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...
 A: For the first part, you can conclude using the "ML-inequality" (you may know it under a different name):
$$
|f(0)| = \Big| \frac{1}{2\pi i} \int_{|z|=r} \frac{f(z)}{z}\,dz \Big|
\le \frac{1}{2\pi} \cdot 2\pi r \cdot \sup_{|z|=r} \Big|\frac{f(z)}{z} \Big| \le r \cdot \frac{M_r}{r} = M_r.
$$
A: $$
\underbrace{\,\,\left| \frac{1}{2\pi i} \int_{|z|=r} \frac{f(z)}{z} \, dz \right| = |f(0)|\leq \frac{1}{2\pi} \int_{|z|=r} \left|\frac{f(z)}{z} \, dz \right|\,\,}_{\huge \text{?}\vphantom{\displaystyle\int}}
$$
Instead of the above, I'd write
$$
|f(0)| = \left| \frac{1}{2\pi i} \int_{|z|=r} \frac{f(z)}{z} \, dz \right| \leq \frac{1}{2\pi} \int_{|z|=r} \left|\frac{f(z)}{z} \, dz \right|.
$$
The point is we know $\text{“}=\text{''}$ is true because of Cauchy's formula, and we know $\text{“}\le\text{''}$ is true for a different reason, and we initially put $\text{“}=\text{''}$ and $\text{“}\le\text{''}$ only where we already know they're true rather than where we're trying to prove they're true.  Then we can say we've proved
$$
|f(0)| \leq \frac{1}{2\pi} \int_{|z|=r} \left|\frac{f(z)}{z} \, dz \right|.
$$
We can go on to say
$$
\frac{1}{2\pi} \int_{|z|=r} \left|\frac{f(z)}{z} \, dz \right| = \frac 1 {2\pi r} \int_{|z|=r} |f(z)| \, |dz|.
$$
What does it mean to write $|dz|$ instead of $dz$ (which you omitted)? It means we're integrating with respect to arc length.
That last integral is the average value of $|f(z)|$ on the circle of radius $r$ centered at $0$.  So what we're trying to show is that the average value is less than or equal to the largest value. We can say
$$
\int_{|z|=r} |f(z)| \, |dz| \le \int_{|z|=r} M_r \, |dz| = 2\pi r M_r.
$$
Notice that it would be grave error to write $\displaystyle\int_{|z|=r} M_r \,dz$ instead of $\displaystyle\int_{|z|=r} M_r \, |dz|$, since that integral (with $dz$ instead of $|dz|$) evaluates to $0$ rather than to $2\pi rM_r \vphantom{\displaystyle\int}$.
