# Newton's and Maclaurin's Inequalities

### Symmetric functions and averages

Symmetric (or elementary symmetric) polynomials arise in Viète's formulas for the roots of polynomial equations. The formulas relate the coefficients of the polynomial to its roots. An $n\text{-th}$ degree polynomial has $n$ roots and $n+1$ coefficients. A polynomial is said to be *monic* when its leading coefficient equals $1.$

For a set $X_n=\{x_1,x_2,\ldots,x_n\}$ of $n$ variables, denote by $X_{n}^{k}$ the set of all **products** of the $n$ variables taken $k$ at a time. There are $n\choose k$ such products: $\left|X_{n}^{k}\right|={n\choose k}.$

$\displaystyle\begin{align} e_0(x_1,x_2,\ldots ,x_n) &= 1,\\ e_1(x_1,x_2,\ldots ,x_n) &= \sum_{t\in X_{n}^{1}}t = x_1+x_2+\ldots+x_n,\\ e_2(x_1,x_2,\ldots ,x_n) &= \sum_{t\in X_{n}^{2}}t = x_1x_2+x_1x_3+\ldots+x_{n-1}x_n,\\ &\cdots \cdots\\ e_n(x_1,x_2,\ldots ,x_n) &= \sum_{t\in X_{n}^{n}}t = x_1x_2\cdots x_n, \end{align}$

and their averages as

$\displaystyle\begin{align} E_0(X) &= E_0(x_1,x_2,\ldots ,x_n) = \frac{1}{{n\choose 0}}e_0(x_1,x_2,\ldots ,x_n),\\ E_1(X) &= E_1(x_1,x_2,\ldots ,x_n) = \frac{1}{{n\choose 1}}e_1(x_1,x_2,\ldots ,x_n),\\ E_2(X) &= E_0(x_1,x_2,\ldots ,x_n) = \frac{1}{{n\choose 2}}e_0(x_1,x_2,\ldots ,x_n),\\ &\cdots \cdots\\ E_n(X) &= E_n(x_1,x_2,\ldots ,x_n) = \frac{1}{{n\choose n}}e_n(x_1,x_2,\ldots ,x_n). \end{align}$

Newton's and Maclaurin's inequalities deal with those averages, although they can be rewritten in terms of the symmetric functions themselves.

### Newton's and Maclaurin's Inequalities

To formulate both Newton's inequality and Maclaurin's inequality we assume that all $x_i,$ $i=1,\ldots,n$ are non-negative.

**Newton's inequality**

$\left(E_k(X)\right)^2\gt E_{k-1}(X)E_{k+1}(X),$ $1\le k\le n-1.$

**Maclaurin's inequality**

$E_1(X)\gt \left(E_{2}(X)\right)^\frac{1}{2}\gt \left(E_{3}(X)\right)^\frac{1}{3}\gt\ldots\gt \left(E_{n}(X)\right)^\frac{1}{n}.$

The inequalities become equalities only when all $x_i$ are equal.

Note that $E_1(X)\gt \left(E_{n}(X)\right)^\frac{1}{n}$ is none other than the better known AM-GM inequality.

### Newton's inequality implies that of Maclaurin

Indeed, multiply Newton's inequalities in a sequence, as below:

$\displaystyle (E_0E_2)(E_1E_3)^2(E_2E_3)^3\cdots (E_{k-1}E_{k+1})^k\lt (E_1)^2(E_2)^4\cdots (E_k)^{2k}$

which reduces to $(E_k)^{k+1}\gt (E_{k+1})^k,$ or $\displaystyle (E_k)^{\frac{1}{k}}\gt (E_{k+1})^\frac{1}{k+1}.$

### Preliminaries for a proof of Newton's inequality

We'll prove a somewhat more general statement:

Suppose that $\alpha,\beta\gt 0$ are real and $j,k\ge 0$ integers such that

$\alpha+\beta=1$ and $j\alpha+k\beta\in\{0,1,\ldots,n\}.$

Then

$E_{j\alpha+k\beta}(X)\ge(E_j(X))^{\alpha}(E_k(X))^{\beta},$

for every $n$-tuple $X$ of non-negative real numbers. Moreover, equality occurs if and only if all $x_i$ are equal.

The proof is by induction on the number of variables $n.$

According to Rolle's theorem, if all roots of a polynomial $P(x)$ are real, or real and distinct, then the same is true for its derivative $P'(x).$ This remark will be applied to the polynomial

$\displaystyle P_X(x)=\prod_{i=1}^{n}(x-x_i)=\sum_{k=0}^{n}(-1)^{k}e_{k}(x_1,x_2,\ldots,x_n)x^{n-k}.$

If $y_1,y_2,\ldots,y_{n-1}$ the roots of the derivative of $P_X(x)$ then the $(n-1)-\text{tuple}$ $X'=\{y_1,y_2,\ldots,y_{n-1}\}$ is denoted $X'$ and is said to be derived from the $n-\text{tuple}$ $X.$

Observe that

$\displaystyle\prod_{i=1}^{n-1}(x-y_i)=\sum_{k=0}^{n_1}(-1)^ke_{k}(y_1,\ldots,y_{n-1})x^{n-1-k}$

and also that

$\displaystyle\begin{align} \prod_{i=1}^{n-1}(x-y_i) &=\frac{1}{n}\frac{P_X(x)}{dx}\\ &= \sum_{k=0}^{n_1}(-1)^k\frac{n-k}{n}e_{k}(x_1,\ldots,x_{n})x^{n-1-k}. \end{align}$

By comparing the two expressions we conclude that $\displaystyle e_{k}(y_1,\ldots,y_{n-1})=\frac{n-k}{n}e_{k}(x1,\ldots,x_{n}).$ Which gives the following

**Lemma 1**

$E_k(X)=E_k(X'),$ for every $k\in\{0,1,\ldots,|X|-1\}.$

Another simple and useful result is the following

**Lemma 2**

Assume that $X$ is an $n-\text{tuple}$ of real numbers that does not contain $0.$ Define $\displaystyle X^{-1}=\{\frac{1}{a}:\;a\in X\}.$ Then

$\displaystyle E_k(X^{-1})=\frac{E_{n-k}(X)}{E_n(X)},$

for every $k\in\{0,1,\ldots,n\}.$

### The induction

Now, for $|X|=2,$ we have to prove just one inequality:

$\displaystyle x_1x_2\le\left(\frac{x_1+x_2}{2}\right)^2,$

which the *AM-GM inequality* for $n=2,$ with the equality only when $x_1=x_2.$

Assume that the assertion holds for all $k-\text{tuples},$ with $k\le n-1.$ Let $|X|=n\ge 3,$ all $x_i$ non-negative and let $j,k$ non-negative integers, while $\alpha,\beta$ positive real numbers such that

$\alpha+\beta=1$ and $j\alpha+k\beta\in\{0,1,\ldots,n\}.$

According to Lemma 1,

$E_{j\alpha+k\beta}(X)\ge\left(E_j(X)\right)^{\alpha}\cdot\left(E_k(X)\right)^{\beta},$

except possibly for the case where $j\lt k=n,$ or $k\lt j=n.$ Assume, for example, that $j\lt k=n;$ then necessarily $j\alpha+n\beta\lt n.$ We have to show that

$E_{j\alpha+n\beta}(X)\ge\left(E_j(X)\right)^{\alpha}\cdot\left(E_n(X)\right)^{\beta}.$

If $0\in X,$ then $E_n(X)=0,$ and the inequality is obvious. The equality occurs only if $E_{j\alpha+n\beta}(X')=E_{j\alpha+n\beta}(X)=0,$ i.e. (according to the induction hypothesis), when all entries of $X$ coincide.

If none of $x_i$ is $0,$ then, by Lemma 2, we have to prove that

$E_{n-j\alpha-n\beta}(X^{-1})\ge\left(E_{n-j}(X^{-1})\right)^{\alpha},$

or equivalently (due to Lemma 1),

$E_{n-j\alpha-n\beta}((X^{-1})')\ge\left(E_{n-j}((X^{-1})')\right)^{\alpha}.$

The latter is true by virtue of our induction hypothesis.

### References

Constantin P. Niculescu, A NEW LOOK AT NEWTON’S INEQUALITIES, Journal of Inequalities in Pure and Applied Mathematics, Volume 1, Issue 2, Article 17, 2000

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