Maximum value of a complex polynomial on the unit disk The polynomial is $p(z)=\sum^n_{k=0} a_kz^k$. And I want to prove the following inequality on the unit disk$$\max_{B_1(0)}|p(z)|\geq |a_n|+|a_0|$$
By the maximum modulus principle, the maximum must be on the unit circle and greater than $|a_0|$ by considering $p(0)$. However, I cannot make further conclusions from this, since any attempt of using the triangle inequality will result in the opposite direction of the wanted result.
I have also seen a similar problem, although I can conclude $\max_{|z|=1}|p(z)|$is greater than any of the two on RHS, but since there is no relation of $\max_{k\in\{0,\ldots,n\}}|a_k|\geq|a_0|+|a_k|$, a tighter bound is needed.
I also tried expanding it into trig functions, and consider the roots, but it didn't work as expected.
 A: This was trickier than I first thought. Perhaps there is a simpler solution, but here is one approach: Let $\omega = \exp(2\pi i/n)$. Then 
$$
\sum_{j=0}^{n-1} {(\omega^k)}^j = \begin{cases} 
n, & k \in \{ 0, n \} \\ 
0, & 1 \le k \le n-1. 
\end{cases}
$$
(This is a well-known property of roots of unity, or follows from computing the geometric sum if you prefer.)
Hence
\begin{align}
\frac1n \sum_{j=0}^{n-1} f(\omega^j z) 
&= \frac1n \sum_{j=0}^{n-1} \sum_{k=0}^{n} a_k {(\omega^j)}^k z^k \\
&= \frac1n \sum_{k=0}^{n} a_k z^k \sum_{j=0}^{n-1} {(\omega^k)}^j = a_0 + a_nz^n.
\end{align}
Consequently, we get
\begin{align}
|a_0| + |a_n| &= \max_{|z|=1} |a_0 + a_nz^n | \\
&=  \max_{|z|=1} \Big| \frac1n \sum_{j=0}^{n-1} f(\omega^j z) \Big| \\
&\le    \frac1n \sum_{j=0}^{n-1} \max_{|z|=1} \big| f(\omega^j z) \big|
= \max_{|z|=1} |f(z)|.
\end{align}
A: One gets that
$$\max_{B_1(0)}|p(z)|\geq \sqrt{|a_n|^2+|a_{n-1}|^2+ \cdots+|a_0|^2}$$
by calculating
$$\mathbb{E} \left[ |p(e^{i \Theta})|^2\right],$$
where $\Theta$ is a uniform random variable on $[0,2\pi]$.
Even though it does not answer precisely the question, it could be of interest. 
