2 misprint corrected

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: $$|F(z)|=H^{2n}(z),\quad |F(z)|=H^{2/n}(z),\quad |z|=1,$$ 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.

1

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: $$|F(z)|=H^{2n}(z),\quad |z|=1,$$ 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.