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Let $2 \leq m \leq n$ be integers and let $\mathbf{x} \in \mathbb{R}^n$ (importantly, I am not assuming that the entries of $\mathbf{x}$ are non-negative). The elementary symmetric polynomials are defined by $$ e_k(\mathbf{x}) := \sum_{1 \leq i_1 < \cdots < i_k \leq n} \left(\prod_{j=1}^k x_{i_j}\right). $$ for $1 \leq k \leq n$. Is the following conjecture true?

Conjecture: If $\max_j(x_j) \leq 1/(m-1)$, $e_1(\mathbf{x}) = 1$, and $e_k(\mathbf{x}) \geq 0$ for $1 \leq k \leq m$, then $e_k(\mathbf{x}) \geq \binom{m-1}{k} / (m-1)^k$ for $1 \leq k < m$.

Background/motivation: It is not too hard to show that if the hypotheses hold (i.e., if $\max_j(x_j) \leq 1/(m-1)$, $e_1(\mathbf{x}) = 1$, and $e_k(\mathbf{x}) \geq 0$ for $1 \leq k \leq m$) then in fact $e_k(\mathbf{x}) > 0$ for $1 \leq k < m$ (though it may be the case that $e_m(\mathbf{x}) = 0$). The conjecture's goal is to establish a precise bound on how small $e_k(\mathbf{x})$ can be.

Special cases that are known: It's trivially true that $e_k(\mathbf{x}) \geq \binom{m-1}{k} / (m-1)^k$ when $k = 1$, since the quantity on the right in this case is equal to $1$, and one of the hypotheses was that $e_1(\mathbf{x}) = 1$. So we just need to prove this bound for $1 < k < m$.

I can prove that the conjecture is true if the entries of $\mathbf{x}$ are all non-negative (so it's true when $m = n$). However, numerics suggest that it might be true even if $\mathbf{x}$ has negative entries (the hypothesis $e_k(\mathbf{x}) \geq 0$ for $1 \leq k \leq m$ allows entries of $\mathbf{x}$ to be slightly negative; just not "too" negative).

Finally, the conjecture (if true) is tight: for the vector $(1,\ldots,1,0,\ldots,0)/(m-1)$, where there are $m-1$ entries equal to $1/(m-1)$, all of the hypotheses are satisfied and we have $e_k(\mathbf{x}) = \binom{m-1}{k} / (m-1)^k$ for $1 \leq k < m$.

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    $\begingroup$ If I am not mistaken, then the non-negativity condition on elementary symmetric polynomials should imply a certain bound on the number of non-negative $x_i$'s (something like Descartes' rule of signs?). Also, this whole set-up reminds me Newton's and Maclaurin's inequalities, see e.g. here: en.m.wikipedia.org/wiki/Maclaurin%27s_inequality Perhaps, a similar approach via induction and Rolle's theorem would work in your case? $\endgroup$
    – richrow
    Commented Jul 30 at 22:40
  • $\begingroup$ @richrow - Yep, that non-negativity condition implies that at least $m$ of the $x_i$'s are non-negative. I've played around with Newton's and Maclaurin's inequalities, but unfortunately without luck. For example, in Maclaurin's inequality we might have $e_m = 0$, which then just gives us the trivial bound $e_k \geq 0$ when $k \leq m$. $\endgroup$
    – Nathaniel Johnston
    Commented Jul 31 at 0:03
  • $\begingroup$ In fact if any of $e_2,\ldots,e_{m-1}$ is at its minimum, then $e_m=0$, right? $\endgroup$
    – Brendan McKay
    Commented Aug 1 at 13:50
  • $\begingroup$ @BrendanMcKay - Good observation, yes that’s true. $\endgroup$
    – Nathaniel Johnston
    Commented Aug 1 at 16:01

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