Means Moments and Newton's Inequalities

TL;DR

This paper explores how Newton's and Maclaurin's inequalities refine classical mean inequalities, providing new bounds involving means, variance, and higher moments of positive real numbers.

Contribution

It introduces new inequalities involving third and fourth central moments, extending classical mean inequalities with refined bounds.

Findings

Newton's inequalities refine AM-GM-HM inequalities.

New inequalities involving third and fourth moments are derived.

Results provide tighter bounds on means and moments.

Abstract

It is shown that Newton's inequalities and the related Maclaurin's inequalities provide several refinements of the fundamental Arithmetic mean - Geometric mean - Harmonic mean inequality in terms of the means and variance of positive real numbers. We also obtain some inequalities involving third and fourth central moments of real numbers.

Equations

$$m_r^{\prime }=\frac{1}{n}\stackrel{n}{\sum _{i=1}}{x}_i^r$$

$$m_r=\frac{1}{n}\stackrel{n}{\sum _{i=1}}{\left(x_i-m_1^{\prime }\right)}^r,$$

$$G={\left(\stackrel{n}{\prod _{i=1}}{x}_i\right)}^{\frac{1}{n}}\text{ and }H={\left(\frac{1}{n}\underset{i=1}{\sum ^n}\frac{1}{x_i}\right)}^{-1}.$$

$$S_k=\frac{1}{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}{C}_k.$$

$$S_k^2\ge S_{k+1}{S}_{k-1}\text{ ,}$$

$$S_1\ge S_2^{\frac{1}{2}}\ge \dots \ge S_n^{\frac{1}{n}}.$$

$${\alpha }_k=\stackrel{n}{\sum _{i=1}}{x}_i^k\text{ ,}$$

$$k{C}_k=\underset{i=1}{\sum ^k}{\left(-1\right)}^{i-1}{C}_{k-i}{\alpha }_i.$$

$$H\le H{\left(\frac{A}{H}\right)}^{\frac{1}{n}}\le G\le {\left(\frac{H}{A}\right)}^{\frac{1}{n}}A\le A.$$

$${\left(\frac{G}{H}\right)}^{\frac{1}{n-1}}G\le A\le {\left(\frac{G}{H}\right)}^{n-1}G.$$

$${\left(\frac{G}{A}\right)}^{n-1}G\le H\le {\left(\frac{G}{A}\right)}^{\frac{1}{n-1}}G.$$

$$G\le \sqrt{\frac{n{A}^2-m_2^{\prime }}{n-1}}\le A.$$

$$\frac{2}{n\left(n-1\right)}\stackrel{n}{\sum _{i<j}}{x}_i{x}_j\ge G^2,$$

$$2\underset{i<j}{\sum ^n}{x}_i{x}_j={\left(\underset{i=1}{\sum ^n}{x}_i\right)}^2-\stackrel{n}{\sum _{i=1}}{x}_i^2=n\left(n{A}^2-m_2^{\prime }\right).$$

$$G\le A\sqrt{1-\frac{1}{n-1}{\left(\frac{s}{A}\right)}^2}$$

$$m_2^{\prime }\le n{A}^2-\left(n-1\right)G^2$$

$$s^2\le \left(n-1\right)\left(A^2-G^2\right).$$

$$G\le {\left(\frac{G^n}{H}\right)}^{\frac{1}{n-1}}\le \sqrt{\frac{n{A}^2-m_2^{\prime }}{n-1}}\le A.$$

$$\frac{2}{n\left(n-1\right)}\stackrel{n}{\sum _{i<j}}{x}_i{x}_j\ge {\left(\frac{C_{n-1}}{n}\right)}^{\frac{2}{n-1}},$$

$$C_{n-1}=\stackrel{n}{\underset{i=1}{\Pi }}{x}_i\underset{i=1}{\sum ^n}\frac{1}{x_i}=n\frac{G^n}{H}.$$

$$G\le {\left(\frac{G^n}{H}\right)}^{\frac{1}{n-1}}\le A\sqrt{1-\frac{1}{n-1}{\left(\frac{s}{A}\right)}^2}$$

$$m_2^{\prime }\le n{A}^2-\left(n-1\right){\left(\frac{G^n}{H}\right)}^{\frac{2}{n-1}}.$$

$$s^2\le \left(n-1\right)\left(A^2-{\left(\frac{G^n}{H}\right)}^{\frac{2}{n-1}}\right).$$

$$A{m}_3^{\prime }\le \frac{1}{2\left(n-1\right)}\left[\left(n-2\right)m_2^{\prime 2}+n\left(n+1\right)m_2^{\prime }{A}^2\right].$$

$${\left(\frac{2}{n\left(n-1\right)}\underset{i<j}{\sum ^n}{x}_i{x}_j\right)}^2\ge \frac{6A}{n\left(n-1\right)\left(n-2\right)}\underset{i<j<k}{\sum ^n}{x}_i{x}_j{x}_k.$$

$$3\underset{i<j<k}{\sum ^n}{x}_i{x}_j{x}_k=\underset{i=1}{\sum ^n}{x}_i^3+\frac{1}{2}{\left(\underset{i=1}{\sum ^n}{x}_i\right)}^3-\frac{3}{2}\underset{i=1}{\sum ^n}{x}_i\underset{i=1}{\sum ^n}{x}_i^2$$

$$=n\left(m_3^{\prime }+\frac{n^2}{2}{A}^3-\frac{3}{2}nA{m}_2^{\prime }\right).$$

$$A{m}_3\le \frac{n-2}{2\left(n-1\right)}{s}^4+\frac{\left(n-2\right)}{2}{A}^2{s}^2$$

$$3\underset{i<j<k}{\sum ^n}{x}_i{x}_j{x}_k=n\left(m_3-\frac{3\left(n-2\right)}{2}A{s}^2+\frac{\left(n-1\right)\left(n-2\right)}{2}{A}^3\right)$$

$$m_3\le \frac{n-2}{2\left(n-1\right)}\frac{s^4}{A}+\frac{\left(n-2\right)}{2}A{s}^2$$

Full text

**Means Moments and Newton’s Inequalities **

R. Sharma, A. Sharma, R. Saini and G. Kapoor

Department of Mathematics & Statistics

Himachal Pradesh University

Shimla - 171005,

India

email: rajesh.sharma.hpn@nic.in

**Abstract. **It is shown that Newton’s inequalities and the related Maclaurin’s inequalities provide several refinements of the fundamental Arithmetic mean - Geometric mean - Harmonic mean inequality in terms of the means and variance of positive real numbers. We also obtain some inequalities involving third and fourth central moments of real numbers.

**AMS classification.  **60E15

Key words and phrases.  Arithmetic mean, Geometric mean, Moments, Newton’s identities.

1

Introduction

Let $x_{1},x_{2},...,x_{n}$ denote $n$ real numbers. Their rth moment and rth central moment are respectively the numbers

[TABLE]

and

[TABLE]

where $r=1,2,\ldots$     . Note that $m_{1}=0,$ $m_{1}^{\prime}$ is the arithmetic mean and $m_{2}$ is the variance of $n$ real numbers $x_{1},x_{2},...,x_{n}.$ It is customary to denote $m_{1}^{\prime}$ and $m_{2}$ respectively by $A$ and $s^{2}$. The Geometric mean $\left(G\right)$ and Harmonic mean $\left(H\right)$ of $n$ positive real numbers $x_{1},x_{2},...,x_{n}$ are respectively the numbers

[TABLE]

The $k^{\text{th}}$ elementary symmetric function $C_{k\text{ }}$ of $x_{1},x_{2},...,x_{n}$ is the  sum of the products taken $k$ at a time of different $x_{i}$’s. The $k^{\text{th}}$ elementary symmetric mean is the number

[TABLE]

We write $S_{0}=C_{0}=1.$

A well known theorem of Newton (1707) says that elementary symmetric means satisfy the inequality

[TABLE]

for all $k=1,2,\ldots,n,$  and equality holds if and only if the numbers $x_{i}$’s are all equal.

A related  theorem due to Maclaurin (1729) asserts that the elementary symmetric means of $n$ positive real numbers satisfy the following inequalities string:

[TABLE]

Equality occurs if and only if $x_{i}$’s are all equal. One can easily see that the inequalities (1.5) also follow from the inequalities (1.4). For more detail, see Hardy et al (1952).

Newton’s identities give the relations between the elementary symmetric functions and the sum of the powers of $x_{1},x_{2},...,x_{n}.$ Let

[TABLE]

and write $\alpha_{0}=1.$ Then Newton’s identities say that

[TABLE]

The well known Arithmetic mean - Geometric mean - Harmonic mean inequality (A-G-H inequality) says that $H\leq G\leq A$ . Alternative proofs, refinements and generalizations of A-G-H inequality have been studied extensively in literature over roughly the last two centuries. See Bullen (2003). Note that $S_{1}=A$ and $S_{n}^{\frac{1}{n}}=G.$ The Maclaurin inequalities (1.5) therefore provide the refinements of the A-G inequality. These refinements are in terms of the symmetric means $S_{k}.$ On using Newton’s identity we can find the relation between symmetric means and moments of the $n$ real numbers. Then, the Maclaurin inequalities can be used to find the refinements of the A-G inequality in terms of the expressions involving means and moments. The purpose of the present note is to point out some such inequalities which remains unnoticed in the literature in the present explicit forms.

Our first theorem gives a refinement of the A-G-H inequality in terms  of expressions involving arithmetic mean and harmonic mean (see Theorem 2.1, below). A refinement of the A-G inequality involving second moment is given. This also provides an upper bound for the variance, and Brunk inequalities (Theorem 2.2). A further refinement of this inequality involves harmonic mean (Theorem 2.3). We derive bounds for the third and fourth moments and central moments of statistical interest (Theorem 2.4 and 2.7). The refinements of A-G inequality in terms of first four moments are also given (Theorem 2.5, 2.6 and 2.8).

2 Main results

**Theorem 2.1. **For $n$ positive real numbers $x_{1},x_{2},...,x_{n}$ and with notations as above, we have

[TABLE]

**Proof. **The third inequality (2.1) follows from the inequality $S_{1}\geq S_{n-1}^{\frac{1}{n-1}}$ in (1.5) and the fact that from (1.3) we have $S_{1}=A$ and $S_{n-1}H=G^{n}.$ The second inequality (2.1) follows on applying third inequality (2.1) to the numbers $\frac{1}{x_{1}},\frac{1}{x_{2}},\ldots,\frac{1}{x_{n}}.$ The extreme inequalities are immediate. $\blacksquare$

It may be noted here that from (2.1) we also have

[TABLE]

and

[TABLE]

**Theorem 2.2. **For $n$ positive real numbers $x_{1},x_{2},...,x_{n}$ and with notations as above, we have

[TABLE]

**Proof. **It follows from the inequality $S_{2}\geq\left(S_{n}\right)^{\frac{2}{n}}$ in (1.5) that for $n$ positive real numbers $x_{1},x_{2},...,x_{n}$ the inequality

[TABLE]

holds true. From (1.6), $2C_{2}=C_{1}\alpha_{1}-C_{0}\alpha_{2}$ where $C_{0}=1,$ $C_{1}=\alpha_{1}=\sum x_{i}$ , $\alpha_{2}=\sum x_{i}^{2}$ and $C_{2}=\underset{i<j}{\sum}x_{i}x_{j}$** .** Therefore, on using (1.1), we get that

[TABLE]

Insert (2.4) in (2.3); a little computation leads to first inequality (2.2). The second inequality (2.2) follows from the fact that  $m_{2}^{\prime}\geq A^{2}.\blacksquare$

The inequality (2.1) can be written in several different ways, for example the inequality

[TABLE]

follows from (2.1) on using $m_{2}^{\prime}=s^{2}+A^{2}.$ We also have

[TABLE]

and

[TABLE]

We note that Brunk’s inequalities (1959) follow from the inequality (2.6). Let $a\leq x_{i}\leq b$ for all $i=1,2,\ldots,n.$ From (2.6), $s^{2}\leq\left(n-1\right)A^{2}$. Apply this to $n$ positive numbers $x_{i}-a$, we get Brunk’s inequality, $s\leq\sqrt{n-1}\left(A-a\right)$. Likewise, the inequality $s\leq\sqrt{n-1}\left(b-A\right)$ follows from (2.6). For the related variance upper bounds, see Bhatia and Davis (2000), Sharma (2008) and Sharma et al (2010).

**Theorem 2.3. **With conditions as in above theorem and for $n\geq 3$,  we have

[TABLE]

**Proof. **From (1.5) , $S_{2}\geq\left(S_{n-1}\right)^{\frac{2}{n-1}}.$ Therefore, for $n$ positive real numbers $x_{1},x_{2},...,x_{n}$ the inequality

[TABLE]

holds true. Also

[TABLE]

Insert (2.4) and  (2.9) in (2.8); a little computation leads to the second inequality (2.7)  $\blacksquare$

From (2.7) , we also have

[TABLE]

[TABLE]

and

[TABLE]

The inequalities (2.10) and (2.11) provide refinements of the inequalities (2.5) and (2.6) , respectively.

Theorem 2.4. For $n$ real numbers $x_{1},x_{2},...,x_{n}$ and $n\geq 3$, we have

[TABLE]

Proof. For $k=2,$ the inequality (1.4) gives $S_{2}^{2}\geq S_{1}S_{3}.$ Therefore, for $n$ real numbers $x_{1},x_{2},...,x_{n},$ we have

[TABLE]

Further, from (1.6), $3C_{3}=C_{2}\alpha_{1}-C_{1}\alpha_{2}+\alpha_{3}$ where $\alpha_{3}=\sum x_{i}^{3}$ and $C_{3}=\underset{i<j<l}{\sum}x_{i}x_{j}x_{l}$ . Therefore, on using (2.3) and (1.1), we find that

[TABLE]

[TABLE]

On inserting (2.4) and (2.14) in (2.13) and simplifying the resulting expressions; we immediately get (2.12).  $\blacksquare$

The twin inequality

[TABLE]

follows from (2.12) on using the relations $m_{2}^{\prime}=s^{2}+A^{2}$ and $m_{3}^{\prime}=m_{3}+3As^{2}+A^{3}$ .It may be noted here that

[TABLE]

and the inequality

[TABLE]

holds true for positive real numbers $x_{1},x_{2},...,x_{n}.$

**Corollary 2.1. **Under the conditions of Theorem 2.2, we have

[TABLE]

**Proof. **The inequality (2.16) follows from (2.15) and the fact that for positive real numbers $x_{1},x_{2},...,x_{n},$ we have $A>0$ and $s\leq\sqrt{n-1}$ $A.$  $\blacksquare$

Note that the second inequality (2.16) implies that for $n$ real numbers $a\leq x_{i}\leq b$, $i=1,2,\ldots,n$,  we have $m_{3}\leq\left(n-1\right)\left(n-2\right)\left(b-A\right)^{3}$ and $m_{3}\leq\left(n-1\right)\left(n-2\right)\left(A-a\right)^{3}.$ Thus, we get inequalities analogous to Brunk’s inequalities (1959) for third central moment. Likewise, from the first inequality (2.16), we have $m_{3}\leq\left(n-2\right)\left(b-A\right)s^{2}$ and $m_{3}\leq\left(n-2\right)\left(A-a\right)s^{2}$. Also, see Sharma et al (2012).

**Theorem 2.5. **With conditions as in Theorem 2.3 ,  we have

[TABLE]

Proof. It follows from the inequality $S_{3}\geq S_{n}^{\frac{3}{n}}$ in (1.5) that for $n$ positive real numbers $x_{1},x_{2},...,x_{n},$ we have

[TABLE]

Insert (2.14) in (2.18); a little computation leads to (2.17).  $\blacksquare$

From (2.17), we also have

[TABLE]

[TABLE]

and

[TABLE]

**Theorem 2.6.**With conditions as in above theorem and for $n\geq 4$,  we have

[TABLE]

Proof. It follows from inequality $S_{3}\geq S_{n-1}^{\frac{3}{n-1}}$ in (1.5) that for $n$ positive real numbers $x_{1},x_{2},...,x_{n},$ we have

[TABLE]

Insert (2.14) in (2.20); a little computation leads to (2.19).  $\blacksquare$

From (2.19), we also have

[TABLE]

[TABLE]

and

[TABLE]

Recently , Sharma and Bhandari (2015) have shown that the Newton inequality provides an inequality involving skewness and kurtosis,

[TABLE]

Sharma et al (2015) have obtained bounds for the fourth central moment. Here we show that Newton inequality provide several other inequalities involving first four moments.

Theorem 2.7. For $n$ real numbers $x_{1},x_{2},...,x_{n}$ and $n\geq 4$, we have

[TABLE]

**Proof. The inequality (2.21) follows from the inequality $S_{3}^{2}\geq S_{2}S_{4}$ in (1.4). Note that from (1.6), $4C_{4}=C_{3}\alpha_{1}-C_{2}\alpha_{2}+C_{1}\alpha_{3}-C_{0}\alpha_{4}$. **Therefore, on using (2.3), (2.14) and (1.1), we get that

[TABLE]

$\blacksquare$

Theorem 2.8. For $n$ positive real numbers $x_{1},x_{2},...,x_{n}$  and $n\geq 4$, we have

[TABLE]

**Proof. **From (1.5), $S_{4}\geq\left(S_{n}\right)^{\frac{4}{n}}.$ Therefore, for $n$ positive real numbers $x_{1},x_{2},...,x_{n}$ the inequality

[TABLE]

holds true.

$\blacksquare$

The inequality (2.22) also gives

[TABLE]

Acknowledgements. The authors are grateful to Prof. Rajendra Bhatia for the useful discussions and suggestions, and I.S.I. Delhi for a visit in February 2017 when this work was done. The support of the UGC-SAP is also acknowledged.