For a polynomial $p(z)=a_0+\ldots+a_nz^n$, prove there exist $R>0$, such that if $|z|=R$, then $|p(z)|\geq |a_n|R^n/2$.
I get $|p(z)|\geq |a_n|R^n$, which makes that factor of $1/2$ useless. So I must have done something wrong.
$$|\frac{\partial^n{f}}{\partial z^n}(P)|\leq \frac{\sup_{z\in \overline{D}(P,r)} |f(z)|k!}{r^k}$$. This is the Cauchy estimate.
$$|\frac{\partial^n{p}}{\partial z^n}(z)| = |a_n n!|$$ $$|\frac{\partial^n{p}}{\partial z^n}(0)|\leq \sup_{\zeta\in \overline{D}(0,|z|)} |p(\zeta)|n!/R^n$$ $$|a_n n!|\leq \sup_{\zeta\in \overline{D}(0,|z|)} |p(\zeta)|n!/R^n$$ $$R^n |a_n|\leq \sup_{\zeta\in \overline{D}(0,|z|)} |p(\zeta)|$$
There exist a $z_0$, such that $|z_0|>0$ and $$\sup_{\zeta\in \overline{D}(0,|z_0|)} |p(\zeta)| = |p(z_0)|$$.
This is saying there exist a disk centered at the origin, such that $|p|$ is maximum on some point on the boundary.
Let $R=|z_0|$, we have $R^n |a_n|\leq |p(z_0)|$
Edit: I know what I got wrong. $z_0$ is just one point on the circle $|z| = R$. One can chose another point $z_1$ on the circle, such that $p(z_1)