India — Team Selection Test, 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
$$P(z) = \sum_{k=0}^{n} a_k z^k, \quad a_n \neq 0, a_0 \neq 0,$$
and we have to prove that
$$\max_{|z|=1}\{|P(z)|\} \ge \sqrt{\sum_{k=0}^{n} |a_k|^2 + 2|a_0 a_n|}.$$
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 $k-j \equiv 0 \pmod{n}$ 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
$$|P(\omega^l z_0)|^2 \geq \sum_{k=0}^{n} |a_k|^2 + 2|a_0 a_n|.$$
The result follows.