Newton's Inequality

Background

For $x_1, \ldots, x_n$ , we define the symmetric sum $\displaystyle s_k$ to be the coefficient of $\displaystyle t^{n-k}$ in the polynomial $\prod_{i=1}^{n}(t+x_i)$ (see Viete's sums). We define the symmetric average $\displaystyle d_k$ to be $\textstyle s_k/{n \choose k}$ .

Statement

For non-negative $x_1, \ldots, x_n$ and $\displaystyle 0 < k < n$ ,

$\displaystyle d_k^2 \ge d_{k-1}d_{k+1}$ ,

with equality exactly when all the $\displaystyle x_i$ are equal.

Proof

Lemma. For real $x_1 , \ldots, x_m$ , there exist real $x'_1, \ldots, x'_{m-1}$ with the same symmetric averages $d_0, \ldots, d_{m-1}$ .

Proof. We consider the derivative of $P(t) = \prod_{i=1}^n (t+x_i)$ . The roots of $\displaystyle P$ are $-x_1, \ldots, -x_n$ . Without loss of generality, we assume that the $\displaystyle x_i$ increase as $\displaystyle i$ increases. Now for any $\displaystyle i \in [1, m-1]$ , $\displaystyle P'(t)$ must have a root between $\displaystyle x_i$ and $\displaystyle x_{i+1}$ by Rolle's theorem if $\displaystyle x_i \neq x_{i+1}$ , and if $x_i = x_{i+1} = \cdots = x_{i+k}$ , then $\displaystyle x_{i}$ is a root of $\displaystyle P$ $\displaystyle k+1$ times, so it must be a root of $\displaystyle P'$ $\displaystyle k$ times. It follows that $\displaystyle P'$ must have $\displaystyle m-1$ non-positive, real roots, i.e., for some non-negative reals $x'_1, \ldots, x'_{m-1}$ ,

${} P'(t) = m \prod_{i=1}^{m-1}(t+x'_i)$ .

It follows that the symmetric sum $\displaystyle s'_k$ for $x'_1, \ldots, x'_{m-1}$ is $\frac{m-k}{m}s_k$ , so the symmetric average $\displaystyle d'_k = \frac{s'_k}{{m-1 \choose k}} = \frac{m-k}{m} \cdot \frac{s_k}{{m-1 \choose k}} = \frac{s_k}{{m \choose k}} = d_k$ .

Thus to prove Newton's theorem, it is sufficient to prove

$\displaystyle d_{n-1}^2 \ge d_{n-2}d_n$

for any $\displaystyle n$ . Since this is a homogenous inequality, we may normalize it so that $d_n = \prod_{i=1}^{n}x_i = 1$ . The inequality then becomes