Newton's Inequality: Difference between revisions

Revision as of 07:40, 22 January 2018

Background

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

Statement

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

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

with equality exactly when all the $x_i$ 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 $P(t) = \prod_{i=1}^m (t+x_i)$ . The roots of $P$ are $-x_1, \ldots, -x_m$ . Without loss of generality, we assume that the $x_i$ increase as $i$ increases. Now for any $i \in \{1, m-1\}$ , $P'(t)$ must have a root between $x_i$ and $x_{i+1}$ by Rolle's theorem if $x_i \neq x_{i+1}$ , and if $x_i = x_{i+1} = \cdots = x_{i+k}$ , then $x_{i}$ is a root of $P$ $k+1$ times, so it must be a root of $P'$ $k$ times. It follows that $P'$ must have $m-1$ 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 $s'_k$ for $x'_1, \ldots, x'_{m-1}$ is $\frac{m-k}{m}s_k$ , so the symmetric average $d'_k = \frac{s'_k}{{m-1 \choose k}} = \frac{m-k}{m} \cdot \frac{s_k}{{m-1 \choose k}} = \frac{s_k}{{m \choose k}} = d_k$ .

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

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

for any $n$ . 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

$(n-1)\left(\sum_{i=1}^{n}\frac{1}{x_i} \right)^2 \ge 2n \sum_{0\le i<j \le n} \frac{1}{x_ix_j}$ .

Expanding the left side, we see that this is

$(n-1)\sum_{i=1}^{n}\frac{1}{x_i^2} + (n-1)\sum_{0\le i<j \le n}\frac{2}{x_ix_j} \ge 2n \sum_{0\le i<j \le n} \frac{1}{x_ix_j}$ .

But this is clearly equivalent to

$\sum_{\rm sym}\frac{1}{x_1^2} \ge \sum_{\rm sym}\frac{1}{x_1x_2}$ ,

which holds by the rearrangement inequality.

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

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Newton's Inequality: Difference between revisions

Revision as of 07:40, 22 January 2018

Contents

Background

Statement

Proof

See Also