f(\xi)}{(\xi - z)^2} = \frac{1}{i} \cdot (1 - |z|) \sup_{\xi \in D_r(z)} \frac{f(\xi)}{(\xi - z)^2}$$
Next we use $|\cdot|$ as you're interested in $|f'(z)|$ and by that get rid of the $i$.
$$|f'(z)| \le \left|\frac{1}{i}\right| \cdot \left|1 - |z|\right| \sup_{\xi \in D_r(z)} \left|\frac{f(\xi)}{(\xi - z)^2}\right| = \left|1 - |z|\right| \sup_{\xi \in D_r(z)} \left|\frac{f(\xi)}{(\xi - z)^2}\right|$$
We know that $\frac{1}{|\xi - z|} \le \frac{1}{1 - |z|}$ thus we simplify.
$$|f'(z)| \le \left|1 - |z|\right| \sup_{\xi \in D_r(z)} \left|\frac{f(\xi)}{(\xi - z)^2}\right| \le \left|1 - |z|\right| \sup_{\xi \in D_r(z)} \left|\frac{f(\xi)}{(1 - |z|)^2}\right|$$
Next, we see that the denominator now is independent of $\xi$, we pull it out. We also know that $|1 - |z|| = 1 - |z|$ because of $1 - |z| \ge 0$ since |z| is bounded by $1$.
$$|f'(z)| \le \left|1 - |z|\right| \sup_{\xi \in D_r(z)} \left|\frac{f(\xi)}{(1 - |z|)^2}\right| = \left|1 - |z|\right| \cdot \frac{1}{(1 - |z|)^2}\sup_{\xi \in D_r(z)} \left|f(\xi)\right| = \frac{1}{1 - |z|} \sup_{\xi \in D_r(z)} \left|f(\xi)\right|$$
Right factor to the left and we receive:
$$(1 - |z|) |f'(z)| \le \sup_{\xi \in D_r(z)} \left|f(\xi)\right|$$
Almost finished, we need $D$ and not $D_r(z)$. We already noticed $D_r(z) \subseteq D$. Thus, we simply can expand the expression to $D$ and would only receive a bigger value.
$$(1 - |z|) |f'(z)| \le \sup_{\xi \in D_r(z)} \left|f(\xi)\right| \le \sup_{\xi \in D} \left|f(\xi)\right|$$
And we are finished, hope you understood everything :)
A: The simplest proof of the inequality you ask about is as in @Kelener's answer; apply the Cauchy Integral Formula on the circle $|w-z|=r$ for $0<r<1-|z|$. (Or one could apply the inequalities often known as "Cauchy's Estimates", which is really the same thing, since those inequalities follow from CIF.)
Let's say $M=\sup_{|w|<1}|f(w)|$.
In fact a stronger inequality is true: $$(1-|z|^2)|f'(z)|\le M.$$(This is stronger since $1-|z|^2=(1+|z|)(1-|z|)\ge1-|z|$.) This follows from CIF on the circle $|w|=r$, $|z|<r<1$. We have $$f'(z)=\frac{1}{2\pi i}\int_{|w|=r}\frac{f(w)}{(w-z)^2}\,dw.$$Parametrizing that circle as $w=re^{it}$ shows that $$|f'(z)|\le \frac{Mr}{2\pi}\int_0^{2\pi}\frac{1}{|re^{it}-z|^2}\,dt\le \frac{M}{2\pi}\int_0^{2\pi}\frac{1}{|re^{it}-z|^2}\,dt,$$and now we need to calculate that integral. A fun way to do that is using Fourier series.
Note first that $|re^{it}-z|=|r-e^{-it}z|$. Since $|e^{-it}z|<r$ we can apply the formula for the sum of a geometric series:
$$\frac{1}{r-e^{-it}z}=\frac1r\sum_{n=0}^\infty\left(\frac{e^{-it}z}{r}\right)^n=\frac1r\sum_{n=0}^\infty\left(\frac{z}{r}\right)^ne^{-int}.$$So Parseval shows that $$\frac1{2\pi}\int_0^{2\pi}\frac1{|r-e^{-it}z|^2}=\frac1{r^2}\sum_{n=0}^\infty\frac{|z|^{2n}}{r^{2n}}=
\frac{1}{r^2}\frac1{1-\left(\frac{|z|}{r}\right)^2}=\frac{1}{r^2-|z|^2}.$$
Now let $r\to1$ and insert the result above: $$|f'(z)|\le\frac{M}{1-|z|^2}.$$ 

The Last Word: In fact the best possible bound on $|f'(z)|$ in terms of $M$ and $|f(z)|$ is $$|f'(z)|\le\frac1M\frac{M^2-|f(z)|^2}{1-|z|^2}.$$Hint: First show wlog $M=1$. Now compose $f$ on the left and on the right by an appropriate linear-fractional transformation and apply the Schwarz Lemma. (Martin R is right when he says I should mention that this is the Schwarz-Pick theorem.)
