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

http://jipam.vu.edu.au/

Volume 5, Issue 4, Article 108, 2004

A GENERALIZATION OF AN INEQUALITY INVOLVING THE GENERALIZED ELEMENTARY SYMMETRIC MEAN

ZHI-HUA ZHANG AND ZHEN-GANG XIAO ZIXINGEDUCATIONALRESEARCHSECTION

CHENZHOU, HUNAN423400, CHINA. zxzh1234@163.com

HUNANINSTITUTE OFSCIENCE ANDTECHNOLOGY

YUEYANG, HUNAN414006, CHINA. xiaozg@163.com

Received 23 May, 2003; accepted 09 October, 2004 Communicated by F. Qi

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

Key words and phrases: Generalized elementary symmetric mean, Inequality.

2000 Mathematics Subject Classification. Primary 26D15.

1. INTRODUCTION

Leta = (a₁, a₂,· · · , a_n)andrbe a nonnegative integer, wherea_i for1≤i≤nare nonneg- ative 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 therth 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

.

ISSN (electronic): 1443-5756 c

2004 Victoria University. All rights reserved.

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.

070-03

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+· · ·+x_n−1)and the integral is taken overx_k ≥0fork = 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

.

In 1968, K.V. Menon [2] proved that forn = 2orn ≥ 3andr = 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 arbitraryn, 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).

2. PROOF OFTHEOREM1.1

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 +an

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].

Proof. It will be verified by induction. It is clear that identity (2.4) holds trivially for n = 1.

Suppose identity (2.4) is true forn−1and nonnegative integersr.

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]=anE_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].

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

r+1

X

j=0

a^j_nE_n−1^[r+1−j]

(2.6)

=

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





=

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

= (av−au)

j−k−1

X

t=0

a^{j−k−1−t}_v a^t_u.

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^j−1−t_v a^k+t_u

!

(av−au)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 Property 2, 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].

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^j−1−t_v a^k+t_u

!

(av−au)²

#

−

s

X

k=0 k

X

j=0

"

X

1≤v<u≤n

E_n^[s−k]E_n^[r−j]

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

E_n^[s−k]E_n^[r−j]

j−k−1

X

t=0

a^j−1−t_v a^k+t_u

!

(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^j−1−t_v a^k+t_u

!

(a_v−a_u)²

#

=

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

# ,

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 leastn−1numbers equal zero among{a₁, a₂, . . . , a_n}.

Proof. From Property 1, 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]

!

=

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 Property 4, 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 Theorem 1.1. Combination of Property 3 and Property 5 easily leads to Theorem 1.1.

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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