14
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Nov 2 '17 at 3:30
  • 1
    $\begingroup$ 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 Nov 2 '17 at 6:58
4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 Nov 3 '17 at 5:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.