Let $f:D\to \mathbb{C}$ be a holomorphic function where $D$ is the open unit disk.
Then prove
$ 2|f'(0)| \leq \sup_{z_1,z_2\in D} |f(z_1)-f(z_2)| $
I can show that $2f'(0) = \frac{1}{2\pi i} \int_{\gamma_R} \frac{f(w)-f(-w)}{w^2} dw$
where $\gamma_R$ is a circle with radius $R<1$.
Then by using the standard tools I can arrive that $2|f'(0)| \leq \frac{1}{R} \sup_{\gamma_R} |f(w)-f(-w)|$
$\sup_{\gamma_R} |f(w)-f(-w)| \leq \sup_{z_1,z_2\in D} |f(z_1)-f(z_2)|$ Then I'm stuck, as I can't just let $R$ be 1, and it is not obvious why $\frac{1}{R} \sup_{\gamma_R} |f(w)-f(-w)|$ can't be a constant larger than $ \sup_{\gamma_R} |f(w)-f(-w)|$ for any $R$.