I have been thinking about the validity of the following inequality:  
 if $P(z)=\sum\_{k=0}^na\_kz^k, a\_n\neq 0$ and $P(z)$ is non-zero in $|z|<1, $ then for $\theta \in [0, 2\pi],$ and $p>0$
\begin{align\*}
 &\int\_{0}^{2\pi}\left|n\left(1-\frac{|a\_0|-|a\_n|}{n(|a\_0|+|a\_n|)}\right)P(e^{i\theta})+(\eta-e^{i\theta})P'(e^{i\theta})\right|^{p}d\theta\\
 &\qquad\qquad\quad \leq \left[n\left(1-\frac{|a\_0|-|a\_n|}{n(|a\_0|+|a\_n|)}\right)\right]^{p}\int\_{0}^{2\pi}|P(e^{i\theta})|^{p}d\theta,
 \end{align\*}
for any $\eta$ lying in the closed unit disc.  

In fact the above inequity is motivated by the inequality from Melas and Rubinstein [2] who proved that
if $P(z) $ is a polynomial of degree $n$ then for $\theta \in [0, 2\pi],$
\begin{align\*}
 &\int\_{0}^{2\pi}\left|nP(e^{i\theta})+(1-e^{i\theta})P'(e^{i\theta})\right|^{p}d\theta\\
 &\qquad\qquad\quad \leq n^{p}\int\_{0}^{2\pi}|P(e^{i\theta})|^{p}d\theta.
 \end{align\*}
Alternatively the above result [due to Melas] can be established directly through an inequality due to Arestov [1] involving admissible operators on the class of polynomials. Is my above claim right? Your suggestions are of great help.


**References**


[1] Vitaliĭ Vladimirovich Arestov, "[On integral inequalities for trigonometric polynomials and their derivatives](https://doi.org/10.1070/IM1982v018n01ABEH001375)" (Russian original)
(English translation) Mathematics of the USSR, Izvestiya 18, 1-17 (1982), [Zbl 0517.42001](https://zbmath.org/0517.42001); Izvestiya Akademii Nauk SSSR. Seriya Matematicheskaya, 45, 3-22 (1981), [MR607574](http://www.ams.org/mathscinet-getitem?mr=607574), [Zbl 0538.42001](https://zbmath.org/0538.42001).


[2] Antonios D. Melas and Zalman Rubinstein, "[Problem 10255 and solution](https://doi.org/10.1080/00029890.1996.12004751)" American Mathematical Monthly, 103, 177-181 (1996).



An inequality for polynomials
$\newcommand\ep\varepsilon$This inequality does not hold in general.


Indeed, by continuity/denseness, the condition $a\_n\ne0$ can be dropped (in response to a comment by the OP, details on this are now given at the end of this answer).


Now just take $p=2$, $\eta=-1$, $n=2$, $P(z)=1-z+0z^2$ (so that $a\_0=1$ and $a\_n=a\_2=0$). Then the left- and right-hand sides of the conjectured inequality are $8\pi$ and $4\pi$, respectively. $\quad\Box$




---


**Details on the mentioned continuity/denseness:** Let again $p=2$, $\eta=-1$, $n=2$, but now $P(z)=1-\ep-z+\ep z^2$ with $\ep<0$ -- so that $a\_0=1-\ep$ and $a\_n=a\_2=\ep\ne0$). Then the roots $1$ and $1/\ep-1$ of $P(z)$ are outside the open unit disk, so that $P(z)\ne0$ if $|z|<1$. Moreover, with $\ep\uparrow0$, the left- and right-hand sides of the conjectured inequality will converge to $8\pi$ and $4\pi$, respectively. More specifically, the difference between the left- and right-hand sides of the conjectured inequality will then be $4(1-\ep)\pi\to4\pi>0$.
$\quad\Box$


