We present a generalization of Newton’s inequality, i.e., an inequality of mixed form connecting symmetric functions and weighted means

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Volume 10 (2009), Issue 2, Article 44, 4 pp.

A NOTE ON NEWTON’S INEQUALITY

SLAVKO SIMIC

MATHEMATICALINSTITUTESANU, KNEZAMIHAILA36 11000 BELGRADE, SERBIA.

Received 25 March, 2009; accepted 12 April, 2009 Communicated by S.S. Dragomir

ABSTRACT. We present a generalization of Newton’s inequality, i.e., an inequality of mixed form connecting symmetric functions and weighted means. Two open problems are also stated.

Key words and phrases: Symmetric functions, Weighted means.

2000 Mathematics Subject Classification. 26E60.

1. INTRODUCTION

A well-known theorem of Newton [1] states the following:

Theorem 1.1. If all the zeros of a polynomial

(1.1) P_n(x) = e₀xⁿ+e₁xⁿ⁻¹+· · ·+e_kx^n−k+· · ·+e_n, e₀ = 1, are real, then its coefficients satisfy

(1.2) ek−1e_k+1 ≤A⁽ⁿ⁾_k e²_k, k= 1,2, . . . , n−1;

whereA⁽ⁿ⁾_k := _k+1^k _n+1−k^n−k .

For a sequencea={a_i}ⁿ_i=1of real numbers, by putting

(1.3) P_n(x) =

n

Y

i=1

(x+a_i) =

n

X

k=0

e_kx^n−k,

we see that the coefficiente_k = e_k(a) represents thekth elementary symmetric function ofa, i.e. the sum of all the products,kat a time, of differenta_i ∈a.

There are several generalizations of Newton’s inequality [2], [3]. In this article we give another one. For this purpose define the sequences a⁰_i := a/{a_i}, i = 1,2, . . . , n, and by e_k(a⁰_i)denote thek-th elementary symmetric function overa⁰_i. We have:

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2 SLAVKOSIMIC

Theorem 1.2. Letc={c_i}ⁿ_i=1 be a weight sequence of non-negative numbers satisfying

(1.4)

n

X

i=1

c_i = 1,

and, for an arbitrary sequencea={ai}ⁿ_i=1of real numbers, define

(1.5) E_k^(c):=

n

X

i=1

ciek(a⁰_i), E₀^(c) = 1,

or equivalently,

(1.6) E_k^(c) =e_k−e_k−1f₁+e_k−2f₂²− · · ·+ (−1)^re_k−rf_r^r+· · ·+ (−1)^kf_k^k, where

(1.7) f_s :=

n

X

i=1

c_ia^s_i

!¹_s .

Then

(1.8) E_k−1^(c) E_k+1^(c) ≤A⁽ⁿ⁻¹⁾_k E_k^(c)2

, k = 1,2, . . . , n−1.

Proof. We shall give an easy proof supposing that the sequencecconsists of arbitrary positive rational numbers. Since aand care independent of each other, the truthfulness of the above theorem follows by the continuity principle.

Therefore, letp={p_k}ⁿ_k=1 be an arbitrary sequence of positive integers and put

(1.9) c_i = p_i

Pn

1 p_k, i= 1,2, . . . , n; p∈N.

Now, for a given real sequencea, consider the polynomialQ(x)defined by

(1.10) Q(x) :=

n

Y

i=1

(x+a_i)^pⁱ.

Since all its zeros are real, by the well-known Gauss theorem, the zeros ofQ⁰(x),

(1.11) Q⁰(x) = Q(x)

n

X

i=1

p_i x+a_i,

are also real.

In particular, the same is valid for the polynomialR(x)defined by

(1.12) R(x) :=

n

Y

i=1

(x+a_i)

n

X

i=1

c_i (x+a_i).

Since

(1.13) R(x) = xⁿ⁻¹+E₁^(c)xⁿ⁻²+· · ·+E_k^(c)x^n−1−k+· · ·+E_n−1^(c) , the result follows by simple application of Theorem 1.1.

Remark 1. SincePn

i=1e_k(a⁰_i) = (n−k)e_k(a), puttingc_i = 1/n, i = 1,2, . . . , nin (1.5) and (1.8), we obtain the assertion from Theorem 1.1. Hence our result represents a generalization of Newton’s theorem.

J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 44, 4 pp. http://jipam.vu.edu.au/

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NEWTON’SINEQUALITY 3

Also, denoting byfs^(c)(a) = f_s := (Pn

i=1c_ia^s_i)^1/s, s >0, the classical weighted mean (with weightsc) of orders, and using the identity

(1.14) e_k(a⁰_i) = e_k(a)−a_iek−1(a⁰_i), an equivalent form ofE_k^(c)arises, i.e.,

(1.15) E_k^(c) =e_k−ek−1f₁+ek−2f₂²− · · ·+ (−1)^rek−rf_r^r+· · ·+ (−1)^kf_k^k.

Putting this in (1.8), we obtain a mixed inequality connecting elementary symmetric func-

tions with weighted means of integer order.

Problem 1.1. An interesting fact is that non-negativity of c is not a necessary condition for (1.8) to hold. We shall illustrate this point by an example. Fork = 1, n= 3, we have

(E₁^(c))²−4E₀^(c)E₂^(c)

= (c₁(a₂+a₃) +c₂(a₁+a₃) +c₃(a₁+a₂))²−4(c₁a₂a₃+c₂a₁a₃ +c₃a₁a₂)

= (1−c₂)²(a₁−a₂)² + 2(c₁−c₂c₃)(a₁−a₂)(a₃−a₁) + (1−c₃)²(a₃−a₁)², and this quadratic form is positive semi-definite wheneverc₁c₂c₃ ≥0.

Hence, in this case the inequality (1.8) is valid for all real sequencesawithcsatisfying (1.16) c₁+c₂+c₃ = 1, c₁c₂c₃ ≥0.

Therefore there remains the seemingly difficult problem of finding true bounds for the sequence c satisfying (1.4), such that the inequality (1.8) holds for an arbitrary real sequence a.

Problem 1.2. There is an interesting application of Theorem 1.2 to the well known Turan’s problem. Under what conditions does the sequence of polynomials {Q_n(x)} satisfy Turan’s inequality

(1.17) Qn−1(x)Qn+1(x)≤(Qn(x))², for eachx∈[a, b]andn∈[n₁, n₂]?

This problem is solved for many classes of polynomials [4]. We shall consider here the following question [5].

An arbitrary sequence{d_i}, i= 1,2, . . . of real numbers generates a sequence of polynomials {P_n(x)}, n= 0,1,2, . . . defined by

(1.18) P_n(x) := xⁿ+d₁xⁿ⁻¹+d₂xⁿ⁻²+· · ·+d_n−1x+d_n, P₀(x) := 1.

Denote also byA_nthe set of zeros ofP_n(x).

Now, if for somem >1the setA_m consists of real numbers only, then from Theorem 1.2, it follows that

(1.19) Pn−1(a)P_n+1(a)≤P_n²(a),

for eacha∈A_mandn∈[1, m−1].

Is it possible to establish some simple conditions such that the Turan inequality

(1.20) Pn−1(x)P_n+1(x)≤P_n²(x),

holds for eachx∈[mina,maxa]a∈Amandn∈[1, m−1].

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4 SLAVKOSIMIC

REFERENCES

[1] G.H. HARDY, J.E. LITTLEWOOD ANDG. POLYA, Inequalities, Camb. Univ. Press, Cambridge, 1978.

[2] J.N. WHITELEY, A generalization of a theorem of Newton, Proc. Amer. Math. Soc., 13 (1962), 144–151.

[3] K.V. MENON, Inequalities for symmetric functions, Duke Math. J., 35 (1968), 37–45.

[4] S. SIMIC, Turan’s inequality for Appell polynomials, J. Ineq. Appl., 1 (2006), 1–7.

[5] S. SIMIC, A log-concave sequence, Siam Problems and Solutions, [ONLINE: http//:www.

siam.org].