An inequality for polynomials

Question

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" (Russian original) (English translation) Mathematics of the USSR, Izvestiya 18, 1-17 (1982), Zbl 0517.42001; Izvestiya Akademii Nauk SSSR. Seriya Matematicheskaya, 45, 3-22 (1981), MR607574, Zbl 0538.42001.

[2] Antonios D. Melas and Zalman Rubinstein, "Problem 10255 and solution" American Mathematical Monthly, 103, 177-181 (1996).

Answer 1

$\newcommand\ep\varepsilon$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 $4(1-\ep)\pi\to4\pi>0$. $\quad\Box$