My answer:

Er.. There's something wrong in my answer. Maybe I'll modify it soon. Update:

Suppose $\omega$ be the $n$th root of unity. Now so $|z\omega ^i|=|z|=1$ for all $i\in\mathbb N$. Now suppose $P(z)=z^mQ(z)\implies |P(z)|=|Q(z)|$. Now note $|P(z)|^2=P(z)\overline {P(z)}=\sum _{k=0}^{n}\sum_{j=0}^{n} a_k\overline {a_j}z^{k-j}$.Now putting $z=\omega^i z_0$ for all $i\in\{0,1,2,.....n-1\}$ and summing those we obtain $\sum_{i=0}^{n-1}|P(\omega^iz)|^2=n(\sum_{k=0}^{n}|a_k|^2+z^na_n\overline{a_0}+z^{-n}a_0\overline{a_n}$.Now put $z=(\frac {a_0|a_n|}{a_n|a_o|})^{\frac {1}{n}}$. And now so we get $\sum_{i=0}^{n-1}|P(\omega^i(\frac {a_0|a_n|}{a_n|a_o|})^{\frac {1}{n}})|^2=n(\sum_{k=0}^{n}|a_k|^2+2|a_0a_n|$. Thus taking that $i$ for that $P(\omega^i (\frac {a_0|a_n|}{a_n|a_o|})^{\frac {1}{n}})$ we're done, because $|(\frac {a_0|a_n|}{a_n|a_o|})^{\frac {1}{n}}|=1$.

consider $\{\varepsilon , \varepsilon^2, \cdots , \varepsilon^{n-m}\}$ be the roots of the equation $x^{n-m}=1$. Also , $P(z)=z^{m}(a_{n}z^{n-m}+a_{n-1}z^{n-m-1}+\cdots+a_{m})$ and $ P(z)\cdot \overline{P(z)}=|P(z)|^2$ , so we transform the polynomial by $ z \mapsto \varepsilon^{k} z$ as $|\varepsilon^{k} z|=1$, where $ k \in \{ 1, 2 , \cdots , n-m\}$ , so we have $\sum_{k=1}^{n-m} |P(\varepsilon^{k}z)|^2= (n-m)\cdot \left(\sum_{k=m}^{n} |a_{k}|^2+a_{n}\overline{a_{m}}z^{n-m}+a_{m}\overline{a_{m}z^{n-m}}\right)$ (since , $\sum_{k=1}^{n-m} \varepsilon^{k \cdot i}=0$, where $i \in \mathbb{Z^{+}}$.) So , plug $z=\left(\frac{a_{m}|a_{n}|}{a_{n}|a_{m}|}\right)^{\frac{1}{n-m}}$ to get $\frac{\sum_{k=1}^{n-m} |P(\varepsilon^{k} z)|^2}{n-m}=\sum_{k=m}^{n} |a_{k}|^2+2|a_{m}a_{n}|$. and also $\sqrt{\frac{\sum_{k=1}^{n-m} |P(\varepsilon^{k} z)|^2}{n-m}} \leqslant \max_{|z|=1}\{|P(z)|\}$ \[\implies \text{max}_{|z|=1}\{|P(z)|\}\geqslant \sqrt{2|a_ma_n|+\sum_{k=m}^{n} |a_k|^2}\]. $\blacksquare$