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 $\max_{|z|=1}|P(z)|=2$.
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
$$
2|P'(1)| \le n \max_{|z|=1}|P(z)| \, .
$$
I don't think this can hold in general, take $P(z) = z^2-1$, this satisfies $P(\pm1)=0$, so that $\max_{|z|=1}|P(z)|=\max_{z\in\{-1,1\}}|P(z)|$. However, $P'(1) = 2$ which is a contradiction.