Fix integers $1 \leq k \leq n$ and suppose $\mathbf{x} \in \mathbb{R}^n$ is such that $e_j(x_1,x_2,\ldots,x_n) \geq 0$ for all $1 \leq j \leq k$, where $e_j$ is the $j$-th elementary symmetric polynomial.

Question 1: Is it true that $x_1 + x_2 + \cdots + x_{n-k+1} \geq 0$?

The answer is trivially "yes" in the edge cases $k = 1$ and $k = n$, but the intermediate cases seem much less obvious.

While Question 1 is the one that I'm really interested in, there is a natural generalization of it that is perhaps known, so I'll ask it now:

Question 2: Is it true that $e_j(x_1,x_2,\ldots,x_{n-1}) \geq 0$ for all $1 \leq j \leq k-1$?

In other words, can we use non-negativity of elementary symmetric polynomials in $n$ variables to infer non-negativity of elementary symmetric polynomials in $n-1$ variables? If the answer to Question 2 is "yes" then we can use it $k-1$ times to see that the answer to Question 1 is "yes" too.


1 Answer 1


The answer to question 2, and therefore question 1, is "yes". We abbreviate $e_k(x_1, x_2, \ldots, x_{n-1})$ to $b_k$ and $e_k(x_1, x_2, \ldots, x_{n-1}, x_n)$ to $a_k$, so $a_k = x_n b_{k-1} + b_k$.

We are trying to show that, if $a_1$, $a_2$, ..., $a_k \geq 0$ then $b_1$, $b_2$, ..., $b_{k-1} \geq 0$. We prefer to prove the contrapositive: If $b_j<0$ then one of $a_1$, $a_2$, ..., $a_j$, $a_{j+1} <0$. We may assume that $j$ is minimal with $b_j<0$, so $b_1$, $b_2$, ..., $b_{j-1} \geq 0$.

Case 1: $x_n<0$. Then $a_j = x_n b_{j-1} + b_j < 0$.

Case 2: $x_n \geq 0$. First of all, if $b_{j-1} \leq 0$, then $a_j<0$ and, if $b_{j+1} \leq 0$, then $a_{j+1} < 0$. So we may assume that $b_{j-1}$ and $b_{j+1}>0$.

Now, Newton's inequalities give $$\frac{b_{j-1} b_{j+1}}{\binom{n-1}{j-1} \binom{n-1}{j+1}} \leq \frac{b_j^2}{\binom{n-1}{j}^2}$$ or $$b_{j-1} b_{j+1} \leq \frac{j(n-j-1)}{(j+1)(n-j)} b_j^2 < b_j^2.$$ So $$0<\frac{b_{j+1}}{- b_j} < \frac{-b_j}{b_{j-1}}.$$

We must either have $x_n > \tfrac{b_{j+1}}{- b_j}$ or $x_n < \tfrac{-b_j}{b_{j-1}}$. In the first case, $0 > x_n b_j + b_{j+1} = a_{j+1}$; in the second case, $a_j = x_n b_{j-1} + b_j < 0$. Either way, we have found a negative $a_j$.

  • $\begingroup$ This is very like a standard proof of Descartes' rule of signs, where one checks that multiplying by $f(t)$ by $t+a$ adds one sign alternation if $a<0$ and, if $a>0$ and all roots of $f$ are real, keeps the number of sign alternatations the same. I couldn't find this version of the proof quickly enough to link to though. $\endgroup$ Mar 7, 2022 at 15:17
  • $\begingroup$ The equality $a_j=x_n b_{j-1}+b_j$ does not seem to be correct. E.g., when $j=2$ and $x_1=\cdots=x_{n-1}=0$, then $a_j=0\ne x_n^2=x_n b_{j-1}+b_j$ if $x_n<0$. $\endgroup$ Mar 7, 2022 at 16:34
  • $\begingroup$ $a$'s and $b$'s switched from the first paragraph to the rest; try now. $\endgroup$ Mar 7, 2022 at 16:49
  • $\begingroup$ Why can't $x_n$ be equal to $\frac{b_{j+1}}{- b_j}$ or $\frac{-b_j}{b_{j-1}}$? $\endgroup$ Mar 7, 2022 at 17:09
  • $\begingroup$ Regarding my latter comment, it appears that the things can be fixed by a density-continuity argument, which allows one to assume that all the $a_j$'s and $b_j$'s are nonzero. $\endgroup$ Mar 7, 2022 at 18:37

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