TITLE: The maximum of a polynomial on the unit circle
QUESTION [45 upvotes]: Encouraged by the progress made in a recently posted MO problem, here is a "conceptually related" problem originating from a 2003 joint paper of Sergei Konyagin and myself. 
Suppose we are given $n$ points $z_1,...,z_n$ on the unit circle $U=\{z\colon |z|=1\}$ and $n$ weights $p_1,...,p_n\ge 0$ such that $p_1+....+p_n=n$, and we want to find yet another point $z\in U$ to maximize the product
  $$ \prod_{i=1}^n |z-z_i|^{p_i}. $$
How large can we make this product by the optimal choice of $z$?

Conjecture. For any given $z_1,...,z_n\in U$ and $p_1,...,p_n\ge 0$ with $p_1+...+p_n=n$, there exists $z\in U$ with
    $$ \prod_{i=1}^n |z-z_i|^{p_i} \ge 2. $$

Here are some comments. 

If true, the estimate of the conjecture is best possible, as evidenced by the situation where the points are equally spaced on $U$ and all weights are equal to $1$.
We were able to resolve a number of particular cases; say, that where the points $z_i$ are equally spaced on $U$, and also that where all weights are equal to $1$.
The case $n=2$ is almost trivial, but already the case $n=3$ is wide open.
In the general case we have shown that the maximum  is larger than some absolute constant exceeding $1$.
Although this is not obvious at first glance, this conjecture is actually about the maxima of polynomials on the unit circle.   

I would be very interested to see any further progress!

REPLY [17 votes]: Here is a second proof. Let
$$H(z)=\prod_j|z-z_j|^{p_j}.$$
Let
$$F(z)=\prod_j\left((z-z_j)(1/z-\bar{z}_j)\right)^{p_j/n},\quad |z|\geq 1.$$
It is easy to see that:
$$|F(z)|=H^{2/n}(z),\quad |z|=1,$$
and that $F$ maps conformally the exterior of the unit disc on the exterior of a
"star" consisting of $n$ straight segments $[0,a_j]$ with some complex $a_j$.
Moreover, $F(z)\sim z$ as $z$ tends to infinity.
Now we see that the result is equivalent to the following theorem of Dubinin:
Let $K$ be the union of some intervals of the form $[0,a_j]$, and suppose
that capacity of $K$ is $1$. Then $\max_j|a_j|$ takes its minimal value when
the star is symmetric: all |a_j| are equal and the angles between adjacent
intervals are equal.
The reference is Dubinin, Uspekhi Mat. Nauk (=Russian Math. Surveys), 49 (1994).
Fedja's proof is much simpler, so it is a new proof of  Dubinin's theorem.