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](https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial).



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> **Question 1:** Is it true that $x\_1 + x\_2 + \cdots + x\_{n-k+1} \geq 0$?
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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:



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> **Question 2:** Is it true that $e\_j(x\_1,x\_2,\ldots,x\_{n-1}) \geq 0$ for all $1 \leq j \leq k-1$?
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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.



Dimension reduction for non-negativity of elementary symmetric polynomials
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](https://en.wikipedia.org/wiki/Newton%27s_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 $a\_j$.


