$$Let \ N \in \mathbb{N^+}, S = [1,N]\cap \mathbb{N^+}, \forall n \in S, S_n = \sum_{sym}^n a_i > 0. \\ Prove \ that \ \forall i\in S, a_i > 0.$$
$$Proof: \ f(x) = \prod^N_{i=1} (x-a_i) = \sum_{s \subseteq S} \prod_{i\in s} a_i(-x)^{N-|s|} = \sum^N_{n=1} S_n(-x)^{N-n}. \\ f(a_i) = 0 \Rightarrow a_i > 0.$$