Problem statement: Let $P(z)$ be a polynomial of degree $n$ with complex coefficients, $P(0) = 1$, and $|P(z)| \leq M$ for $|z| \leq 1$. Prove that every root of $P(z)$ in the closed unit disc has multiplicity at most $c \cdot \sqrt{n}$, where $c = c(M) > 0$ is a constant depending only on $M$.
attempt(this is my friends attempt because I don’t know where to start)
It is sufficient to examine the multiplicity of the number 1. In fact, if we prove something for 1, then we may apply the result to the polynomial $p(z) = P(\alpha z)$ with $|\alpha| \leq 1$, and in this way, we obtain the same estimate for all roots lying in the unit disc.
The idea of the solution is the following. We consider the integral $$F(P) = \int_{0}^{2\pi} \log(|P(e^{i\phi})|) \, d\phi$$
and show that it exists and is nonnegative. Then we estimate it from above, once in the neighborhood of 1 with the aid of the multiplicity of 1 and the degree of $P$, and once at other points using the condition $|P(z)| \leq M$. It is sufficient to prove the existence of the integral for polynomials of the form $z - z_0$. If
$$P(z) = c \cdot \prod_{i=1}^{n} (z - z_i)$$
then $$\log(|P(z)|) = \log(|c|) + \sum_{i=1}^{n} \log(|z - z_{i}|).$$
The existence of $$\int_{0}^{2\pi} \log(|e^{i\phi} - z_{0}|) \, d\phi$$ is ensured since $|e^{i\phi} - z_{0}| \geq 0$ for all $\phi$, and the integrand is continuous on the interval $[0, 2\pi]$.
Here is my attempt.
Let $P(z)=\sum_{k=0}^n a_kz^k$, then $a_0=P(0)=1$.
Let $m+1$ be the multiplicity of the root 1, we have $P^{k}(1)=0$ for any $0\le k\le m$.
Hence$\sum_{k}a_kf(k)=0$ for any polynomial f of degree does not exceed m, since $\left\{ x(x-1)\cdots(x-k+1)\right\}_{k\le m}$ is a base.
Now we choose $f(x)=T_m(\frac{2x-n-1}{n-1})$, where $T_m(x)$ is the Chebyshev polynomial of the first kind. Then $|Tm(k)|\le 1$ for $1\le k \le n $.
By triangle inequality, we deduce that \begin{align} \sum_{k=1}^n |a_k|&\ge \sum_{k=1}^n |a_kf(k)|\ge|\sum_{k=1}^n a_kf(k)|=|a_0f(0)|\\&=T_m(1+\frac{2}{n-1})\\ &=\cosh(m\log(1+\frac{2}{\sqrt{n}-1}))\\ &>\cosh(\frac{2m}{\sqrt{n}})\\ &>\frac{2m^2}{n} \end{align}
So if we choose c sufficiently large, we can ensure that when $m>c\sqrt{n}$, $\sum_{k} |a_k|$could be very large. But I still can't figure out the relationship between $\sum |a_k|$ and $\max_{|z|=1} |P(z)|$.