# Newton's Inequality

## Contents

## Background

For , we define the symmetric sum to be the coefficient of in the polynomial (see Viete's sums). We define the *symmetric average* to be .

## Statement

For non-negative and ,

,

with equality exactly when all the are equal.

## Proof

**Lemma.**
For real , there exist real with the same symmetric averages .

*Proof.*
We consider the derivative of . The roots of are . Without loss of generality, we assume that the increase as increases. Now for any , must have a root between and by Rolle's theorem if , and if , then is a root of times, so it must be a root of times. It follows that must have non-positive, real roots, i.e., for some non-negative reals ,

.

It follows that the symmetric sum for is , so the symmetric average .

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

for any . Since this is a homogenous inequality, we may normalize it so that . The inequality then becomes

.

Expanding the left side, we see that this is

.

But this is clearly equivalent to

,

which holds by the rearrangement inequality.

*Proof: without calculus*

We will proceed by induction on .

For , the inequality just reduces to AM-GM inequality. Now suppose that for some positive integer the inequality holds.

Let , , , be non-negative numbers and be the symmetric averages of them. Let be the symmetric averages of , , . Note that .

By induction this completes the proof.