Proving an analytic function $f$ is bounded on $|z|\le1/2$ independent of $f$ subject to certain conditions 
Let $f:D(0,1) \to \mathbb C$ be analytic. Show that there is a constant $C$ independent of $f$ such that if $f(0)=1$ and $f(z) \notin (-\infty,0]$ for all $z \in D(0,1)$, then $|f(z)| \le C$ whenever $|z| \le 1/2$.

I have (finally) figured out how to prove this, and I ended up with $C=9$. I am curious what the “best” bound is though, and what the best approach would be for proving this. In other words, what is the supremum of all analytic functions $f$ on $|z|\le 1/2$, subject to the two conditions above?

Take $f(z) = \left( \dfrac{1 + z}{1 - z} \right)^2\ (|z| < 1)$, then $f(0) = 1$ and$$
\left| f\left( \frac{1}{2} \right) \right| = \left| \frac{1 + \dfrac{1}{2}}{1 - \dfrac{1}{2}} \right|^2 = 9.
$$
Thus $9$ is indeed the tightest bound.
