Let $a_1,\ldots,a_n \in \mathbb{C}$ be complex numbers that are the zeros of a real polynomial (meaning that the non-real ones come in complex conjugate pairs). Suppose that these numbers are such that $$ s_\lambda(a_1,\ldots,a_n) \ge 0 $$ for every partition $\lambda = (\lambda_1 \ge \ldots \ge \lambda_n)$, where $s_\lambda$ denotes the Schur polynomial associated to $\lambda$. Do these inequalities imply that $a_i \in \mathbb{R}_+$ for all $i$?
Here's what I know so far:
- The converse is obvious: if $a_i \in \mathbb{R}_+$ for all $i$, then the sum over Young tableaux formula immediately shows $s_\lambda(a_1,\ldots,a_n) \ge 0$.
- It's enough to show that the $a_i$ are real. Nonnegativity then follows by Descartes' rule of signs upon noting that the elementary symmetric functions $s_{(1,\ldots,1)}$ are, up to alternating signs, the coefficients of the polynomial $\prod_i (x-a_i)$.
- The statement is true and easy to prove for $n = 2$: it's enough to postulate nonnegativity merely for $\lambda = (n,0,\ldots)$, i.e. on the complete homogeneous symmetric polynomials. On these, $$ s_{(n,0,\ldots)}(r e^{i\theta},r e^{-i\theta}) = \frac{(r e^{i\theta}))^{n+1} - (r e^{-i\theta})^{n+1}}{re^{i\theta}-re^{-i\theta}} = r^n \frac{\sin((n+1)\theta)}{\sin(\theta)}, $$ which for every nonzero $\theta$ is obviously negative for some $n$, proving the contrapositive statement. This suggests that some Fourier analysis may be useful for the general case, or perhaps the Harish-Chandra-Itzykson-Zuber formula, but I don't know how to go about it.
