This is a homework problem, so hints or rough outlines are strongly preferred to a full solution.
Problem. Let $C$ be the unit circle. Suppose the continuous function $f : C \rightarrow \mathbb{C}$ on the unit circle satisfies $|f(z)| \leq M$ and $\left|\int_{C} f(z)\;dz\right| = 2\pi M$. Show that $f(z) = c\bar{z}$ for some constant $c$ with modulus $|c| = M$.
I've been able to argue $|f(z)| = M$ for all $z \in C$ by assuming this is not the case and finding a contradiction using continuity, the given identity, and the $ML$-inequality. This is clearly necessary, but may not be useful to solve the problem.