# 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?

• Could you give your thoughts on why you think such an inequality must hold? – Kavi Rama Murthy Jun 22 '18 at 6:04
• For the polynomial having all zeros on the unit circle, the inequality becomes equality! – user159888 Jun 22 '18 at 6:27

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 $P$).

• Very nice approach. Thank you – user159888 Jun 22 '18 at 9:16
• @user159888 What about the base case? – Robert Z Jun 22 '18 at 11:32
• Base case is correct. Thank you – user159888 Jun 23 '18 at 4:31
• Did you prove it? – Robert Z Jun 23 '18 at 4:36
• Yes. We need to show $Re \left(\frac{1}{1-Ke^{i\phi}}\right)\geq \frac{1}{1+K}.$ Finding the real part of L.H.S explicitly we can verify that this holds whenever $K\leq 1.$ – user159888 Jun 23 '18 at 5:08