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I have recently come across the following result.

Let $0 < d \leq n$. Given any vector $x \in \mathbb{R}^n$ that satisfies $e_{d-1}(x) = 0$, show that $$|x_1 \cdots x_d| \leq |e_d(x)|$$ where $e_k$ is the $k$-th elementary symmetric polynomial.

I think the inequality is tight exactly when $x_{d+1} = \cdots = x_{n} = 0$. I believe this result is true, but I am having problems making progress. Any references would be appreciated.

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  • $\begingroup$ The case $d=n$ is trivial. The case $d=2$ is easy. I have also checked (with Mathematica) the cases when $d=3$ and $n\in\{4,5\}$ -- by squaring both sides of the inequality and replacing $x_n$ by the equivalent expression $-e_{d-1}(y)/e_{d-2}(y)$ (given $e_{d-1}(x)=0$), where $y:=(x_1,\dots,x_{n-1})$. The general case is a nice mystery! $\endgroup$
    – Iosif Pinelis
    Commented Nov 2, 2017 at 3:30
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    $\begingroup$ We may substitute $x_n$ via other variables and get a condition-free equivalent inequality $|x_1\dots x_d\sigma_{d-2}|\leqslant \sigma_{d-1}^2-\sigma_d\sigma_{d-2}$, where $\sigma$'s now denote elementary symmetric polynomials for $x_1,\dots,x_{n-1}$. This is true for $n-1=d$ (for $y_i=1/x_i$ this means just $\sigma_1^2-\sigma_2\geqslant |\sigma|$ that is obvious.) $\endgroup$
    – Fedor Petrov
    Commented Nov 2, 2017 at 6:58

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It looks false to me and the counterexample is the second one that comes to one's head: three $1$'s and plenty (say, $m$) of $-x$ where $x$ is a small number.

$e_2=3-3mx+\frac{m(m-1)}{2}x^2=0$ means that $x=y/m+o(1/m)$ where $y$ is the root of $3-3y+y^2/2=0$, i.e., $y=3-\sqrt 3$.

Then

$e_3=1-3mx+3\frac{m(m-1)}{2}x^2-\frac{m(m-1)(m-2)}{6}x^3 \\ \approx 1-3y+3y^2/2-y^3/6\approx -0.73$,

which is smaller than $1\cdot 1\cdot 1=1$ in absolute value.

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    $\begingroup$ One case: For $m,x=6,\frac15$ one has $e_2=0$ and $e_3=-0.96$ which is smaller in absolute value than $1\cdot 1 \cdot 1$ $\endgroup$
    – Aaron Meyerowitz
    Commented Nov 3, 2017 at 5:50

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