volume 5, issue 3, article 69, 2004.
Received 03 March, 2004;
accepted 30 March, 2004.
Communicated by:D. Stefanescu
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Journal of Inequalities in Pure and Applied Mathematics
SOME INEQUALITIES FOR A CLASS OF GENERALIZED MEANS
KAI-ZHONG GUAN
Department of Mathematics and Physics Nanhua University, Hengyang, Hunan 421001 The Peoples’ Republic of China
EMail:[email protected]
c
2000Victoria University ISSN (electronic): 1443-5756 054-04
Some Inequalities for a Class of Generalized Means
Kai-Zhong Guan
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Abstract
In this paper, we define a symmetric function, show its properties, and establish several analytic inequalities, some of which are "Ky Fan" type inequalities. The harmonic-geometric mean inequality is refined.
2000 Mathematics Subject Classification:05E05, 26D20.
Key words: Symmetric function ; Ky Fan inequality ; Harmonic-geometric mean in- equality.
Contents
1 Introduction. . . 3 2 Lemmas . . . 5 3 Refinement of the Harmonic-Geometric Mean Inequality . . . 8 4 Schur-convexity of the FunctionHnr(x). . . 11 5 Some “Ky Fan” Type Inequalities . . . 16
References
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1. Introduction
Let x = (x1, x2, . . . , xn) be ann-tuple of positive numbers. The un-weighted arithmetic, geometric and harmonic means of x, denoted by An(x) , Gn(x), Hn(x), respectively, are defined as follows
An(x) = 1 n
n
X
i=1
xi, Gn(x) =
n
Y
i=1
xi
!n1
, Hn(x) = n Pn
i=1 1 xi
.
Assume that0≤xi <1,1≤i≤nand define1−x= (1−x1,1−x2, . . . ,1− xn).Throughout the sequel the symbolsAn(1−x), Gn(1−x), Hn(1−x)will stand for the un-weighted arithmetic, geometric, harmonic means of1−x.
A remarkable new counterpart of the inequalityGn ≤Anhas been published in [1].
Theorem 1.1. If0< xi ≤ 12, for alli= 1,2, . . . , n,then
(1.1) Gn(x)
Gn(1−x) ≤ An(x) An(1−x) with equality if and only if all thexi are equal.
This result, commonly referred to as the Ky Fan inequality, has stimulated the interest of many researchers. New proofs, improvements and generaliza- tions of the inequality (1.1) have been found. For more details, interested read- ers can see [2], [3] and [4].
W.-L. Wang and P.-F. Wang [5] have established a counterpart of the classical inequalityHn ≤Gn ≤An.Their result reads as follows.
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Theorem 1.2. If0< xi ≤ 12, for alli= 1,2, . . . , n,then
(1.2) Hn(x)
Hn(1−x) ≤ Gn(x)
Gn(1−x) ≤ An(x) An(1−x).
All kinds of means about numbers and their inequalities have stimulated the interest of many researchers. Here we define a new mean, that is:
Definition 1.1. Letx∈Rn+={x|x= (x1, x2, . . . , xn)|xi >0, i= 1,2, . . . , n}, we define the symmetric function as follows
Hnr(x) =Hnr(x1, x2, . . . , xn) =
"
Y
1≤i1<···ir≤n
r Pr
i=1x−1ij
!#(n1 r)
.
ClearlyHnn(x) =Hn(x),Hn1(x) = Gn(x), where nr
= r!(n−r)!n! .
The Schur-convex function was introduced by I. Schur in 1923 [7]. Its defi- nition is as follows:
Definition 1.2. f :In→R(n >1)is called Schur-convex ifx≺y, then
(1.3) f(x)≤f(y)
for allx, y ∈In=I×I×· · ·×I(n copies).It is called strictly Schur-convex if the inequality is strict;f is called Schur-concave (resp. strictly Schur-concave) if the inequality (1.3) is reversed. For more details, interested readers can see [6], [7] and [8].
The paper is organized as follows. A refinement of harmonic-geometric mean inequality is obtained in Section3. In Section4, we investigate the Schur- convexity of the symmetric function. Several “Ky Fan” type inequalities are
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2. Lemmas
In this section, we give the following lemmas for the proofs of our main results.
Lemma 2.1. ([5]) If0< xi ≤ 12, for alli= 1,2, . . . , n,then (2.1)
Pn i=1
1 1−xi
Pn i=1
1 xi
≤
" Qn i=
1 1−xi
Qn i=
1 xi
#1n
or Hn(x)
Hn(1−x) ≤ Gn(x) Gn(1−x). Lemma 2.2. If 0 < xi ≤ 12,for alli = 1,2, . . . , n+ 1,andSn+1 =Pn+1
i=1 1 xi, Sn+1 =Pn+1
i=1 1
1−xi, then
(2.2)
Pn+1
i=1
Sn+1− 1−x1
i
Pn+1
i=1
Sn+1− x1
i
n
≤
Qn+1
i=1
Sn+1− 1−x1
i
Qn+1
i=1
Sn+1− x1
i
1 n+1
.
Proof. Inequality (2.2) is equivalent to the following
nlnSn+1
Sn+1 ≤ 1 n+ 1 ln
Qn+1
i=1
Sn+1−1−x1
i
Qn+1
i=1(Sn+1− x1
i)
Since0< xi ≤ 12, and1−xi ≥xi, it follows that
Sn+1−1−x1
j
Sn+1− x1
j
=
1
1−x1 +· · ·+ 1−x1
j−1 +1−x1
j+1 +· · ·+1−x1
n+1
1
x1 +· · ·+ x1
j−1 +x1
j+1 +· · ·+ x1
n+1
≥
1
1−x1 · · ·1−x1
j−1
1
1−xj+1· · ·1−x1
n+1
1 x1 · · ·x1
j−1
1
xj+1· · ·x1
n+1
.
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By the above inequality and Lemma2.1, we have 1
n+ 1 ln
n+1
Y
i=1
Sn+1−1−x1
i
Sn+1− x1
i
≥ 1 n+ 1ln
n+1
Y
i=1
1 1−xi
1 xi
n
=nln
n+1
Y
i=1
1 1−xi
1 xi
n+1
≥nln
1
1−x1 +· · ·+1−x1
n+1
1
x1 +· · ·+x1
n+1
,
or
nlnSn+1
Sn+1 ≤ 1 n+ 1 ln
Qn+1
i=1
Sn+1− 1−x1
i
Qn+1
i=1
Sn+1− x1
i
.
Lemma 2.3. [6, p. 259]. Let f(x) = f(x1, x2, . . . , xn) be symmetric and have continuous partial derivatives on In, where I is an open interval. Then f :In→Ris Schur-convex if and only if
(2.3) (xi −xj)
∂f
∂xi − ∂f
∂xj
≥0
on In. It is strictly Schur-convex if (2.3) is a strict inequality for xi 6= xj, 1≤i, j ≤n.
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Sincef(x)is symmetric, Schur’s condition can be reduced as [7, p. 57]
(2.4) (x1 −x2)
∂f
∂x1 − ∂f
∂x2
≥0,
and f is strictly Schur-convex if (2.4) is a strict inequality for x1 6= x2. The Schur condition that guarantees a symmetric function being Schur-concave is the same as (2.3) or (2.4) except the direction of the inequality.
In Schur’s condition, the domain of f(x) does not have to be a Cartesian productIn.Lemma2.3remains true if we replaceInby a setA⊆Rnwith the following properties ([7, p. 57]):
(i) Ais convex and has a nonempty interior;
(ii) A is symmetric in the sense that x ∈ A implies P x ∈ Afor any n×n permutation matrixP.
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3. Refinement of the Harmonic-Geometric Mean Inequality
The goal of this section is to obtain the basic inequality of Hnr(x), and give a refinement of the Harmonic-Geometric mean inequality.
Theorem 3.1. Letx∈ Rn+={x|x= (x1, x2, . . . , xn)|xi >0, i= 1,2, . . . , n}, then
(3.1) Hnr+1(x)≤Hnr(x), r= 1,2, . . . , n−1.
Proof. By the arithmetic-geometric mean inequality and the monotonicity of the functiony= lnx, we have
n r+ 1
lnHnr+1(x)
= X
1≤i1<···<ir+1≤n
ln r+ 1 Pr+1
j=1x−1ij
= X
1≤i1<···<ir+1≤n
ln
"
(r+ 1)r (r+ 1)Pr+1
k=1x−1i
k −Pr+1
j=1x−1ij
#
= X
1≤i1<···<ir+1≤n
ln
r hPr+1
j=1
Pr+1 k=1x−1i
k −x−1i
j
i.
(r+ 1)
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≤ X
1≤i1<···<ir+1≤n
ln
r Qr+1
j=1
Pr+1 k=1x−1i
k −x−1ij r+11
= X
1≤i1<···<ir+1≤n
ln
"r+1 Y
j=1
r Pr+1
k=1x−1i
k −x−1ij
#r+11
= 1
r+ 1
X
1≤i1<···<ir+1≤n
"r+1 X
j=1
ln r
Pr+1 k=1x−1i
k −x−1i
j
#
= 1
r+ 1
n
X
j=1
i1,...,ir6=j
X
1≤i1<···<ir≤n
ln r Pr
k=1x−1ik .
Let
Sj =
i1,...,ir6=j
X
1≤i1<···<ir≤n
ln r Pr
k=1x−1i
k
, j = 1,2, . . . , n.
We can easily get
n
X
j=1
Sj = (n−r) X
1≤i1<···<ir≤n
ln r Pr
k=1x−1i
k
= (n−r)n r
lnHnr(x).
Thus n
r+ 1
lnHnr+1(x)≤ n−r r+ 1
n r
lnHnr(x) = n
r+ 1
lnHnr(x),
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or
Hnr+1(x)≤Hnr(x), r= 1,2, . . . , n−1.
Corollary 3.2. Letx∈Rn+={x|x= (x1, x2, . . . , xn)|xi >0, i= 1,2, . . . , n}, then
(3.2) Hn(x)≤Hnn−1(x)≤ · · · ≤Hn2(x)≤Hn1(x) = Gn(x).
Remark 3.1. The corollary refines the harmonic-geometric mean inequality.
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4. Schur-convexity of the Function H
nr(x)
In this section, we investigate the Schur-convexity of the function Hnr(x), and establish several analytic inequalities by use of the theory of majorization.
Theorem 4.1. LetRn+ ={x|x= (x1, x2, . . . , xn)|xi >0, i= 1,2, . . . , n}, then the functionHnr(x)is Schur-concave inRn+.
Proof. It is clear thatHnr(x)is symmetric and has continuous partial derivatives onRn+. By Lemma2.3, we only need to prove
(x1−x2)
∂Hnr(x)
∂x1 −∂Hnr(x)
∂x2
≤0.
As matter of fact, we can easily derive lnHnr(x) = 1
n r
X
2≤i1<···<ir≤n
ln r Pr
j=1x−1i
j
+ X
2≤i1<···<ir−1≤n
ln r
x−11 +Pr−1 j=1x−1ij . DifferentiatinglnHnr(x)with respect tox1, we have
∂Hnr(x)
∂x1 = Hnr(x)
n r
X
2≤i1<···<ir−1≤n
1 x−11 +Pr−1
j=1x−1ij
· 1 x21
= Hnr(x)
n r
· 1 x21
X
3≤i1<···<ir−1≤n
1 x−11 +Pr−1
j=1x−1i
j
+ X
3≤i1<···<ir−2≤n
1
x−11 +x−12 +Pr−2 j=1x−1i
j
.
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Similar to the above, we can also obtain
∂Hnr(x)
∂x2 = Hnr(x)
n r
X
2≤i1<···<ir−1≤n
1 x−12 +Pr−1
j=1x−1i
j
· 1 x22
= Hnr(x)
n r
· 1 x22
X
3≤i1<···<ir−1≤n
1 x−12 +Pr−1
j=1x−1ij
+ X
3≤i1<···<ir−2≤n
1
x−11 +x−12 +Pr−2 j=1x−1i
j
.
Thus (x1−x2)
∂Hnr(x)
∂x1
−∂Hnr(x)
∂x2
= (x1−x2)Hnr(x)
n r
X
2≤i1<···<ir−1≤n
1 x−11 +Pr−1
j=1x−1i
j
· 1 x21
−
X
2≤i1<···<ir−1≤n
1 x−12 +Pr−1
j=1x−1i
j
· 1 x22
+ X
3≤i1<···<ir−2≤n
1
x−11 +x−12 +Pr−2 j=1x−1i
1 x21 − 1
x22
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=−(x1−x2)2
(x1+x2) x21x22
X
3≤i1<···<ir−2≤n
1 x−11 +x−12 +Pr−2
j=1x−1i
j
+ X
2≤i1<···<ir−1≤n
1 + (x1+x2)Pr−1 j=1x−1i
j
x21x22
x−11 +Pr−1 j=1x−1i
j x−12 +Pr−1
j=1x−1i
j
≤0.
Corollary 4.2. Let xi > 0, i = 1,2, . . . , n, n ≥ 2 ,andPn
i=1xi = s, c > 0, then
(4.1) Hnr(c+x)
Hnr(x) ≥nc
s + 1(n1 r)
, r= 1,2, . . . , n, wherec+x= (c+x1, c+x2, . . . , c+xn).
Proof. By [9], we have c+x nc+s =
c+x1
nc+s, . . . ,c+xn nc+s
≺x1
s , . . . , xn s
= x s.
Using Theorem4.1, we obtain Hnr
c+x nc+s
≥Hnrx s
.
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Or
Hnr(c+x)
Hnr(x) ≥nc
s + 1(1n r)
.
Corollary 4.3. Let xi > 0, i = 1,2, . . . , n, n ≥ 2, and Pn
i=1xi = s, c ≥ s, then
(4.2) Hnr(c−x)
Hnr(x) ≥nc
s −1(1n r)
, r= 1,2, . . . , n, wherec−x= (c−x1, c−x2, . . . , c−xn).
Proof. By [9], we have c−x nc−s =
c−x1
nc−s, . . . , c−xn nc−s
≺x1
s , . . . ,xn s
= x s.
Using Theorem4.1, we obtain Hnr
c−x nc−s
≥Hnrx s
,
or
Hnr(c−x)
Hnr(x) ≥nc s −1
1
(nr)
.
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Remark 4.1. Letc=s= 1, we can obtain Hnr(1−x)
Hnr(x) ≥(n−1)
1
(nr), r= 1,2, . . . , n.
In particular,
Hn(1−x)
Hn(x) ≥(n−1), Gn(1−x) Gn(x) ≥ √n
n−1.
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5. Some “Ky Fan” Type Inequalities
In this section, some “Ky Fan” type inequalities are established, the Ky Fan inequality is generalized.
Theorem 5.1. Assume that0< xi ≤ 12, i= 1,2, . . . , n,then
(5.1) Hnr+1(x) Hnr+1(1−x) ≤
Hnr(x) Hnr(1−x)
1r
, r= 1,2, . . . , n−1.
Proof. Set
ϕr = Hnr(x)
Hnr(1−x) = Y
1≤i1<···<ir≤n
Pr
j=1 1 1−xij
Pr j=1
1 xij
1
(nr)
.
By Lemma2.2and the monotonicity of the functiony= lnx, we have n
r+ 1
lnφr+1 = X
1≤i1<···<ir+1≤n
ln Pr+1
j=1 1 1−xij
Pr+1 j=1
1 xij
= X
1≤i1<···<ir+1≤n
ln Pr+1
j=1
Pr+1 k=1
1
1−xik − 1−x1
ij
Pr+1
j=1
Pr+1 k=1
1 xik − x1
ij
≤ X
ln
r+1
Y Pr+1
k=1 1
1−xik −1−x1
ij
Pr+1 1 − 1
1 r(r+1)
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= 1
r(r+ 1)
n
X
j=1
i1,...,ir6=j
X
1≤i1<···<ir≤n
ln Pr
k=1 1 1−xik
Pr k=1
1 xik
.
Similar to Theorem3.1, we can derive n
r+ 1
lnφr+1 ≤ 1
r(r+ 1)(n−r) n
r
lnφr = 1 r
n r+ 1
lnφr. Thus
(φr)1r ≥φr+1, or
Hnr+1(x) Hnr+1(1−x) ≤
Hnr(x) Hnr(1−x)
1r
, r= 1,2, . . . , n−1.
Remark 5.1. By Theorem5.1, we can obtain (5.2) Hn2(x)
Hn2(1−x) ≤ Hn1(x)
Hn1(1−x) = Gn(x)
Gn(1−x) ≤ An(x) An(1−x). This is a generalization of the “Ky Fan” inequality.
By Lemma2.1and the proof of Theorem3.1, we have the following Theorem 5.2. If0< xi ≤ 12, i = 1,2, . . . , n,then
(5.3)
Qn i=1(xi) Qn
i=1(1−xi) ≤ Hn(x)
Hn(1−x) ≤ Gn(x)
Gn(1−x) ≤ An(x) An(1−x).
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The inequality (5.3) generalizes the inequality (1.2).
Theorem 5.3. If0< xi ≤ 12, i = 1,2, . . . , n,then Hnr(x)
Hnr(1−x) ≤ Hn1(x) Hn1(1−x) (5.4)
= Gn(x)
Gn(1−x) ≤ An(x)
An(1−x), r = 2, . . . , n.
Proof. Set
ϕr = Hnr(x)
Hnr(1−x) = Y
1≤i1<···<ir≤n
Pr
j=1 1 1−xij
Pr j=1
1 xij
1
(nr)
.
By Lemma2.1and the monotonicity of the functiony= lnx, we have n
r
lnφr = X
1≤i1<···<ir+1≤n
ln Pr
j=1 1 1−xij
Pr j=1
1 xij
≤ X
1≤i1<···<ir+1≤n
ln
Qr
j=1 1 1−xij
Qr j=1
1 xij
1 r
= 1 r
X
1≤i1<···<ir≤n r
X
j=1
ln
1 1−xij
1 xij
.
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By knowledge of combinations, we can easily find n
r
lnφr≤ 1 r ln
" n Y
i=1 1 1−xi
1 xi
#(n−1r−1)
= 1 r
n−1 r−1
ln
" n Y
i=1 1 1−xi
1 xi
#
= 1 r
n−1 r−1
lnφ1 =n r
lnφ1. Thus
φr ≤φ1, r = 2, . . . , n, or
(5.5) Hnr(x)
Hnr(1−x) ≤ Gn(x)
Gn(1−x), r = 2, . . . , n.
The inequality (5.5) generalizes the “Ky Fan” inequality.
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[3] J. SÁNDOR, On an inequaliy of Ky Fan, Babe¸s-Bolyai Univ., Fac. Math.
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[4] J. SÁNDOR AND T. TRIF, A new refinement of the Ky Fan inequality, Math. Inequal. Appl., 2 (1999), 529–533.
[5] W.-L. WANGANDP.-F. WANG, A class of inequalities for symmetric func- tions (in chinese), Acta. Math .Sinica, 27 (1984), 485–497.
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