Volume 2010, Article ID 509323,8pages doi:10.1155/2010/509323
Research Article
A Note on Mixed-Mean Inequalities
Peng Gao
Division of Mathematical Sciences, School of Physical and Mathematical Sciences, Nanyang Technological University, Singapore 637371
Correspondence should be addressed to Peng Gao,[email protected]
Received 25 February 2010; Accepted 29 June 2010
Academic Editor: Ram N. Mohapatra
Copyrightq2010 Peng Gao. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We give a simpler proof of a result of Holland concerning a mixed arithmetic-geometric mean inequality. We also prove a result of mixed-mean inequality involving the symmetric means.
1. Introduction
Let Mn,rx be the generalized weighted power means:Mn,rq,x ni1qixri1/r, where
q q1, q2, . . . , qn,x x1, x2, . . . , xn, andqi >0,1 ≤i≤n, with n
i1qi 1. HereMn,0q,x denotes the limit ofMn,rq,xasr → 0. Unless specified, we always assume thatxi>0,1≤ i≤n. When there is no risk of confusion, we will writeMn,rforMn,rq,xand we also define
AnMn,1, GnMn,0,and HnMn,−1.
The celebrated Hardy’s inequalitysee1, Theorem 326 asserts that, forp >1, an≥0,
∞
n1 n
k1ak n
p
≤
p
p−1 p∞
n1
apn. 1.1
Among the many different proofs of Hardy’s inequality as well as its generalizations and extensions in the literature, one novel approach is via the mixed-mean inequalitiessee, e.g.,2, Theorem 7. By mixed-mean inequalities, we will mean the following inequalities:
m
n1 am,n
m
k1 bn,kxk
p1/p
≤m
n1 bm,n
m
k1 an,kxpk
1/p
, 1.2
whereai,j,bi,jare twom×mmatrices with nonnegative entries and the above inequality
The meaning of mixed mean becomes more clear whenai,j,bi,jare weighted mean
matrices. Here we say that a matrixA an,kis a weighted mean matrix ifan,k0 forn < k and
an,k Wwk
n, 1≤k≤n, Wn n
i1
wi, wi≥0, w1>0. 1.3
Now we focus our attention to the case of1.2forai,j bi,jbeing weighted mean
matrices given in1.3. In this case, for fixedx x1, . . . , xn,w w1, . . . , wn, we definexi
x1, . . . , xi,wi w1, . . . , wi, Wi ij1wj,Mi,r Mi,rxi Mi,rwi/Wi,xi,andMi,r
M1,r, . . . , Mi,r. Then we have the following mixed-mean inequalities of Nanjundiah3 see
also4.
Theorem 1.1. Letr > sandn≥2. If, for2≤k≤n−1,Wnwk−Wkwn>0, then
Mn,sMn,r≥Mn,rMn,s, 1.4
with equality holding if and only ifx1· · ·xn.
A very elegant proof of Theorem 1.1for the caser 1, s 0 is given by Kedlaya in5. In fact, the following Popoviciu-type inequalities were established in5 see also4, Theorem 9.
Theorem 1.2. Letn≥2. If, for2≤k≤n−1,Wnwk−Wkwn>0, then
Wn−1lnMn−1,0Mn−1,1−lnMn−1,1Mn−1,0≤WnlnMn,0Mn,1−lnMn,1Mn,0, 1.5
with equality holding if and only ifxnMn−1,0Mn−1,1Mn−1,0.
It is easy to see that the caser 1, s 0 ofTheorem 1.1follows fromTheorem 1.2. As was pointed out by Kedlaya that the method used in5can be applied to establish both Popoviciu-type and Rado-type inequalities for mixed means for a general pair r > s, the details were worked out in6and the following Rado-type inequalities were established in
6.
Theorem 1.3. Let1> sandn≥2. If, for2≤k≤n−1,Wnwk−Wkwn>0, then
Wn−1Mn−1,sMn−1,1−Mn−1,1Mn−1,s≤WnMn,sMn,1−Mn,1Mn,s, 1.6
with equality holding if and only ifx1· · ·xnand the above inequality reverses whens >1.
A different proof ofTheorem 1.1for the caser1, s0 was given in7and Bennett used essentially the same approach in8,9to study1.2for the casesai,j,bi,jbeing lower
triangular matrices, namely,ai,j bi,j 0 ifj > i. Among other things, he showed8that inequalities1.2hold whenai,j,bi,jare Hausdorffmatrices.
Theorem 1.4. Letn≥2. If, for2≤k≤n−1,Wk2≥wk1ki−11Wi, then
Wn−1Mn−1,0Mn−1,1−Mn−1,1Mn−1,0≤WnMn,0Mn,1−Mn,1Mn,0, 1.7
with equality holding if and only ifx1· · ·xn.
It is our goal in this paper to first give a simpler proof of the above result by modifying Holland’s own approach. This is done in the next section and, inSection 3, we will prove a result of mixed-mean inequality involving the symmetric means.
2. A Proof of
Theorem 1.4
First, we recast1.7as
GnAn−WWn−1
n Gn−1An−1− wn
WnGn≥0. 2.1
We now note that
GnAn Gn−1An−1Wn−1/WnAwnn/Wn,
Gn−1An−1 An
n−1
i1 A
i Ai1
Wi/Wn−1
. 2.2
We may assume thatxk>0,1≤k≤n, and the casexk0 for somekwill follow by continuity. Thus on dividingGnAnon both sides of2.1and using2.2, we can recast2.1as
Wn−1 Wn
n−1
i1 A
i Ai1
Wiwn/Wn−1Wn
Wwn
n n
i1 x
i Ai
wi/Wn
≤1. 2.3
We now express thatxi WiAi−Wi−1Ai−1/wi,1≤i≤n, withW0A00 to recast2.3as
Wn−1 Wn
n−1
i1 A
i Ai1
Wiwn/Wn−1Wn
Wwn
n n
i1 W
iAi−Wi−1Ai−1 wiAi
wi/Wn
≤1. 2.4
We now setyi Ai/Ai1, 1≤i≤2, to further recast the above inequality as
Wn−1 Wn
n−1
i1
yWiwn/Wn−1Wn
i Wwnn
n−1
i1 W
i1 wi1 −
Wi wi1yi
wi1/Wn
≤1. 2.5
It now follows from the assumption ofTheorem 1.4that
cn1− n−1
i1 Wiwn Wn−1Wn ≥
so that by the arithmetic-geometric mean inequality we have
n−1
i1
yWiwn/Wn−1Wn
i 1cn
n−1
i1
yWiwn/Wn−1Wn
i ≤
n−1
i1
Wiwnyi Wn−1Wn 1−
n−1
i1 Wiwn
Wn−1Wn. 2.7
Similarly, we have
n−1
i1 W
i1 wi1 −
Wi wi1yi
wi1/Wn
≤n−1
i1 wi1
Wn
W
i1 wi1 −
Wi wi1yi
w1
Wn. 2.8
Now it is easy to see that inequality2.5follows on adding inequalities2.7and2.8, and this completes the proof ofTheorem 1.4.
3. A Discussion on Symmetric Means
Let 0≤r≤n; we recall that therth symmetric functionEn,rofxand its meanPn,rare defined
by
En,rx
1≤i1<···<ir≤n r
j1
xij, Pn,rr x
En,rx
n
r , 1≤r≤n, En,0Pn,01. 3.1
It is well known that, for fixed x of dimensionn,Pn,r is a nonincreasing function of r for
1 ≤ r ≤ nwithPn,1 An, Pn,n Gn with weightswi 1, 1 ≤i≤ n. In view of the
mixed-mean inequalities for the generalized weighted power mixed-meansTheorem 1.1, it is natural to ask whether similar results hold for the symmetric means. Of course one may have to adjust the notion of such mixed means in order for this to make sense for alln. For example, when
r 3,n 2, the notion ofP2,3is not even defined. From now on, we will only focus on the extreme cases of the symmetric means; namely,r 2 orr n−1. In these cases it is then natural to defineP1,2 x1, and, on recastingPn,n−1 Gnn/n−1/Hn1/n−1, we see that it is also
natural for us to defineP1,0 x1 note that this is not consistent with our definition ofPn,0 above.
We now prove a mixed-mean inequality involving Pn,2 and An. We first note the
following result of Marcus and Lopes11 see also12, pages 33–35.
Theorem 3.1. Let0< r≤nandxi, yi>0fori1,2, . . . , n. Then
Pn,rxy≥Pn,rx Pn,ry, 3.2
with equality holding if and only ifr 1or there exists a constantλsuch thatxλy.
Lemma 3.2. Letf :R1 → R1be a convex function and suppose that forn≥2,1≤k≤n−1, that Wnwk−Wkwn>0. Then
1
Wn−1
n−1
k1
wkfWn−1Ak≥ W1 n
n
k1
wkfWnAk−wnxk. 3.3
The equality holds if and only ifn 2orx1 · · ·xnwhenfxis strictly convex. Whenfxis
concave, then the above inequality is reversed.
We now applyLemma 3.2to obtain the following.
Lemma 3.3. Forn≥2andwi1,1≤i≤n,
Pn−1,2n−1An−1≤Pn,2nAn−xn, 3.4
with equality holding in both cases if and only ifn2orx1· · ·xn.
Proof. The casen 2 yields an identity, so we may assume that n ≥ 3 here. Write ai
n−1Ai,1 ≤ i ≤ n−1;bj nAj −xj,1 ≤ j ≤ n. Note that nni−11ai n−1ni1bi, and nowLemma 3.2withfx x2implies thatn−1n
i1b2i ≤ n
n−1
i1 a2i. On expanding
nn−1
i1 ai2 n−1 n
i1bi2, we obtain
n2n−1
i1 a2
i 2n2
1≤i /j≤n−1
aiaj n−12 n
i1 b2
i 2n−12
1≤i /j≤n bibj
≤nn−1
n−1
i1 a2
i 2n−12
1≤i /j≤n bibj.
3.5
Hence,
nn−1 i1
a2
i 2n2
1≤i /j≤n−1
aiaj≤2n−12
1≤i /j≤n
bibj. 3.6
UsingMn,2≥AnPn,1≥Pn,2, we obtain
1
n−1
n−1
i1 a2
i ≥ n1−1 2
1≤i /j≤n−1
aiaj. 3.7
So by3.6,
1
n−1 2
1≤i /j≤n−1
aiaj≤ 1n
2
1≤i /j≤n
bibj, 3.8
We now prove the following mixed-mean inequality involving the symmetric means.
Theorem 3.4. Letn≥1and definePn,2 P1,2, . . . , Pn,2. Then
n−1Pn−1,2Pn−1,1−Pn−1,1Pn−1,2≤nPn,2Pn,1−Pn,1Pn,2, 3.9
with equality holding if and only ifx1· · ·xn. It follows that
Pn,1Pn,2≤Pn,2Pn,1, 3.10
with equality holding if and only ifx1· · ·xn.
Proof. It suffices to prove3.9here. We may assume thatn≥2 here and we will use the idea
in6.Lemma 3.3implies that
Pn,2 n−1Pn−1,2Pn−1,1≤Pn,2Pn,2nAn−xn
≤Pn,2nAn−xnxn nPn,2Pn,1,
3.11
where the last inequality follows fromTheorem 3.1for the caser2. It is easy to see that the above inequality is equivalent to3.9and this completes the proof.
Now we letn ≥ 1 and definePn,n−1 P1,0, . . . , Pn,n−1withP1,0 x1here. Then it is interesting to see whether the following inequality holds or not:
Pn,1Pn,n−1≤Pn,n−1Pn,1. 3.12
We note here that, if the above inequality holds, then it is easy to deduce from it via the approach in2the following Hardy-type inequality:
n
i1 Gi/i−1
i H1/i−1
i
xi ≤e n
i1
xi, 3.13
where we defineG11/0/H11/0x1. We now end this paper by proving the following result.
Theorem 3.5. Letn≥1andx≥0. Then
n
i1
Gi/i−1
i H1/i−1
i
xi ≤3 n
i1
xi, 3.14
Proof. We follow an approach of Knopp13,14heresee also15. Fori≥1, we define
ai i
k1 kxk
ii1. 3.15
It is easy to check by partial summation that
n
i1 ai≤
n
i1
xi. 3.16
Certainly, we havea1 x1/2P1,0x1/2 and, fori≥2, we apply the inequalityPi,1 ≥ Pi,i−1 to the numbersx1/i1,2x2/i1, . . . , ixi/i1to see that
ai≥
i−1!
i1i−1
1/i−1
Pi,i−1xi:γiPi,i−1xi. 3.17
We now show by induction thatγi≥1/3 fori≥2; equivalently, this is
3i−1i−1!≥i1i−1. 3.18
Note first that the above inequality holds wheni 2,3 and suppose now that it holds for someik≥3. Then by induction,
3kk!≥3kk1k−1. 3.19
Now using11/nn< e, we have
3kk1k−1
k2k
3kk2
k12 k
1
k2 k1
≥ 3kk2
ek12. 3.20
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