An inequality involving the logarithmic derivative of a polynomial If $P(z)=a_0+a_1z+\cdots+a_{n-1}z^{n-1}+z^n$ is a polynomial of degree $n\geq 1$ having all its zeros in $|z|\leq 1,$  then I was trying to verify the question, is it true that for all $z$ on $|z|=1$  for which $P(z)\neq 0$ 
$$\text{Re}\left(\frac{zP'(z)}{P(z)}\right)\geq \frac{n-1}{2}+\frac{1}{1+|a_0|}?$$
I think this is true and some properties of reciprocal polynomial might help us in solving this. I request you to help me in this.
I am also thinking in the direction of adding the information on arithmetic average of zeros $|a_n|/n$ of $P(z)$ in the R.H.S of the above inequality, just like we brought the term $|a_0|$ which is the product of the zeros of $P(z).$ Whether $|a_n|/n$ reveal extra information?
 A: We show that the inequality holds by induction on the degree $n$.
Base step. If $n=1$ then $P(z)=z-w$ with $|w|\leq 1$ and we have to show that for $|z|=1$ and $z\not=w$,
$$\text{Re}\left(\frac{zP'(z)}{P(z)}\right)
=\text{Re}\left(\frac{z}{z-w}\right)\stackrel{?}{\geq} \frac{1}{1+|w|}.$$
Left to the reader.
Inductive step.  Let $Q(z)=(z-w)P(z)$ with $|w|\leq 1$, and let $n$ be the degree of the monic polynomial $P$. Hence for $|z|=1$, such that $Q(z)\not=0$,
$$\begin{align}\text{Re}\left(\frac{zQ'(z)}{Q(z)}\right)&=\text{Re}\left(\frac{z}{z-w}\right)+\text{Re}\left(\frac{zP'(z)}{P(z)}\right)\\&\geq  \frac{1-1}{2}+\frac{1}{1+|w|}+ \frac{n-1}{2}+\frac{1}{1+|P(0)|}\\
&\stackrel{?}{\geq} \frac{n}{2}+\frac{1}{1+|w||P(0)|}\end{align}$$
where the last inequality holds if and only if
$$\frac{1}{1+|w|}- \frac{1}{2}+\frac{1}{1+|P(0)|}
-\frac{1}{1+|w||P(0)|}=
\frac{(1-|w|)(1-|P(0)|)(1-|w||P(0)|)}{2(1+|w|)(1+|P(0)|)(1+|w||P(0)|)}\geq 0$$
which is satisfied because $|w|\leq 1$ and $|P(0)|\leq 1$ (recall that $P(0)$ is the product of the roots of $P$).
