$P$ is an $n$-degree polynomial . For $r>0$ let's define: $$M(r,P)=\sup_{|z|=r} |P(z)| $$ Prove that function $r\mapsto M(r,P)$ is increasing and function $r\mapsto\frac{M(r,P)}{r^n}$ is decreasing.
I think I should use the principle of maximum. Have you got any hints?
I will answer the first part of the question, and give a hint for the second.
By the maximum modulus principle, $\left|P(z)\right|$ achieves no maximum in the domain $|z|<r$ (I'm assuming $n>0$, so $P$ isn't constant). However, as a continuous function, it does a achieve a maximum in the compact domain $|z|\le r$. This implies that $\max_{|z|\leq r}\left|P(z)\right| = \max_{|z|=r}\left|P(z)\right| = M(r,P)$, and it isn't achieved anywhere in $|z|<r$.
In particular, for all $|z_0| < r$ we have $$\left|P(z_0)\right| \leq \max_{|z|\leq |z_0|}\left|P(z)\right| < \max_{|z|=r}\left|P(z)\right| = M(r,P)$$ Put differently, for $r<R$ and all $|z|=r$ we have $\left|P(z)\right|<M(R,P)$. A direct consequence is that $M(r,P)<M(R,P)$.
For the second part, as mentioned in comments, we can develop $$\frac{M(r,P)}{r^n} = \sup_{|z|=r}\left|\frac{1}{z^n}P(z)\right| = \sup_{\left|\frac{1}{z}\right|=\frac{1}{r}}\left|Q\left(\frac{1}{z}\right)\right|$$ for a different polynomial $Q(z)$ (which?). We can now use part (a) to conclude (how?).