Indija TS 2012

Note that we may assume $m=0$, since $P(z)=z^{m}Q(z)$ for some polynomial $Q(z)$ where $|P(z)|=|Q(z)|$ on the unit circle. Thus we may write $$P(z)=\sum _{k=0}^n{a}_k{z}^k,a_n\ne 0,a_0\ne 0,$$ and we have to prove that $$\underset{|z|=1}{\max }\{|P(z)|\}\ge \sqrt{\sum _{k=0}^n|a_k{|}^2+2|a_0{a}_n|}.$$ Observe that $|P(z){|}^2={\sum }_{k=0}^n{\sum }_{j=0}^n{a}_k{\bar{a}}_j{z}^{k-j}$. Let $\omega$ be a primitive $n$-th root of unity. Then $$\sum _{l=0}^{n-1}|P({\omega }^{l}z){|}^2=\sum _{l=0}^{n-1}\left(\sum _{k=0}^n\sum _{j=0}^n{a}_k{\bar{a}}_j{\omega }^{l(k-j)}{z}^{k-j}\right)=\sum _{k=0}^n\sum _{j=0}^n{a}_k{\bar{a}}_j{z}^{k-j}\sum _{l=0}^{n-1}{\omega }^{l(k-j)}.$$ The last sum is $n$ if $k-j\equiv 0(\mathrm{mod}n)$ and zero otherwise. Hence $$\frac{1}{n}\sum _{l=0}^{n-1}|P({\omega }^{l}z){|}^2=\sum _{k=0}^n|a_k{|}^2+a_n{\bar{a}}_0{z}^n+{\bar{a}}_n{a}_0{z}^{-n}.$$ Choosing $z_0$ such that $z_0^n=\frac{a_0|a_n|}{|a_0|a_n}$, we obtain $$\frac{1}{n}\sum _{l=0}^{n-1}|P({\omega }^l{z}_0){|}^2=\sum _{k=0}^n|a_k{|}^2+2|a_0{a}_n|.$$ Hence we can find $l$ such that $$|P({\omega }^l{z}_0){|}^2\ge \sum _{k=0}^n|a_k{|}^2+2|a_0{a}_n|.$$ The result follows.