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Abstract and Applied Analysis
Volume 2011, Article ID 520648, 9 pages
doi:10.1155/2011/520648
Research Article
Optimal Lower Power Mean Bound for the Convex
Combination of Harmonic and Logarithmic Means
Yu-Ming Chu,1 Shan-Shan Wang,2 and Cheng Zong2
1
2
Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China
School of Science, Hangzhou Normal University, Hangzhou 310012, China
Correspondence should be addressed to Yu-Ming Chu, chuyuming2005@yahoo.com.cn
Received 1 February 2011; Accepted 15 May 2011
Academic Editor: Irena Lasiecka
Copyright q 2011 Yu-Ming Chu et al. 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 find the least value λ ∈ 0, 1 and the greatest value p pα such that αHa, b1−αLa, b >
Mp a, b for α ∈ λ, 1 and all a, b > 0 with a /
b, where Ha, b, La, b, and Mp a, b are the
harmonic, logarithmic, and p-th power means of two positive numbers a and b, respectively.
1. Introduction
For p ∈ R, the p-th power mean Mp a, b and logarithmic mean La, b of two positive
numbers a and b are defined by
⎧
p
p 1/p
⎪
⎨ a b
, p/
0,
Mp a, b √ 2
⎪
⎩ ab,
p 0,
⎧
b−a
⎪
⎨
, a/
b,
log
b − log a
La, b ⎪
⎩a,
a b,
1.1
1.2
respectively.
It is well known that Mp a, b is continuous and strictly increasing with respect to
p ∈ R for fixed a, b > 0 with a /
b. In the recent past, both mean values have been the
subject of intensive research. In particular, many remarkable inequalities for Mp a, b and
La, b can be found in the literature 1–17. It might be surprising that the logarithmic
2
Abstract and Applied Analysis
mean has applications in physics, economics, and even in meteorology 18–20. In 18,
the authors study a variant of Jensen’s functional equation involving L, which appears in
a heat conduction problem. A representation of L as an infinite product and an iterative
algorithm for computing the logarithmic mean as the common limit of two sequences of
special geometric and arithmetic means are given in 8. In 21, 22, it is shown that L can be
expressed in terms of Gauss’s hypergeometric function 2 F1 . And, in 21, the authors prove
that the reciprocal of the logarithmic mean is strictly totally positive, that is, every n × n
determinant with elements 1/Lai , bi , where 0 < a1 < a2 < · · · < an and 0 < b1 < b2 < · · · < bn ,
is positive for all n ≥ 1.
1/b−a
b /
a, Ia, b a b a, Ga, b Let Aa, b 1/2a b, Ia, b 1/ebb /aa √
ab, and Ha, b 2ab/a b be the arithmetic, identric, geometric, and harmonic means
of two positive numbers a and b, respectively, then it is well known that
min{a, b} < Ha, b M−1 a, b < Ga, b M0 a, b
< La, b < Ia, b < Aa, b M1 a, b < max{a, b}
1.3
for all a, b > 0 with a /
b.
In 23, Alzer and Janous established the following best possible inequality:
Mlog 2/ log 3 a, b <
1
2
Aa, b Ga, b < M2/3 a, b
3
3
1.4
for all a, b > 0 with a / b.
In 8, 11, 24, the authors presented bounds for L in terms of G and A
G2/3 a, bA1/3 a, b < La, b <
1
2
Ga, b Aa, b
3
3
1.5
for all a, b > 0 with a /
b.
The following companion of 1.3 provides inequalities for the geometric and
arithmetic means of L and I. A proof can be found in 25
G1/2 a, bA1/2 a, b < L1/2 a, bI 1/2 a, b <
1
1
1
1
La, b Ia, b < Ga, b Aa, b 1.6
2
2
2
2
for all a, b > 0 with a /
b.
The following sharp bounds for L, I, LI1/2 , and LI/2 in terms of the power means
are proved in 4, 5, 7, 9, 16, 25, 26:
M0 a, b < La, b < M1/3 a, b,
M2/3 a, b < Ia, b < Mlog 2 a, b,
M0 a, b < L1/2 a, bI 1/2 a, b < M1/2 a, b,
1
1
La, b Ia, b < M1/2 a, b
2
2
for all a, b > 0 with a /
b.
1.7
Abstract and Applied Analysis
3
Alzer and Qiu 27 found the sharp bound of 1/2La, b Ia, b in terms of the
power mean as follows:
Mc a, b <
1
La, b Ia, b
2
1.8
for all a, b > 0 with a /
b, with the best possible parameter c log 2/1 log 2.
The main purpose of this paper is to find the least value λ ∈ 0, 1 and the greatest
value p pα such that αHa, b 1 − αLa, b > Mp a, b for α ∈ λ, 1 and all a, b > 0
with a /
b.
2. Lemmas
In order to establish our main result we need three lemmas, which we present in this section.
Lemma 2.1. Let α ∈ 1/4, 1, p 1 − 4α/3 ∈ −1, 0, and ft −4αpp 12 p 2tp−1 21 −
αp2 1 − p2 tp−2 21 − αp1 − p2 2 − ptp−3 121 − α1 − p. Then ft > 0 for t ∈ 1, ∞.
Proof. Simple computations lead to
64
1 − α2 56α2 23α 11 > 0,
81
lim ft 121 − α 1 − p 81 − α1 2α > 0,
f1 t → ∞
f t −2p 1 − p tp−4 f1 t,
2.1
2.2
2.3
where
2 f1 t −2α p 1 p 2 t2 1 − αp p 1 2 − p t 1 − α 1 − p 2 − p 3 − p ,
f1 1 4
1 − α 148α2 − 11α 25 > 0,
27
lim f1 t −∞,
t → ∞
2 f 1 t −4α p 1 p 2 t 1 − αp p 1 2 − p
−
4
1 − α2 16α7 − 4αt 4α − 14α 5 < 0
27
2.4
2.5
2.6
for t ∈ 1, ∞.
Inequality 2.6 implies that f1 t is strictly decreasing in 1, ∞, then from 2.4 and
2.5 we know that λ1 > 1 exists such that f1 t > 0 for t ∈ 1, λ1 and f1 t < 0 for t ∈ λ1 , ∞.
Hence, equation 2.3 leads to the conclusion that ft is strictly increasing in 1, λ1 and
strictly decreasing in λ1 , ∞.
Therefore, Lemma 2.1 follows from 2.1 and 2.2 together with the piecewise
monotonicity of ft.
4
Abstract and Applied Analysis
Lemma 2.2. Let α ∈ 1/4, 1, p 1 − 4α/3 ∈ −1, 0, and gt −1 − αp 1p 22 p 3tp p 1p3 − αp3 − 19αp2 3p2 − 34αp 2p − 8αtp−1 1 − αpp3 − 8p2 − p 4tp−2 1 −
α1 − pp3 5p2 − 14p 4tp−3 41 − α7 − 4p − 4p1 − αt−1 4α1 pt−2 , then gt > 0 for
t ∈ 1, ∞.
Proof. Let g1 t −1 − αp 1p 22 p 3t3 p 1p3 − αp3 − 19αp2 3p2 − 34αp 2p −
8αt2 1 − αpp3 − 8p2 − p 4t 1 − α1 − pp3 5p2 − 14p 4 41 − α7 − 4pt3−p −
4p1 − αt2−p 4α1 pt1−p . Then simple computations lead to
gt tp−3 g1 t,
g1 1 16
1 − α 80α2 110α − 1 > 0,
27
2.7
2.8
g1 t −31 − αp 1p 22 p 3t2 2p 1p3 − αp3 − 19αp2 3p2 − 34αp 2p − 8αt 1 − αpp3 − 8p2 − p 4 41 − α7 − 4p3 − pt2−p − 4p1 − α2 − pt1−p 4α1 − p2 t−p ,
g1 1 32
1 − α −16α3 38α2 176α − 9 > 0,
27
2.9
g1 t −61 − αp 1p 22 p 3t 2p 1p3 − αp3 − 19αp2 3p2 − 34αp 2p − 8α 41 −
α7 − 4p3 − p2 − pt1−p − 4p1 − α2 − p1 − pt−p − 4αp1 − p2 t−p−1 ,
g1 1 8
1 − α −128α4 896α3 288α2 5294α − 437
81
2.10
> 0,
g1 t −61 − αp 1p 22 p 3 41 − α7 − 4p3 − p2 − p1 − pt−p 4p2 1 − α2 −
p1 − pt−p−1 4αp1 p2 1 − pt−p−2
8
1 − α 576α4 3872α3 660α2 6612α − 785
81
> 0,
g
1 1 4
g1 t −4p 1 − p t−p−3 g2 t,
2.11
2.12
where
2 g2 t 1 − α 7 − 4p 3 − p 2 − p t2 1 − αp 2 − p p 1 t α p 1 p 2 ,
4
1 − α 96α3 232α2 388α 175 > 0,
27
g2 t 21 − α 7 − 4p 3 − p 2 − p t 1 − αp 2 − p p 1
4
≥ g2 1 1 − α5 4α 12α2 31α 23 > 0
9
g2 1 for t ∈ 1, ∞.
2.13
2.14
Abstract and Applied Analysis
5
From 2.13 and 2.14, we clearly see that g2 t > 0 for t ∈ 1, ∞, then 2.12 leads to
the conclusion that g1 t is strictly in 1, ∞.
Therefore, Lemma 2.2 follows from 2.7–2.11 and the monotonicity of g1 t.
Lemma 2.3. Let α ∈ 1/4, 1, p 1 − 4α/3 ∈ −1, 0, and ht 2α1 − tp1 tlog2 t 1 − α1 tp−1 1 t2 t log t 1 − α1 t2 1 − ttp 1, then ht > 0 for t ∈ 1, ∞.
Proof. Let h1 t t−p h t and h2 t tp2 h1 t, then simple computations lead to
h1 0,
2.15
h t 2α1 − p 2tp1 log2 t p 2 − αp − 6αtp1 21 − αp 1tp 1 − αptp−1 31 −
αt2 41 − αt 3α 1 log t − 1 − αp 3tp2 p 1tp1 − p 3tp − p 1tp−1 2t2 − 2,
h 1 0,
2.16
h1 t −2αp 1p 2log2 t p2 − αp2 3p − 11αp − 14α 2 21 − αpp 1t−1 − 1 −
αp1 − pt−2 61 − αt1−p 41 − αt−p 4αt−1−p log t − 1 − αp 2p 3t 1 − αp2 5p 2t−1 1 − αp2 p − 1t−2 − 1 − αt1−p 41 − αt−p 1 3αt−1−p − 1 − αp2 − 1 − αp − 5α 1,
h1 1 0,
2.17
h2 t −4αp 1p 2tp1 21 − αpp 1tp − 21 − αp1 − ptp−1 − 61 − α1 − pt2 41 − αpt 4α1 p log t − 1 − αp 2p 3tp2 p2 − αp2 3p − 11αp − 14α 2tp1 1 −
αp2 − 3p − 2tp − 1 − αp2 3p − 2tp−1 1 − αp 5t2 41 − α1 − pt α − 3αp − p − 1,
h2 1 0,
2.18
h2 t −4αp 12 p 2tp 21 − αp2 p 1tp−1 21 − αp1 − p2 tp−2 − 121 − α1 − pt 41 − αp log t − 1 − αp 22 p 3tp1 p 1p2 − αp2 − 15αp 3p − 22α 2tp 1 − αpp2 −
5p − 4tp−1 1 − α1 − pp2 5p − 2tp−2 41 − α4 − pt − 4αp 1t−1 41 − α1 − 2p,
h2 1 0,
h2 t ft log t gt,
2.19
where ft and gt are defined as in Lemmas 2.1 and 2.2, respectively.
From 2.19 and 2.10 together with Lemmas 2.1 and 2.2, we clearly see that h2 t is
strictly increasing in 1, ∞.
Therefore, Lemma 2.3 follows from 2.15–2.18 and the monotonicity of h2 t.
6
Abstract and Applied Analysis
3. Main Result
Theorem 3.1. Inequality
αHa, b 1 − αLa, b > M1−4α/3 a, b
3.1
holds for α ∈ 1/4, 1 and all a, b > 0 with a /
b, and M1−4α/3 a, b is the best possible lower power
mean bound for the sum αHa, b 1 − αLa, b.
Proof. We divide the proof of inequality 3.1 into two cases.
Case 1 α 1/4. Without loss of generality, we assume that a > b and put t from 1.1 and 1.2, we have
a/b > 1, then
αHa, b 1 − αLa, b − M1−4α/3 a, b
1
Ha, b 3La, b − ab
4
3t4 − 4 2t3 − t2 2t log t − 3
b.
8t2 1 log t
3.2
Let
Ft 3t4 − 4 2t3 − t2 2t log t − 3,
3.3
then simple computations lead to
F1 0,
F t 4 3t3 − 2t2 t − 2 − 8 3t2 − t 1 log t,
F 1 0,
F t 3.4
4
F1 t,
t
where F1 t 9t3 − 10t2 3t − 2 − 26t − 1t log t,
F 1 F1 1 0,
F1 t 27t2 − 32t 5 − 212t − 1 log t,
F1 1 0,
F1 t 2
F2 t,
t
3.5
Abstract and Applied Analysis
7
where F2 t 27t2 − 12t log t − 28t 1,
F1 1 F2 1 0,
3.6
F2 t 54t − 12 log t − 40 > 0
for t > 1.
Therefore, inequality 3.1 follows easily from 3.2–3.6.
Case 2 α ∈ 1/4, 1. Without loss of generality, we assume that a > b. Let p 1 − 4α/3 ∈
−1, 0 and t a/b > 1, then from 1.1 and 1.2, one has
αHa, b 1 − αLa, b − M1−4α/3 a, b
αHa, b 1 − αLa, b − Mp a, b
p
t 1 1/p
2αt
1 − αt − 1
.
−
b
t1
log t
2
3.7
Let
Gt log
2αt
1
tp 1
1 − αt − 1
.
− log
t1
log t
p
2
3.8
Then simple computations lead to
lim Gt 0,
3.9
t→1
G t tt 1tp
ht
,
1 log t 2αt log t 1 − αt2 − 1
3.10
where ht is defined as in Lemma 2.3.
From Lemma 2.3 and 3.10, we clearly see that Gt is strictly increasing in 1, ∞.
Therefore, inequality 3.1 follows from 3.7–3.9 and the monotonicity of Gt.
Next, we prove that M1−4α/3 a, b is the best possible lower power mean bound for
the sum αHa, b 1 − αLa, b if α ∈ 1/4, 1.
For any α ∈ 1/4, 1, p > 1 − 4α/3, and x > 0, one has
Mp 1 x, 1 − αH1 x, 1 − 1 − αL1 x, 1 21/p
Jx
,
x 1
log1 x
2
3.11
8
Abstract and Applied Analysis
where Jx 1 x/21 1 xp log1 x − 21/p α1 x log1 x 1 − αx1 x/2.
Letting x → 0 and making use of Taylor expansion, we have
1/p
Jx 1 − 4α 3
21/p
p−
x o x3 .
8
3
3.12
Equations 3.11 and 3.12 imply that for any α ∈ 1/4, 1 and p > 1 − 4α/3 there
exists δ > 0, such that αH1 x, 1 1 − αL1 x, 1 < Mp 1 x, 1 for x ∈ 0, δ.
Remark 3.2. If 0 < α < 1/4, then from 1.1 and 1.2, we have
M1−4α/3 1, x
23/4α−1
x → ∞ αH1, x 1 − αL1, x
lim
3/1−4α
1 x4α−1/3
∞.
x → ∞ 2α/x 1 1 − 1/x1 − α/ log x
3.13
× lim
Equation 3.13 implies that for any 0 < α < 1/4, there exists X > 1, such that
M1−4α/3 1, x > αH1, x 1 − αL1, x for x ∈ X, ∞. Therefore, λ 1/4 is the least
value of λ in 0, 1 such that inequality 3.1 holds for all a, b > 0 with a /
b.
Acknowledgments
This research is supported by the N. S. Foundation of China under Grant 11071069, N.
S. Foundation of Zhejiang province under Grants nos. Y7080106 and Y6100170, and the
Innovation Team Foundation of the Department of Education of Zhejiang Province under
Grant no. T200924.
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