If $P(z)=\sum_{k=0}^na_kz^k, (a_n=1)$$P(z)$ having no zeros in $|z|<1,$ I am trying to provethen $$\frac{\max_{|z|=1}|P'(z)|}{\max_{|z|=1}|P(z)|}\leq \frac{n-1}{2}+\frac{1}{1+|a_0|}.$$$$\frac{\max_{|z|=1}|P'(z)|}{\max_{|z|=1}|P(z)|}\leq \frac{n}{2}.$$ The result is true for $n=1.$ Can I apply induction principle toCan we prove this by induction on $n$? or is there any alternative way?
Attempt at a proof: let us try to show that the inequality holds by induction on the degree $n$ of the polynomial $P(z)$.
If $n=1$ then $P(z)=z-w$ with $|w|\geq 1$, and we have
$$\displaystyle\frac{\max_{|z|=1}|P'(z)|}{\max_{|z|=1}|P(z)|}=\frac{1}{1+|w|}\leq \frac{1}{1+1},$$
which is nothing but the given inequality when $n=1$.
Let $Q(z)=(z-w)P(z)$ with $|w|\geq 1$, where $P(z)$ is a polynomial of degree $n$ having all its zeros in $|z|\geq 1$. $$\displaystyle\frac{\max_{|z|=1}|Q'(z)|}{\max_{|z|=1}|Q(z)|}=\frac{\max_{|z|=1}|(z-w)P'(z)+P(z)|}{\max_{|z|=1}|(z-w)P(z)|}$$ \begin{equation}\label{p1}\displaystyle\leq\frac{\max_{|z|=1}|P'(z)|}{\max_{|z|=1}|P(z)|}+\frac{1}{\max_{|z|=1}|z-w|} (?).\end{equation}
By induction hypothesis we will have then $$\displaystyle\frac{\max_{|z|=1}|Q'(z)|}{\max_{|z|=1}|Q(z)|}\leq \frac{n-1}{2}+\frac{1}{1+|a_0|}+\frac{1}{1+|w|}\leq \frac{n}{2}+\frac{1}{1+|a_0||w|},$$ which is true follows from the fact that $$-\frac{1}{2}+\frac{1}{1+|a_0|}+\frac{1}{1+|w|}-\frac{1}{1+|a_0||w|}=\frac{(1-|a_0|)(1-|w|)(1-|a_0w|)}{2(1+|a_0|)(1+|w|)(1+|a_0w|)}\leq 0.$$
Thus the proof by induction will be complete once we establish (?) part of the above.
