TITLE: Inequalities on elementary symmetric polynomials
QUESTION [14 upvotes]: 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.

REPLY [5 votes]: 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.