complex-analysisroots
I know from here that for a polynomial $p(z)=a_0+a_1z+…+a_nz^n$ with $0<a_0\leq a_1\leq…\leq a_n$ all roots are in the closed unit disk.
What condition do we need to get that all roots are in the open unit disc? I was thinking that maybe some $a_i\neq a_{i+1}$. But I don't know how to prove that?
The thing to do is to look instead at the polynomial $$Q(z) = (1-z)P(z) = (1-z)\left(\sum_{i=0}^n a_iz^i \right) = a_0 -a_n z^{n+1} + \sum_{i=1}^n (a_i-a_{i-1})z^i$$ Now, let $|z|>1$ be a root of $P(z)$, and hence a root of $Q(z)$. Therefore, we have $a_0 + \sum_{i=1}^n (a_i-a_{i-1})z^i = a_n z^{n+1}$ Then, we have \begin{aligned} |a_n z^{n+1}| &= \left|a_0 + \sum_{i=1}^n (a_i-a_{i-1})z^i\right| \\ & \le a_0 + \sum_{i=1}^n (a_i-a_{i-1})|z^i| \\ & < a_0|z^n| + \sum_{i=1}^n (a_i-a_{i-1})|z^n| \\ & = |a_n z^n|\end{aligned} a contradiction.
For a nice article on integer polynomials, see here. (Your problem is Proposition 10)
This is not a answer. I just want to use the space and easy-to-edit feature in the "Answer" section to type the equations.
You did not specify the relation of the modulus of the coefficients $a_n$. If they happen to be cup-shaped, then you may use a theorem by Chen (J. of Math. Anal. and Appl. vol 190, 714-724 (1995)).
By cup-shaped, I mean
$$|a_0|\ge |a_1| \ge \cdots \le |a_{N-2}| \le |a_{N-1}|$$
In your case, we have (assume $N=2n+2$) $$P(z) = \sum_{k=0}^{2n+1}a_kz^k=z^n q(z)+q^*(z)$$
where $$q_n(z)=\sum_{k=0}^na_{n+k+1}z^k$$ $$q_n^*(z)=\sum_{k=0}^n a_kz^k$$
If $N=2n+1$, then we define $Q(z)=(1+z)P(z)$ and treat $Q(z)$ in a similar fashion.
Best Answer
The result you mention is known as the Eneström-Kakeya theorem. Necessary and sufficient conditions for when the roots of the polynomial lie on the boundary of the region are given by Anderson, Saff, and Varga in the paper
The paper is freely available from Varga's website here.