Showing f(0) is bound above by geometric mean of supremum over intervals? So I am working on the following problem.

Suppose that $f$ is entire and $n$ is a fixed positive integer. If
  $$I_k:=\left[\frac{2(k-1)\pi}{n},\frac{2k\pi}{n}\right],$$
  for $k=1,2,\dots,n$ and $$\alpha_k:=\sup_{\theta\in I_k}|f(e^{i\theta})|,$$ prove that $|f(0)|\leq (\alpha_1\alpha_2\cdots\alpha_n)^\frac{1}{n}.$

I began by noting that $m(I_k)=\frac{2\pi}{n}\forall k.$ Then I used the Mean Value Property to get
$$|f(0)|\leq\frac{1}{2\pi}\int_0^{2\pi} |f(e^{i\theta})|d\theta=\frac{1}{2\pi}\sum_{k=1}^n \int_{I_k} |f(e^{i\theta})|d\theta \leq \frac{1}{2\pi}\sum_{k=1}^n \frac{2\pi \alpha_k}{n}=\sum_{k=1}^n\frac{\alpha_k}{n}.$$
Now I can see that what I have on the RHS is the arithmetic mean, but in general $$\left(\prod_{i=1}^n \alpha_i\right)^\frac{1}{n}\leq\sum_{i=1}^n\frac{\alpha_i}{n}.$$ That seems to suggest that my route is not producing the sharpest bound.
Next, I though maybe I could use Jensen's to try and force the geometric mean to appear, but I haven't produced anything useful as of yet.
Any advice is greatly appreciated.

If $f(0)=0$ there is nothing to prove. From now on assume $f(0)\ne0$.
Let $a_1,\dots,a_m$ be the zeroes of $f$ in $\{|z|<1\}$. Suppose first that $f(z)\ne0$ if $|z|=1$. By Jensen's formula
$$
\log|f(0)|=\sum_{i=1}^m\log|a_k|+\frac{1}{2\,\pi}\int_0^{2\pi}\log|f(e^{i\theta})|\,d\theta\le\frac1n\sum_{k=1}^n\log|\alpha_k|,
$$
which gives the desired inequality.
If $f(z)=0$ for some $z$ with $|z|=1$, take a sequence $r_n\to1$ such that $f$ does not vanish on $\{|z|=r_n\}$, apply the above argument on the disk of radius $r_n$ and let $n\to\infty$.
