Journal of Inequalities in Pure and
Applied Mathematics
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
Z IXING E DUCATIONAL R ESEARCH S ECTION
C HENZHOU , H UNAN 423400, C HINA .
zxzh1234@163.com
H UNAN I NSTITUTE OF S CIENCE AND T ECHNOLOGY
Y UEYANG , H UNAN 414006, C HINA .
xiaozg@163.com
Received 23 May, 2003; accepted 09 October, 2004
Communicated by F. Qi
A BSTRACT. 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. I NTRODUCTION
Let a = (a1 , a2 , · · · , an ) and r be a nonnegative integer, where ai for 1 ≤ i ≤ n are nonnegative real numbers. Then
n
X
Y
aikk
(1.1)
En[r] = En[r] (a) =
i1 +i2 +···+in =r,
i1 ,i2 ,··· ,in ≥0 are integers
[0]
[0]
k=1
[r]
with En = En (a) = 1 for n ≥ 1 and En = 0 for r < 0 or n ≤ 0 is called the rth generalized
elementary symmetric function of a.
The rth generalized elementary symmetric mean of a is defined by
(1.2)
[r]
X
n
=
[r]
X
n
[r]
En (a)
(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
2
Z HI -H UA Z HANG AND Z HEN -G ANG X IAO
In 1934, I. Schur [5, p. 182] obtained the following
(1.3)
[r]
X
Z
n
X
Z
(a) = (n − 1)!
···
n
!r
dx1 · · · dxn−1 ,
ai x i
i=1
where xn = 1−(x1 +x2 +· · ·+xn−1 ) and the integral is taken over xk ≥ 0 for k = 1, 2, . . . , n−1.
By using (1.3) and Cauchy integral inequality, he also proved that

2
[r−1]
[r+1]
[r]
X
X
X
(1.4)
(a)
(a) ≥  (a) .
n
n
n
In 1968, K.V. Menon [2] proved that for n = 2 or n ≥ 3 and r = 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.
P [r]
[r]
In this paper, we shall obtain an identity relating
n (a) to En (a) and give an elementary
proof of an inequality which generalizes (1.4). Our main result is as follows.
Theorem 1.1. If r, s ∈ N and r > s, then
[s]
X
(1.5)
[r+1]
(a)
n
X
(a) ≥
n
[r]
X
[s+1]
(a)
X
n
(a).
n
The equality in (1.5) holds if and only if a1 = a2 = · · · = an .
Letting s = r − 1 in inequality (1.5) leads to inequality (1.4).
2. P ROOF OF T HEOREM 1.1
[r]
To prove inequality (1.5), the following properties for En are necessary.
Property 1. If n, r ∈ N, then
[r]
En[r] = En−1 + an En[r−1]
(2.1)
and
En[r]
(2.2)
=
r
X
[r−j]
ajn En−1 .
j=0
Proof. If n = 1 or r = 0, (2.1) holds trivially.
When n > 1 and r ≥ 1, we have
(2.3)
i1 +i2 +···+i
n
X n =r Y
aikk
=
i1 ,i2 ,··· ,in ≥0 k=1
i1 +i2 +···+i
n
X n =r Y
i1 ,i2 ,··· ,in−1 ≥0
in =0
k=1
aikk
+ an
i1 +i2 +···+i
n
Xn =r−1 Y
i1 ,i2 ,··· ,in ≥0
aikk .
k=1
[r]
Combining the definition of En and (2.3), identity (2.1) follows.
Identity (2.2) can be deduced from the recurrence of (2.1).
J. Inequal. Pure and Appl. Math., 5(4) Art. 108, 2004
http://jipam.vu.edu.au/
A G ENERALIZATION OF AN I NEQUALITY I NVOLVING . . .
3
Property 2. If r is an integer, then
(r +
(2.4)
1)En[r+1]
r
n
X
X
=
!
ak+1
i
En[r−k] .
i=1
k=0
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 for n − 1 and nonnegative integers r.
By (2.2), for 0 ≤ k ≤ r, we have
En[r−k] =
r−k
X
[r−k−j]
ajn En−1
,
j=0
and
r
X
[r−j]
[r]
[r−1]
[1]
[0]
2
r
r+1
(j + 1)aj+1
n En−1 = an En−1 + an En−1 + · · · + an En−1 + an En−1
j=0
[0]
[1]
[r−1]
+ a2n En−1 + · · · + arn En−1 + ar+1
n En−1
······
[1]
[0]
+ arn En−1 + ar+1
n En−1
[0]
+ ar+1
n En−1
=
r−k
r X
X
[r−k−j]
aj+k+1
En−1
n
.
k=0 j=0
According to the inductive hypothesis, for nonnegative integers r and 0 ≤ j ≤ r, we have
!
r−j
n−1
X
X
[r+1−j]
[r−k−j]
(2.5)
(r − j + 1)En−1
=
ak+1
En−1 .
i
i=1
k=0
From Property 1 and the above formula, we have
(2.6)
(r + 1)En[r+1] = (r + 1)
r+1
X
[r+1−j]
ajn En−1
j=0
=
r
X
(r − j +
[r−j+1]
1)ajn En−1
+
j=0
=
r
X
j=0
=
=
[r−j+1]
jajn En−1
j=1
ajn
r−j
n−1
X
X
k=0
r X
r−k
X
ajn
k=0 j=0
=
r+1
X
r
X
n−1
X
k=0
i=1
r
X
n
X
k=0
i=1
!
[r−j−k]
En−1
ak+1
i
i=1
n−1
X
[r−j]
(j + 1)aj+1
n En−1
j=0
!
ak+1
i
[r−j−k]
En−1
i=1
+
r X
r−k
X
[r−k−j]
anj+k+1 En−1
k=0 j=0
!
ak+1
+ ak+1
n
i
r−k
X
!
[r−k−j]
ajn En−1
j=0
!
ak+1
i
En[r−k] .
This shows that (2.4) holds for n. The proof is complete.
J. Inequal. Pure and Appl. Math., 5(4) Art. 108, 2004
+
r
X
http://jipam.vu.edu.au/
4
Z HI -H UA Z HANG AND Z HEN -G ANG X IAO
Property 3. If r, s ∈ N and r > s, then
(2.7) (r + 1)(s + 1)
=
j
r X
X
n+r
r+1


X
[s] [r+1]
[r] [s+1]
X
X
X
n+s 

−
s+1
n
n
n
n
j−k−1
"
j=0 k=0
X
X
[r−k]
En[s−k] En[r−j] − En[s−j] En
#
!
(av − au )2 .
ak+t
aj−1−t
u
v
t=0
1≤v<u≤n
Proof. When j > k, we have
n
X
(2.8)
aki
i=1
n
X
aj+1
−
i
i=1
n
X
n
X
aji
i=1
=
i=1
n
n
1 XX
2
ak+1
i
k
j k+1
j
akv aj+1
+ aj+1
− ak+1
u
v au − av au
v au
v=1 u=1
n
n
1 X X k k j−k+1
j−k
av au av
+ auj−k+1 − aj−k
a
−
a
a
=
u
v
v
u
2 v=1 u=1
n
n
1 X X k k j−k
=
av au av − aj−k
(av − au )
u
2 v=1 u=1
X =
akv aku aj−k
− aj−k
(av − au )
v
u
1≤v<u≤n
and
j−k−1
aj−k
v
(2.9)
−
aj−k
u
X
= (av − au )
aj−k−1−t
atu .
v
t=0
Therefore
(2.10)
n
X
aki
i=1
n
X
aj+1
−
i
i=1
n
X
n
X
j
ai
i=1
"
X
1≤v<u≤n
t=0
ak+1
=
i
i=1
k−j−1
X
!
#
aj−1−t
ak+t
(av − au )2 .
v
u
When k > j, we have
(2.11)
n
X
i=1
aki
n
X
i=1
aj+1
−
i
n
X
i=1
n
X
j
ai
"
ak+1
=−
i
i=1
k−j−1
X
X
1≤v<u≤n
t=0
!
#
k−1−t
aj+t
(av − au )2 .
v au
From Property 2, it is deduced that
(2.12)
(r + 1)En[r+1] =
r
n
X
X
j=0
!
aj+1
i
En[r−j]
i=1
and
(2.13)
(n + r)En[r] = nEn[r] + rEn[r] =
r
n
X
X
j=0
J. Inequal. Pure and Appl. Math., 5(4) Art. 108, 2004
!
aji
En[r−j] .
i=1
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A G ENERALIZATION OF AN I NEQUALITY I NVOLVING . . .
[r−k]
Hence, using the above formulas and noting En
(2.14)
(r + 1)(s + 1)
n+r
r+1
5
= 0 for k > r yields


X
[s] [r+1]
[r] [s+1]
X
X
X
n+s 

−
s+1
n
n
n
n
= (n + s)(r + 1)En[s] En[r+1] − (n + r)(s + 1)En[r] En[s+1]
!
!
s
n
r
n
X
X
X
X
j+1
=
aki En[s−k]
ai
En[r−j]
i=1
k=0
−
=
j=0
r
n
X
X
j=0
i=1
s X
r
X
n
X
k=0 j=0
=
j
r X
X
=
aki
j=0 k=0
ak+1
i
En[s−k]
i=1
aji
n
X
i=1
!
ak+1
i
En[s−k] · En[r−j]
i=1
X
En[s−k] En[r−j]
"
X
#
2
(av − au )
k−j−1
!
X
avj+t ak−1−t
(av − au )2
u
En[s−k] En[r−j]
#
t=0
1≤v<u≤n
"
X
!
avj−1−t ak+t
u
t=0
1≤v<u≤n
j−k−1
!
X
avj−1−t ak+t
(av − au )2
u
En[s−k] En[r−j]
#
t=0
1≤v<u≤n
"
X
j=0 k=0
=
−
n
X
!
j−k−1
j
r X
X
j
r X
X
aj+1
i
i=1
X
j=0 k=0
−
s
n
X
X
k=0
k=0 j=0
r X
X
En[r−j]
n
X
i=1
s X
k
X
j
aji
i=1
"
j=0 k=0
−
!
En[s−j] En[r−k]
j−k−1
!
X
aj−1−t
ak+t
(av − au )2
v
u
#
t=0
1≤v<u≤n
"
X
En[s−k] En[r−j] − En[s−j] En[r−k]
1≤v<u≤n
j−k−1
×
X
!
aj−1−t
ak+t
v
u
#
2
(av − au )
,
t=0
which implies the expression (2.7).
Property 4. If r, s ∈ N and r > s, then
En[r−1] En[s] ≥ En[r] En[s−1] .
(2.15)
The equality in (2.15) holds if and only if at least n−1 numbers equal zero among {a1 , a2 , . . . , an }.
Proof. From Property 1, we have
(2.16)
En[r−1] En[s] − En[r] En[s−1]
[s]
[r]
[r−1]
[s−1]
[r−1]
= En
En−1 + an En
− En−1 + an En
En[s−1]
J. Inequal. Pure and Appl. Math., 5(4) Art. 108, 2004
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6
Z HI -H UA Z HANG AND Z HEN -G ANG X IAO
[s]
[r]
= En[r−1] En−1 − En−1 En[s−1]
!
r−1
X
[r−1−j]
[s]
[r]
ajn En−1
En−1 − En−1
=
j=0
=
s−1
X
ajn
s−1
X
!
[s−1−j]
ajn En−1
j=0
[r−1−j] [s]
En−1 En−1
−
[r]
[s−1−j]
En−1 En−1
j=0
+
[s]
En−1
r−1
X
!
[r−1−j]
ajn En−1
.
j=s
Since (2.15) holds for n = 1, it follows by induction that (2.15) holds for n.
Property 5. If r, s, j, k ∈ N and r > s > j > k, then
En[s−k] En[r−j] ≥ En[s−j] En[r−k] .
(2.17)
The equality in (2.17) is valid if and only if at least n−1 numbers equal zero among {a1 , a2 , . . . , an }
Proof. From Property 4, if r −(k +1) > s−(k +1), r −(k +2) > s−(k +2), . . . , r −j > s−j,
then
j
j
Y
Y
[r−m] [s−m+1]
(2.18)
En
En
≥
En[r−m+1] En[s−m] .
m=k+1
m=k+1
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.
[r−1]
Remark 2.1. Finally, we pose an open problem: Give an explicit expression of En
[r] [s−1]
En En
in terms of a1 , a2 , . . . , an .
[s]
En −
R EFERENCES
[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. MITRINOVIĆ, Analytic Inequalities, Springer, 1970.
[4] J.-Ch. KUANG, Chángyòng Búděngshì (Applied Inequalities), 2nd Ed. Hunan Education Press,
Changsha, China, 1993. (Chinese)
[5] G.H. HARDY, J.E. LITTLEWOOD AND G. 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. (Chinese)
J. Inequal. Pure and Appl. Math., 5(4) Art. 108, 2004
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