Supremum of holomorphic function on the unit disk 
Let $f\colon B \to \mathbb{C}$ be holomorphic where $B$ is the unit disk in the complex plane centered at the origin.
  For $0<r<1$, let $M_r=\sup\{|f(z)|:|z|=r\}$.
Show that $|f(0)|≤M_r$ $ \forall r \in (0,1)$.
Now suppose further that $|f(0)|≥|f(z)| \forall z\in B$. Show that $|f(z)|$ is constant on $B$,

For the first bit I use Cauchy's integral formula and got:
$$\left| \frac{1}{2\pi i} \int_{|z|=r}^{} \frac{f(z)}{z} \right| = \left|f(0)\right|\leq \frac{1}{2\pi} \int_{|z|=r}^{} \left|\frac{f(z)}{z} \right| $$ 
Now I'm not sure how to implement the $M_r$
Need help with the second part...

For the first part, you can conclude using the "ML-inequality" (you may know it under a different name):
$$
|f(0)| = \Big| \frac{1}{2\pi i} \int_{|z|=r} \frac{f(z)}{z}\,dz \Big|
\le \frac{1}{2\pi} \cdot 2\pi r \cdot \sup_{|z|=r} \Big|\frac{f(z)}{z} \Big| \le r \cdot \frac{M_r}{r} = M_r.
$$
