The problem:
Let $P(n)$ be a polynomial of degree $n$. Let $M(r):= \underset{|z|\le r}{\mbox{sup}} \hspace{2mm} \left|P(z)\right|.$ I desire to establish that $r\mapsto \frac{M(r)}{r^n}$ for $r>0$ is non-increasing as a function of $r$.
At least I believe this to be true and necessary to conclude a long homework assignment :P.
1st attempt at solution
I think it is reasonable to assume the polynomial has no constant term, for adding a constant term as far as I understand can only change the $M(r)$ term by a constant. So $P(z)$ can be taken to fix the origin. Then considering, say the holomorphic function, $\frac{P(rz)}{M(r)}$ the function should take the disk to the disk, so we can apply Schwartz's Lemma, and get the inequality: $\left|\frac{P(rz)}{M(r)}\right|\le |z|$ From here I would hope to pick a nice $z$ in the disk to establish something useful... I'm at a loss.
2nd attempt at solution
I did start thinking, establishing an inequality between expressions doesn't quite get me that an expression is non-increasing. What would get me there is taking derivative. I think it's clear $M(r)$ is smooth. So we can differentiate the function in question (with respect to $r$) and get that the derivative is:
$\frac{M'(r)r^n - n r^{n-1}M(r)}{r^{2n}}=\frac{M'(r)r - n M(r)}{r^{n+1}}$
We want that this derivative is $\le 0$. That is, that
$M'(r) \le \frac{n M(r)}{r}$
The RHS sort of makes me think of what $M'(r)$ would look like. $P'(z)$ would of course be almost $\frac{n P(z)}{z}$, except of course for the constant term of $P(z)$ which disappears when differentiating.