Volume 2010, Article ID 509323, 8 pagesdoi:10.1155/2010/509323 Research Article A Note on Mixed-Mean Inequalities Peng Gao Division of Mathematical Sciences, School of Physical and Mathe
Trang 1Volume 2010, Article ID 509323, 8 pages
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,penggao@ntu.edu.sg
Received 25 February 2010; Accepted 29 June 2010
Academic Editor: Ram N Mohapatra
Copyrightq 2010 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 M n,r x be the generalized weighted power means: M n,r q, x n
i1 q i x r
i1/r, where
q q1, q2, , q n , x x1, x2, , x n , and q i > 0, 1 ≤ i ≤ n, withn i1 q i 1 Here M n,0 q, x
denotes the limit ofM n,r q, x as r → 0 Unless specified, we always assume thatx i > 0, 1 ≤
i ≤ n When there is no risk of confusion, we will write M n,rforM n,r q, x and we also define
A n M n,1 , G n M n,0 , and H n M n,−1
The celebrated Hardy’s inequalitysee 1, Theorem 326 asserts that, for p > 1, an≥ 0,
∞
n1
n
k1 a k
n
p
p − 1
p ∞
n1
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:
n1
a m,n
k1
b n,k x k
p1/p
≤m
n1
b m,n
k1
a n,k x p k
1/p
wherea i,j , b i,j are two m × m matrices with nonnegative entries and the above inequality
being meant to hold for any vectorx ∈ Rmwith nonnegative entries Herep ≥ 1 and, when
0< p ≤ 1, we want the inequality above to be reversed.
Trang 2The meaning of mixed mean becomes more clear whena i,j , b i,j are weighted mean matrices Here we say that a matrixA a n,k is a weighted mean matrix if a n,k 0 for n < k
and
a n,k W w k
n , 1 ≤ k ≤ n, W nn
i1
w i , w i ≥ 0, w1> 0. 1.3
Now we focus our attention to the case of1.2 for a i,j b i,j being weighted mean matrices given in1.3 In this case, for fixed x x1, , x n , w w1, , w n, we define xi
x1, , x i , w i w1, , w i , W i i
j1 w j,M i,r M i,rxi M i,rwi /W i , x i , and M i,r
M1,r , , M i,r Then we have the following mixed-mean inequalities of Nanjundiah 3 see also4
Theorem 1.1 Let r > s and n ≥ 2 If, for 2 ≤ k ≤ n − 1, W n w k − W k w n > 0, then
with equality holding if and only if x1 · · · x n
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 in 5 see also 4, Theorem 9
Theorem 1.2 Let n ≥ 2 If, for 2 ≤ k ≤ n − 1, W n w k − W k w n > 0, then
W n−1 ln M n−1,0Mn −1,1 − ln M n−1,1Mn−1,0 ≤ W n ln M n,0Mn,1 − ln M n,1Mn,0 , 1.5
with equality holding if and only if x n M n−1,0 M n−1,1 M n−1,0 .
It is easy to see that the caser 1, s 0 of Theorem 1.1follows fromTheorem 1.2
As was pointed out by Kedlaya that the method used in5 can 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 in6 and the following Rado-type inequalities were established in
6
Theorem 1.3 Let 1 > s and n ≥ 2 If, for 2 ≤ k ≤ n − 1, W n w k − W k w n > 0, then
W n−1 M n−1,sMn−1,1 − M n−1,1Mn−1,s ≤ W n M n,sMn,1 − M n,1Mn,s , 1.6
with equality holding if and only if x1 · · · x n and the above inequality reverses when s > 1.
A different proof ofTheorem 1.1for the caser 1, s 0 was given in 7 and Bennett used essentially the same approach in8,9 to study 1.2 for the cases a i,j , b i,j being lower triangular matrices, namely,a i,j b i,j 0 if j > i Among other things, he showed 8 that inequalities1.2 hold when a i,j , b i,j are Hausdorff matrices
In10, Holland further improved the condition inTheorem 1.3for the cases 0 by
proving the following
Trang 3Theorem 1.4 Let n ≥ 2 If, for 2 ≤ k ≤ n − 1, W2
i1 W i , then
W n−1 M n−1,0Mn−1,1 − M n−1,1Mn−1,0 ≤ W n M n,0Mn,1 − M n,1Mn,0 , 1.7
with equality holding if and only if x1 · · · x n
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
First, we recast1.7 as
G nAn −W W n−1
n G n−1An−1 −W w n
We now note that
G nAn G n−1An−1W n−1 /W n A w n /W n
G n−1An−1 A n
n−1
i1
i
A i1
W i /W n−1
We may assume thatx k > 0, 1 ≤ k ≤ n, and the case x k 0 for some k will follow by continuity.
Thus on dividingG nAn on both sides of 2.1 and using 2.2, we can recast 2.1 as
W n−1
W n
n−1
i1
i
A i1
W i w n /W n−1 W n
W w n
n
n
i1
x
i
A i
w i /W n
We now express thatx i W i A i − W i−1 A i−1 /w i , 1 ≤ i ≤ n, with W0 A0 0 to recast 2.3 as
W n−1
W n
n−1
i1
i
A i1
W i w n /W n−1 W n
W w n
n
n
i1
i A i − W i−1 A i−1
w i A i
w i /W n
We now sety i A i /A i1, 1≤ i ≤ 2, to further recast the above inequality as
W n−1
W n
n−1
i1
y W i w n /W n−1 W n
n
n−1
i1
i1
w i1 −w W i
i1 y i
w i1 /W n
It now follows from the assumption ofTheorem 1.4that
c n 1 −n−1
i1
W i w n
Trang 4so that by the arithmetic-geometric mean inequality we have
n−1
i1
y W i w n /W n−1 W n
n−1
i1
y W i w n /W n−1 W n
i1
W i w n y i
W n−1 W n 1 −
n−1
i1
W i w n
W n−1 W n . 2.7
Similarly, we have
n−1
i1
i1
w i1 − W i
w i1 y i
w i1 /W n
≤n−1
i1
w i1
W n
i1
w i1 − W i
w i1 y i
w1
Now it is easy to see that inequality2.5 follows on adding inequalities 2.7 and 2.8, and this completes the proof ofTheorem 1.4
3 A Discussion on Symmetric Means
Let 0≤ r ≤ n; we recall that the rth symmetric function E n,rofx and its mean P n,rare defined by
E n,rx
1≤i 1<···<ir ≤n
r
j1
x i j , P r n,rx E n,rnx
r , 1 ≤ r ≤ n, E n,0 P n,0 1. 3.1
It is well known that, for fixed x of dimension n, P n,r is a nonincreasing function of r for
1 ≤ r ≤ n with P n,1 A n , P n,n G n with weights w i 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 of P2,3is not even defined From now on, we will only focus on the extreme cases of the symmetric means; namely,r 2 or r n − 1 In these cases it is then
natural to defineP1,2 x1, and, on recastingP n,n−1 G n/n−1 n /H1/n−1
n , we see that it is also natural for us to defineP1,0 x1 note that this is not consistent with our definition of P n,0
above
We now prove a mixed-mean inequality involving P n,2 and A n We first note the following result of Marcus and Lopes11 see also 12, pages 33–35
Theorem 3.1 Let 0 < r ≤ n and x i , y i > 0 for i 1, 2, , n Then
with equality holding if and only if r 1 or there exists a constant λ such that x λy.
We also need the following lemma of C D Tarnavas and D D Tarnavas6
Trang 5Lemma 3.2 Let f : R1 → R1be a convex function and suppose that for n ≥ 2, 1 ≤ k ≤ n − 1, that
W n w k − W k w n > 0 Then
1
W n−1
n−1
k1
w k fW n−1 A k ≥ W1
n
n
k1
The equality holds if and only if n 2 or x1 · · · x n when fx is strictly convex When fx is concave, then the above inequality is reversed.
We now applyLemma 3.2to obtain the following
Lemma 3.3 For n ≥ 2 and w i 1, 1 ≤ i ≤ n,
with equality holding in both cases if and only if n 2 or x1 · · · x n
Proof The case n 2 yields an identity, so we may assume that n ≥ 3 here Write a i
n − 1A i , 1 ≤ i ≤ n − 1; b j nA j − x j , 1 ≤ j ≤ n Note that nn−1
i1 a i n − 1n
i1 b i, and nowLemma 3.2withfx x2implies thatn − 1n
i1 b2
i1 a2
i On expanding
nn−1 i1 a i2 n − 1n i1 b i2, we obtain
n2n−1 i1
a2
1≤i / j≤n−1
a i a j n − 12n
i1
b2
i 2n − 12
1≤i / j≤n
b i b j
≤ nn − 1 n−1
i1
a2
i 2n − 12
1≤i / j≤n
b i b j
3.5
Hence,
nn−1
i1
a2
1≤i / j≤n−1
a i a j ≤ 2n − 12
1≤i / j≤n
UsingM n,2 ≥ A n P n,1 ≥ P n,2, we obtain
1
n − 1
n−1
i1
a2
i ≥ n−11 2
1≤i / j≤n−1
So by3.6,
1 n−1 2
1≤i / j≤n−1
a i a j≤ 1n
2
1≤i / j≤n
which is just what we want
Trang 6We now prove the following mixed-mean inequality involving the symmetric means.
Theorem 3.4 Let n ≥ 1 and define P n,2 P1,2 , , P n,2 Then
n − 1P n−1,2Pn−1,1 − P n−1,1Pn−1,2 ≤ nP n,2Pn,1 − P n,1Pn,2 , 3.9
with equality holding if and only if x1 · · · x n It follows that
with equality holding if and only if x1 · · · x n
Proof It suffices to prove 3.9 here We may assume that n ≥ 2 here and we will use the idea
in6.Lemma 3.3implies that
P n,2 n − 1P n−1,2Pn−1,1 ≤ P n,2 P n,2 nA n− xn
≤ P n,2 nA n− xn xn nP n,2Pn,1 , 3.11
where the last inequality follows fromTheorem 3.1for the caser 2 It is easy to see that the
above inequality is equivalent to3.9 and this completes the proof
Now we letn ≥ 1 and define P n,n−1 P1,0 , , P n,n−1 with P1,0 x1here Then it is interesting to see whether the following inequality holds or not:
We note here that, if the above inequality holds, then it is easy to deduce from it via the approach in2 the following Hardy-type inequality:
n
i1
G i/i−1 i
H1/i−1
i
xi ≤ en
i1
where we defineG1/0
1 /H1/0
1 x1 We now end this paper by proving the following result
Theorem 3.5 Let n ≥ 1 and x ≥ 0 Then
n
i1
G i/i−1 i
H1/i−1
i
xi ≤ 3n
i1
where one defines G1/0
1 /H1/0
1 x1.
Trang 7Proof We follow an approach of Knopp13,14 here see also 15 For i ≥ 1, we define
a ii
k1
kx k
It is easy to check by partial summation that
n
i1
a i≤n
i1
Certainly, we havea1 x1/2 P1,0x1/2 and, for i ≥ 2, we apply the inequality P i,1 ≥ P i,i−1
to the numbersx1/i 1, 2x2/i 1, , ix i /i 1 to see that
a i≥
i − 1!
i 1 i−1
1/i−1
We now show by induction thatγ i ≥ 1/3 for i ≥ 2; equivalently, this is
Note first that the above inequality holds wheni 2, 3 and suppose now that it holds for
somei k ≥ 3 Then by induction,
3k k! ≥ 3kk 1 k−1 3.19
Now using1 1/n n < e, we have
3kk 1 k−1
k 2 k 3kk 2
k 12
k 1
k 2
k1
≥ 3kk 2
ek 12. 3.20
It is easy to see that the last expression above is no less than 1 whenk ≥ 3 and this proves
inequality3.18 for the case i k 1 This completes the proof of the theorem.
Trang 81 G H Hardy, J E Littlewood, and G P´olya, Inequalities, Cambridge University Press, Cambridge, UK,
1952
2 A ˇCiˇzmeˇsija and J Peˇcari´c, “Mixed means and Hardy’s inequality,” Mathematical Inequalities & Applications, vol 1, no 4, pp 491–506, 1998.
3 T S Nanjundiah, “Sharpening of some classical inequalities,” Math Student, vol 20, pp 24–25, 1952.
4 P S Bullen, “Inequalities due to T S Nanjundiah,” in Recent Progress in Inequalities, vol 430, pp.
203–211, Kluwer Academic Publishers, Dordrecht, The Netherlands, 1998
5 K S Kedlaya, “A weighted mixed-mean inequality,” The American Mathematical Monthly, vol 106, no.
4, pp 355–358, 1999
6 C D Tarnavas and D D Tarnavas, “An inequality for mixed power means,” Mathematical Inequalities
& Applications, vol 2, no 2, pp 175–181, 1999.
7 K Kedlaya, “Proof of a mixed arithmetic-mean, geometric-mean inequality,” The American Mathematical Monthly, vol 101, no 4, pp 355–357, 1994.
8 G Bennett, “An inequality for Hausdorff means,” Houston Journal of Mathematics, vol 25, no 4, pp.
709–744, 1999
9 G Bennett, “Summability matrices and random walk,” Houston Journal of Mathematics, vol 28, no 4,
pp 865–898, 2002
10 F Holland, “An inequality between compositions of weighted arithmetic and geometric means,”
Journal of Inequalities in Pure and Applied Mathematics, vol 7, article 159, 8 pages, 2006.
11 M Marcus and L Lopes, “Inequalities for symmetric functions and Hermitian matrices,” Canadian Journal of Mathematics, vol 9, pp 305–312, 1957.
12 E F Beckenbach and R Bellman, Inequalities, Springer, Berlin, Germany, 1961.
13 K Knopp, “ ¨Uber Reihen mit positiven Gliedern,” Journal of the London Mathematical Society, vol 3, pp.
205–211, 1928
14 K Knopp, “ ¨Uber Reihen mit positiven Gliedern Zweite Mitteilung,” Journal of the London Mathematical Society, vol 5, pp 13–21, 1930.
15 J Duncan and C M McGregor, “Carleman’s inequality,” The American Mathematical Monthly, vol 110,
no 5, pp 424–431, 2003