An inequality related to the maximum of an analytic function in the disk Let $f(z)$ be an analytic function in the disk $D$, continuous on the closure $\bar D$. Let $L$ be $\{\operatorname{Re}z=\frac{1}{2}\}\cap \bar D$. Let $M$ be the maximum of $\mid f(z)\mid$ in $\bar D$ , and $\mu$ the maximum of $\mid f\mid$ on $L$.
Prove:$\displaystyle \mid f(0)\mid ^3\leq \mu M^2$
My struggle: no idea. Maybe we should normalise the function first? Or maybe a power series expansion would help?

The key idea is geometric: the chord $\operatorname{Re}z=1/2$ subtends the angle of $2\pi/3$, which is one third of the full angle. So, if we take it together with rotations by $\pm 2\pi /3$, it will completely surround the origin, and the maximum principle can be applied to the resulting triangle. 
To formalize the above, let $g(z) = f(z) f(e^{2\pi i/3}z)f(e^{-2\pi i/3}z)$ and observe that $|g|\le \mu M^2$ on every side of the aforementioned triangle. 
