Proving for polynomial: $P (1)=0\implies|P'(1)|\leq \frac{\deg(P)}{2}\max_{|z|=1}|P(z)|$. I wonder if it is true that any polynomial $P(z):\mathbb C\to\mathbb C$ with $P (1)=0$ satisfies
$$|P'(1)|\leq \frac{\deg(P)}{2}\max_{|z|=1}|P(z)|, $$ where the maximum is taken over the unit circle.
At least numerically it seems to be the case and the extremizer seems to be $P(z)=z^n-1$.
If $\deg(P)=1$, then $P(z)=z-1$. Hence, $P'(1)=1$ and $\max_{|z|=1}|P(z)|=2$.
 A: The conjectured inequality is true, and a consequence of Lemma 2 in

*

*Abdul Aziz and Q. G. Mohammad, Simple proof of a theorem of Erdős and Lax, Proc. Amer. Math. Soc. 80 (1980), 119-122.


Lemma 2. If $P(z)$ is a polynomial of degree $n$, then
$$|P'(z)| + |nP(z) - zP'(z)| \le n \max_{|z|=1}|P(z)|$$
for $|z| = 1.$

In our case is $P(1) = 0$, so that
$$
2|P'(1)| \le n \max_{|z|=1}|P(z)| \, .
$$
A: Too long for a comment.
An inductive approach seems very promising to me:
It is true for $n=1$.
Now, let $n\in\mathbb N$, and let $P_{n+1}:\mathbb C\to\mathbb C$ be a polynomial of degree $n+1$ with $P_{n+1}(1)=0$. We can write $$P_{n+1}(z)=P_n(z) (z+c)$$ for some $c\in\mathbb C$ where $P_n$ is a polynomial of degree $n$ satisfying $P_n(1)=0$. We have
$$P_{n+1}'(1)=P_n'(1)(1+c)+P_n(1)=P_n'(1)(1+c).$$
So by the inductive assumption,
$$\lvert P_{n+1}'(1)\rvert\le\frac n2\lvert 1+c\rvert\max_{\lvert z\rvert=1} \lvert P_n(z)\rvert.$$
If we could thus prove that
$$\bbox[15px,border:1px groove navy]{n\lvert1+c\rvert\max_{\lvert z\rvert=1}\lvert P_n(z)\rvert\le(n+1)\max_{\lvert z\rvert=1}\left(\lvert P_n(z)\rvert\lvert z+c\rvert\right)}$$
for every $c\in\mathbb C$ and polynomial $P_n$ of degree $n$ with $P_n(1)=0$, we would be done. I do not know, however, whether the highlighted statement is true and if yes, how to prove it.
