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volume 5, issue 4, article 108, 2004.

Received 23 May, 2003;

accepted 09 October, 2004.

Communicated by:F. Qi

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Journal of Inequalities in Pure and Applied Mathematics

A GENERALIZATION OF AN INEQUALITY INVOLVING THE GENERALIZED ELEMENTARY SYMMETRIC MEAN

ZHI-HUA ZHANG AND ZHEN-GANG XIAO

Zixing Educational Research Section Chenzhou, Hunan 423400, China.

EMail:zxzh1234@163.com

Hunan Institute of Science and Technology Yueyang, Hunan 414006, China.

EMail:xiaozg@163.com

c

2000Victoria University ISSN (electronic): 1443-5756 070-03

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A Generalization of an Inequality Involving the Generalized Elementary

Symmetric Mean

Zhi-Hua Zhang and Zhen-Gang Xiao

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J. Ineq. Pure and Appl. Math. 5(4) Art. 108, 2004

Abstract

A generalization of an inequality involving the generalized elementary symmetric mean and its elementary proof are given.

2000 Mathematics Subject Classification:Primary 26D15 Key words: Generalized elementary symmetric mean, Inequality.

The authors would like to thank Professor Feng Qi and the anonymous referee for some valuable suggestions which have improved the final version of this paper.

1. Introduction

Leta = (a₁, a₂,· · · , a_n)andrbe a nonnegative integer, wherea_ifor1≤i≤n are nonnegative real numbers. Then

(1.1) E_n^[r]=E_n^[r](a) = X

i1+i2+···+in=r, i1,i2,···,in≥0are integers

n

Y

k=1

aⁱ_k^k

withEn^[0] =En^[0](a) = 1forn≥1andEn^[r] = 0forr <0orn ≤0is called the rth generalized elementary symmetric function ofa.

Therth generalized elementary symmetric mean ofais defined by (1.2)

[r]

X

n

=

[r]

X

n

(a) = En^[r](a)

n+r−1 r

.

In 1934, I. Schur [5, p. 182] obtained the following (1.3)

[r]

X

n

(a) = (n−1)!

Z

· · ·

Z ⁿ X

i=1

a_ix_i

!r

dx₁· · ·dxn−1,

wherex_n = 1−(x₁+x₂ +· · ·+xn−1)and the integral is taken overx_k ≥ 0 fork = 1,2, . . . , n−1. By using (1.3) and Cauchy integral inequality, he also proved that

(1.4)

[r−1]

X

n

(a)

[r+1]

X

n

(a)≥





[r]

X

n

(a)





2

.

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In 1968, K.V. Menon [2] proved that forn = 2 orn ≥ 3 andr = 1,2,3, inequality (1.4) is valid.

In [1], the generalized symmetric means of two variables was investigated.

In [3] and [4] a problem was posed: Does inequality (1.4) hold for arbitrary n, r ∈N?

In 1997, Zh.-H. Zhang generalized (1.3) in [7] and also proved (1.4) by a similar proof as in [5].

In [6], some inequalities of weighted symmetric mean were established.

In this paper, we shall obtain an identity relatingP[r]

n (a)toEn^[r](a)and give an elementary proof of an inequality which generalizes (1.4). Our main result is as follows.

Theorem 1.1. Ifr, s∈Nandr > s, then

(1.5)

[s]

X

n

(a)

[r+1]

X

n

(a)≥

[r]

X

n

(a)

[s+1]

X

n

(a).

The equality in (1.5) holds if and only ifa₁ =a₂ =· · ·=a_n. Lettings=r−1in inequality (1.5) leads to inequality (1.4).

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2. Proof of Theorem 1.1

To prove inequality (1.5), the following properties forEn^[r]are necessary.

Property 1. Ifn, r∈N, then

(2.1) E_n^[r]=E_n−1^[r] +a_nE_n^[r−1]

and

(2.2) E_n^[r]=

r

X

j=0

a^j_nE_n−1^[r−j].

Proof. Ifn= 1orr = 0, (2.1) holds trivially.

Whenn >1andr≥1, we have (2.3)

i1+i2+···+in=r

X

i1,i2,···,in≥0 n

Y

k=1

aⁱ_k^k =

i1+i2+···+in=r

X

i1,i2,···,in−1≥0 in=0

n

Y

k=1

aⁱ_k^k +a_n

i1+i2+···+in=r−1

X

i1,i2,···,in≥0 n

Y

k=1

aⁱ_k^k.

Combining the definition ofEn^[r]and (2.3), identity (2.1) follows.

Identity (2.2) can be deduced from the recurrence of (2.1).

Property 2. Ifris an integer, then

(2.4) (r+ 1)E_n^[r+1]=

r

X

k=0 n

X

i=1

a^k+1_i

!

E_n^[r−k].

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Proof. It will be verified by induction. It is clear that identity (2.4) holds triv- ially forn = 1. Suppose identity (2.4) is true forn−1and nonnegative integers r.

By (2.2), for0≤k≤r, we have E_n^[r−k]=

r−k

X

j=0

a^j_nE_n−1^[r−k−j],

and

r

X

j=0

(j + 1)a^j+1_n E_n−1^[r−j] =a_nE_n−1^[r] +a²_nE_n−1^[r−1]+· · ·+a^r_nE_n−1^[1] +a^r+1_n E_n−1^[0]

+ a²_nE_n−1^[r−1]+· · ·+a^r_nE_n−1^[1] +a^r+1_n E_n−1^[0]

· · · ·

+ a^r_nE_n−1^[1] +a^r+1_n E_n−1^[0]

+ a^r+1_n E_n−1^[0]

=

r

X

k=0 r−k

X

j=0

a^j+k+1_n E_n−1^[r−k−j].

According to the inductive hypothesis, for nonnegative integersrand0≤ j ≤ r, we have

(2.5) (r−j+ 1)E_n−1^[r+1−j] =

r−j

X

k=0 n−1

X

i=1

a^k+1_i

!

E_n−1^[r−k−j].

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From Property1and the above formula, we have (r+ 1)E_n^[r+1]

(2.6)

= (r+ 1)

r+1

X

j=0

a^j_nE_n−1^[r+1−j]

=

r

X

j=0

(r−j+ 1)a^j_nE_n−1^[r−j+1]+

r+1

X

j=1

ja^j_nE_n−1^[r−j+1]

=

r

X

j=0

a^j_n

r−j

X

k=0 n−1

X

i=1

a^k+1_i

!

E_n−1^[r−j−k]+

r

X

j=0

(j+ 1)a^j+1_n E_n−1^[r−j]

=

r

X

k=0 r−k

X

j=0

a^j_n

n−1

X

i=1

a^k+1_i

!

E_n−1^[r−j−k]+

r

X

k=0 r−k

X

j=0

a^j+k+1_n E_n−1^[r−k−j]

=

r

X

k=0 n−1

X

i=1

a^k+1_i +a^k+1_n

! _r−k X

j=0

a^j_nE_n−1^[r−k−j]

!

=

r

X

k=0 n

X

i=1

a^k+1_i

!

E_n^[r−k].

This shows that (2.4) holds forn. The proof is complete.

Property 3. Ifr, s∈Nandr > s, then

(2.7) (r+ 1)(s+ 1)

n+r r+ 1

n+s s+ 1





[s]

X

n [r+1]

X

n

−

[r]

X

n [s+1]

X

n





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=

r

X

j=0 j

X

k=0

"

X

1≤v<u≤n

E_n^[s−k]E_n^[r−j]−E_n^[s−j]E_n^[r−k]

×

j−k−1

X

t=0

a^j−1−t_v a^k+t_u

!

(a_v−a_u)²

# .

Proof. Whenj > k, we have

n

X

i=1

a^k_i

n

X

i=1

a^j+1_i −

n

X

i=1

a^j_i

n

X

i=1

a^k+1_i (2.8)

= 1 2

n

X

v=1 n

X

u=1

a^k_va^j+1_u +a^j+1_v a^k_u−a^j_va^k+1_u −a^k+1_v a^j_u

= 1 2

n

X

v=1 n

X

u=1

a^k_va^k_u a^j−k+1_v +a^j−k+1_u −a^j−k_v a_u−a_va^j−k_u

= 1 2

n

X

v=1 n

X

u=1

a^k_va^k_u a^j−k_v −a^j−k_u

(a_v −a_u)

= X

1≤v<u≤n

a^k_va^k_u a^j−k_v −a^j−k_u

(a_v−a_u)

and

(2.9) a^j−k_v −a^j−k_u

= (a_v−a_u)

j−k−1

X

t=0

a^{j−k−1−t}_v a^t_u.

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Therefore (2.10)

n

X

i=1

a^k_i

n

X

i=1

a^j+1_i −

n

X

i=1

a^j_i

n

X

i=1

a^k+1_i

= X

1≤v<u≤n

" _k−j−1 X

t=0

!

(a_v−a_u)²

# .

Whenk > j, we have (2.11)

n

X

i=1

a^k_i

n

X

i=1

a^j+1_i −

n

X

i=1

a^j_i

n

X

i=1

a^k+1_i

=− X

1≤v<u≤n

" _k−j−1 X

t=0

a^j+t_v a^k−1−t_u

!

(a_v−a_u)²

# .

From Property2, it is deduced that (2.12) (r+ 1)E_n^[r+1] =

r

X

j=0 n

X

i=1

a^j+1_i

! E_n^[r−j]

and

(2.13) (n+r)E_n^[r]=nE_n^[r]+rE_n^[r]=

r

X

j=0 n

X

i=1

a^j_i

!

E_n^[r−j].

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Hence, using the above formulas and notingEn^[r−k]= 0fork > ryields (r+ 1)(s+ 1)

n+r r+ 1

n+s s+ 1





[s]

X

n [r+1]

X

n

−

[r]

X

n [s+1]

X

n

 (2.14) 

= (n+s)(r+ 1)E_n^[s]E_n^[r+1]−(n+r)(s+ 1)E_n^[r]E_n^[s+1]

=

s

X

k=0 n

X

i=1

a^k_i

! E_n^[s−k]

r

X

j=0 n

X

i=1

a^j+1_i

! E_n^[r−j]

−

r

X

j=0 n

X

i=1

a^j_i

! E_n^[r−j]

s

X

k=0 n

X

i=1

a^k+1_i

! E_n^[s−k]

=

s

X

k=0 r

X

j=0 n

X

i=1

a^k_i

n

X

i=1

a^j+1_i −

n

X

i=1

a^j_i

n

X

i=1

a^k+1_i

!

E_n^[s−k]·E_n^[r−j]

=

r

X

j=0 j

X

k=0

"

X

1≤v<u≤n

E_n^[s−k]E_n^[r−j]

j−k−1

X

t=0

!

(a_v −a_u)²

#

−

s

X

k=0 k

X

j=0

"

X

1≤v<u≤n

k−j−1

X

t=0

a^j+t_v a^k−1−t_u

!

(a_v−a_u)²

#

=

r

X

j=0 j

X

k=0

"

X

1≤v<u≤n

j−k−1

X

t=0

!

(a_v −a_u)²

#

−

r

X

j=0 j

X

k=0

"

X

1≤v<u≤n

E_n^[s−j]E_n^[r−k]

j−k−1

X

t=0

!

(a_v−a_u)²

#

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=

r

X

j=0 j

X

k=0

"

X

1≤v<u≤n

E_n^[s−k]E_n^[r−j]−E_n^[s−j]E_n^[r−k]

×

j−k−1

X

t=0

!

(a_v−a_u)²

# ,

which implies the expression (2.7).

Property 4. Ifr, s∈Nandr > s, then

(2.15) E_n^[r−1]E_n^[s]≥E_n^[r]E_n^[s−1].

The equality in (2.15) holds if and only if at least n −1 numbers equal zero among{a₁, a₂, . . . , a_n}.

Proof. From Property1, we have E_n^[r−1]E_n^[s]−E_n^[r]E_n^[s−1]

(2.16)

=E_n^[r−1]

E_n−1^[s] +a_nE_n^[s−1]

−

E_n−1^[r] +a_nE_n^[r−1]

E_n^[s−1]

=E_n^[r−1]E_n−1^[s] −E_n−1^[r] E_n^[s−1]

=

r−1

X

j=0

a^j_nE_n−1^[r−1−j]

!

E_n−1^[s] −E_n−1^[r]

s−1

X

j=0

a^j_nE_n−1^[s−1−j]

!

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=

s−1

X

j=0

a^j_n

E_n−1^[r−1−j]E_n−1^[s] −E_n−1^[r] E_n−1^[s−1−j]

+E_n−1^[s]

r−1

X

j=s

a^j_nE_n−1^[r−1−j]

! .

Since (2.15) holds forn= 1, it follows by induction that (2.15) holds forn.

Property 5. Ifr, s, j, k∈Nandr > s > j > k, then (2.17) E_n^[s−k]E_n^[r−j]≥E_n^[s−j]E_n^[r−k].

The equality in (2.17) is valid if and only if at leastn−1numbers equal zero among{a₁, a₂, . . . , a_n}

Proof. From Property4, ifr−(k+ 1)> s−(k+ 1), r−(k+ 2)> s−(k+ 2), . . . , r−j > s−j, then

(2.18)

j

Y

m=k+1

E_n^[r−m]E_n^[s−m+1]

≥

j

Y

m=k+1

E_n^[r−m+1]E_n^[s−m]

. This implies (2.17).

It is easy to see that the equality in (2.17) is valid. The proof is completed.

Proof of Theorem1.1. Combination of Property 3 and Property 5 easily leads to Theorem1.1.

Remark 2.1. Finally, we pose an open problem: Give an explicit expression of En^[r−1]En^[s]−En^[r]En^[s−1] in terms ofa₁, a₂, . . . , a_n.

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References

[1] D.W. DETEMPLE AND J.M. ROBERTSON, On generalized symmetric means of two variables, Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat.

Fiz., 634–672 (1979), 236–238

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

[3] D.S. MITRINOVI ´C, Analytic Inequalities, Springer, 1970.

[4] J.-Ch. KUANG, Chángyòng Búdˇengshì (Applied Inequalities), 2nd Ed. Hu- nan Education Press, Changsha, China, 1993. (Chinese)

[5] G.H. HARDY, J.E. LITTLEWOOD ANDG. POLYA, Inequalities, 2nd Ed.

Cambridge Univ. Press, 1952.

[6] Zh.-G. XIAO, Zh.-H. ZHANG AND X.-N. LU, A class of inequalities for weighted symmetric mean, J. Hunan Educational Institute, 17(5) (1999), 130–134. (Chinese)

[7] Zh.-H. ZHANG, Three classes of new means in n+ 1 variables and their applications, J. Hunan Educational Institute, 15(5) (1997), 130–136. (Chi- nese)