Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2010, Article ID 303286, 13 pages
doi:10.1155/2010/303286
Research Article
Best Possible Inequalities between Generalized
Logarithmic Mean and Classical Means
Yu-Ming Chu1 and Bo-Yong Long2
1
2
Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China
College of Mathematics Science, Anhui University, Hefei 230039, China
Correspondence should be addressed to Yu-Ming Chu, chuyuming2005@yahoo.com.cn
Received 26 November 2009; Revised 16 February 2010; Accepted 16 March 2010
Academic Editor: Lance Littlejohn
Copyright q 2010 Y.-M. Chu and B.-Y. Long. 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 answer the question: for α, β, γ ∈ 0, 1 with α β γ 1, what are the greatest value
p and the least value q, such that the double inequality Lp a, b < Aα a, bGβ a, bH γ a, b <
Lq a, b holds for all a, b > 0 with a / b? Here Lp a, b, Aa, b, Ga, b, and Ha, b denote the
generalized logarithmic, arithmetic, geometric, and harmonic means of two positive numbers a
and b, respectively.
1. Introduction
For p ∈ R the generalized logarithmic mean Lp a, b of two positive numbers a and b with
a
/ b is defined by
⎧
1/p
⎪
p1
p1
⎪
⎪ a −b
⎪
,
⎪
⎪
⎪
p 1a − b
⎪
⎪
⎪
⎪
⎪
⎪
⎨ b 1/b−a
1 b
Lp a, b ,
⎪
e
aa
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
b−a
⎪
⎪
,
⎪
⎪
⎩ log b − log a
p/
0, p /
− 1,
p 0,
1.1
p −1.
It is well known that Lp a, b is continuous and strictly increasing with respect to
p ∈ R for fixed a, b > 0 with a /
b. Recently, the generalized logarithmic mean has been
2
Abstract and Applied Analysis
the subject of intensive research. Many remarkable inequalities and monotonicity results for
the generalized logarithmic mean can be found in the literature 1–9. It might be surprising
that the generalized logarithmic mean has applications in physics, economics, and even in
meteorology 10–13.
Let√Aa, b a b/2, Ia, b 1/ebb /aa 1/b−a , La, b b − a/log b − log a,
Ga, b ab, and Ha, b 2ab/a b be the arithmetic, identric, logarithmic, geometric,
and harmonic means of two positive numbers a and b with a /
b, respectively. Then
min{a, b} < Ha, b < Ga, b L−2 a, b < La, b L−1 a, b
< Ia, b L0 a, b < Aa, b L1 a, b < max{a, b}.
1.2
For p ∈ R, the pth power mean Mp a, b of two positive numbers a and b with a / b is
defined by
Mp a, b ap bp
2
1/p
p/
0 ,
M0 a, b ab1/2 .
1.3
In 14, Alzer and Janous established the following sharp double inequality see also
15, page 350:
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.
For α ∈ 0, 1, Janous 16 found the greatest value p and the least value q such that
Mp a, b < αAa, b 1 − αGa, b < Mq a, b
1.5
for all a, b > 0 with a / b.
In 17–19, the authors presented bounds for La, b and Ia, b in terms of Ga, b and
Aa, b.
Theorem A. For all positive real numbers a and b with a /
b, we have
La, b <
2
1
Aa, b Ga, b,
3
3
1
2
Ga, b Aa, b < Ia, b.
3
3
The proof of the following Theorem B can be found in 20.
1.6
Abstract and Applied Analysis
3
Theorem B. For all positive real numbers a and b with a /
b, we have
Ga, bAa, b <
La, bIa, b
<
1
La, b Ia, b
2
<
1
Ga, b Aa, b.
2
1.7
The following Theorems C–E were established by Alzer and Qiu in 21.
Theorem C. The inequalities
αAa, b 1 − αGa, b < Ia, b < βAa, b 1 − β Ga, b
1.8
hold for all positive real numbers a and b with a /
b if and only if α ≤ 2/3 and β ≥ 2/e 0.73575 · · · .
Theorem D. Let a and b be real numbers with a /
b. If 0 < a, b ≤ e, then
Ga, bAa,b < La, bIa,b < Aa, bGa,b .
1.9
Aa, bGa,b < Ia, bLa,b < Ga, bAa,b .
1.10
And if a, b ≥ e, then
Theorem E. For all positive real numbers a and b with a / b, we have
Mp a, b <
1
La, b Ia, b
2
1.11
with the best possible parameter p log 2/1 log 2 0.40938 · · · .
However, the following problem is still open: for α, β, γ ∈ 0, 1 with α β γ 1, what
are the greatest value p and the least value q, such that the double inequality
Lp a, b < Aα a, bGβ a, bH γ a, b < Lq a, b
1.12
holds for all a, b > 0 with a / b? The purpose of this paper is to give the solution to this open
problem.
2. Lemmas
In order to establish our main result, we need four lemmas, which we present in this section.
4
Abstract and Applied Analysis
Lemma 2.1. If t > 1, then
tt 1 1/3
t−1
−
> 0.
log t
2
2.1
Proof. If t > 1, then it follows from 22, 3.6.16 that
t − 1 t t1/3
>
.
log t 1 t1/3
2.2
t t1/3
tt 1 1/3
>
2
1 t1/3
2.3
It is not difficult to verify that
for t > 1.
Therefore, Lemma 2.1 follows from inequalities 2.2 and 2.3.
Lemma 2.2. If t > 1, then
1
2
1t
t
log t − log t − log
− 1 > 0.
t−1
6
3
2
2.4
Proof. Let ft t/t−1 log t−1/6 log t−2/3 log1t/2−1. Then simple computations
lead to
lim ft 0,
2.5
t→1
f t gt
6tt − 12 t 1
,
2.6
where gt −6tt 1 log t t3 9t2 − 9t − 1,
g1 0,
2.7
g t −62t 1 log t 3t2 12t − 15,
2.8
g 1 0,
2.9
g t ht
,
t
2.10
Abstract and Applied Analysis
5
where ht −12 log t 6t2 − 6,
g 1 h1 0,
2.11
h t −12 log t 12t − 12,
2.12
h 1 0,
h t 12
t − 1 > 0
t
2.13
2.14
for t > 1.
Therefore, Lemma 2.2 follows from 2.5–2.7, 2.9–2.11, 2.13, and 2.14.
Lemma 2.3. Let t > 1 and gt 1−1/2λ3λ−5t3λ−2 λ−13λ−5−1t3λ−3 −1/2λ3λ−
5 1t3λ−4 1/2λ3λ − 5 1t2 − λ − 13λ − 5 − 1t − 1 − 1/2λ3λ − 5. Then
1 gt > 0 for λ ∈ 0, 2/3 ∪ 1, 4/3 ∪ 5/3, 2 and
2 gt < 0 for λ ∈ 2/3, 1 ∪ 4/3, 5/3.
Proof. Simple computations yield
lim gt 0,
t→1
g t 2.15
1
1 − λ 3λ − 53λ − 2t3λ−3 3λ − 1λ − 13λ − 5 − 1t3λ−4
2
1
− 3λ − 4 λ3λ − 5 1 t3λ−5 λ3λ − 5 2t
2
2.16
− λ − 13λ − 5 1,
g 1 0,
1
g t 3 1 − λ λ − 13λ − 53λ − 2t3λ−4
2
3λ − 13λ − 4λ − 13λ − 5 − 1t3λ−5
1
− 3λ − 43λ − 5 λ3λ − 5 1 t3λ−6 λ3λ − 5 2,
2
2.17
2.18
g 1 0,
2.19
g t t3λ−7 ht,
2.20
where ht 31 − 1/2λλ − 13λ − 53λ − 43λ − 2t2 3λ − 13λ − 43λ − 5λ − 13λ −
5 − 1t − 3λ − 23λ − 43λ − 51/2λ3λ − 5 1,
h1 0,
4
5
2
λ−
h t 812 − λ λ −
λ − 1 λ −
t − 1.
3
3
3
2.21
2.22
6
Abstract and Applied Analysis
1 If λ ∈ 0, 2/3 ∪ 1, 4/3 ∪ 5/3, 2, then 2.22 implies
h t > 0
2.23
for t > 1.
Therefore, Lemma 2.31 follows from 2.15, 2.17, 2.19, 2.20, 2.21, and 2.23.
2 If λ ∈ 2/3, 1 ∪ 4/3, 5/3, then 2.22 leads to
h t < 0
2.24
for t > 1.
Therefore, Lemma 2.32 follows from 2.15, 2.17, 2.19, 2.20, 2.21, and 2.24.
Lemma 2.4. Let t > 1 and Gt 1 − 2/λt3−2/λ − 3 − 2/λt2−2/λ 3 − 2/λt 2/λ − 1. Then
1 Gt > 0 for λ ∈ 2/3, 1 and
2 Gt < 0 for λ ∈ 0, 2/3 ∪ 1, 2.
Proof. From the expression of Gt, we clearly see that
lim Gt 0,
G t 1−
2
λ
t→1
2 2−2/λ
1 1−2/λ
2
2
3−
t
1−
t
−2 3−
3− ,
λ
λ
λ
λ
G 1 0,
2
6
G t 3 λ −
λ − 1λ − 2t−2/λ t − 1.
3
λ
2.25
2.26
2.27
2.28
From 2.28, we have
G t > 0
2.29
G t < 0
2.30
for λ ∈ 2/3, 1 and t > 1, and
for λ ∈ 0, 2/3 ∪ 1, 2 and t > 1.
Therefore, Lemma 2.41 follows from 2.25 and 2.27 together with 2.29, and
Lemma 2.42 follows from 2.25 and 2.27 together with 2.30.
3. Main Result
Theorem 3.1. Let a, b > 0 with a /
b and α, β > 0 with α β < 1. Then
Abstract and Applied Analysis
7
1 L6α3β−5 a, b Aα a, bGβ a, bH 1−α−β a, b L−2/2αβ a, b for 2α β ∈ {2/3, 1};
2 L6α3β−5 a, b > Aα a, bGβ a, bH 1−α−β a, b > L−2/2αβ a, b for 2α β ∈ 0, 2/3 ∪
1, 2 and L6α3β−5 a, b < Aα a, bGβ a, bH 1−α−β a, b < L−2/2αβ a, b for 2α β ∈
2/3, 1, and the parameters 6α 3β − 5 and −2/2α β cannot be improved in either case.
Proof. 1 We divide the proof into two cases.
Case 1. If 2α β 2/3, then simple computations lead to
L6α3β−5 a, b L−2/2αβ a, b L−3 a, b
2a2 b2
ab
1/3
3.1
21−2αβ ab1−2αβ/2 a b2αβ−1
Aα a, bGβ a, bH 1−α−β a, b.
Case 2. If 2α β 1, then we clearly see that
L6α3β−5 a, b L−2/2αβ a, b L−2 a, b
3.2
ab1/2 Aα a, bGβ a, bH 1−α−β a, b.
2 Without loss of generality, we assume that a > b and we put t a/b > 1 and
λ 2α β.
Firstly, we compare Aα a, bGβ a, bH 1−α−β a, b with L3λ−5 a, b. We divide the proof
into five cases.
Case 1. If λ 4/3, then 1.1 leads to
L3λ−5 a, b − Aα a, bGβ a, bH 1−α−β a, b
a−b
−
log a − log b
ab
2
α
ab
β/2
2ab
ab
1−α−β
t−1
−
b
log t
tt 1
2
1/3 .
3.3
Therefore, L3λ−5 a, b > Aα a, bGβ a, bH 1−α−β a, b follows from 3.3 and Lemma 2.1.
Case 2. If λ 5/3, then from 1.1 we have
logL3λ−5 a, b − log Aα a, bGβ a, bH 1−α−β a, b bH
t
1
2
1t
log t − log t − log
− 1.
t−1
6
3
2
3.4
From 3.4 and Lemma 2.2, we clearly see that L3λ−5 a, b
a, b.
1−α−β
>
Aα a, bGβ a,
8
Abstract and Applied Analysis
Case 3. If λ ∈ 0, 2/3 ∪ 1, 4/3 ∪ 5/3, 2, then 1.1 yields
logL3λ−5 a, b − log Aα a, bGβ a, bH 1−α−β a, b
1
t3λ−4 − 1
λ
1t
log
− 1−
− λ − 1 log
log t.
3λ − 5
2
2
3λ − 4t − 1
3.5
Let ft 1/3λ−5 logt3λ−4 −1/3λ−4t−1−λ−1 log1t/2−1−λ/2 log t.
Then simple computations lead to
lim ft 0,
t→1
f t gt
,
3λ − 5 t3λ−4 − 1 t2 − 1t
3.6
3.7
where gt 1 − 1/2λ3λ − 5t3λ−2 λ − 13λ − 5 − 1t3λ−3 − 1/2λ3λ − 5 1t3λ−4 1/2λ3λ − 5 1t2 − λ − 13λ − 5 − 1t − 1 − 1/2λ3λ − 5.
We clearly see that
3λ − 5 t3λ−4 − 1 > 0.
3.8
Therefore, L3λ−5 a, b > Aα a, bGβ a, bH 1−α−β a, b follows from 3.5–3.8 and
Lemma 2.31.
Case 4. If λ ∈ 2/3, 1, then 3.5–3.8 again hold, and from 3.5–3.8 and Lemma 2.32, we
know that L3λ−5 a, b < Aα a, bGβ a, bH 1−α−β a, b.
Case 5. If λ ∈ 4/3, 5/3, then 3.5–3.7 hold and
3λ − 5 t3λ−4 − 1 < 0.
3.9
Therefore, L3λ−5 a, b > Aα a, bGβ a, bH 1−α−β a, b follows from 3.5–3.7 and 3.9
together with Lemma 2.32.
Secondly, we compare Aα a, bGβ a, bH 1−α−β a, b with L−2/λ a, b.
From 1.1, we have
logL−2/λ a, b − log Aα a, bGβ a, bH 1−α−β a, b
t1−λ/2 − 1
λ
λ
1t
− 1−
log t.
− log
− λ − 1 log
2
2
2
1 − λ/2t − 1
3.10
Abstract and Applied Analysis
9
Let Ft −λ/2 logt1−λ/2 − 1/1 − λ/2t − 1 − λ − 1 log1 t/2 − 1 − λ/2 log t.
Then simple computations yield
lim Ft 0,
3.11
λGt
,
2t 1 − t1−2/λ t2 − 1
3.12
t→1
F t where Gt 1 − 2/λt3−2/λ − 3 − 2/λt2−2/λ 3 − 2/λt 2/λ − 1.
We clearly see that
1 − t1−2/λ > 0
3.13
for λ ∈ 0, 2.
Therefore, Aα a, bGβ a, bH 1−α−β a, b < L−2/λ a, b for λ ∈ 2/3, 1 follows from
3.10–3.13 and Lemma 2.41, and Aα a, bGβ a, bH 1−α−β a, b > L−2/λ a, b for λ ∈
0, 2/3 ∪ 1, 2 follows from 3.10–3.13 and Lemma 2.42.
At last, we prove that the parameters −2/λ and 3λ − 5 cannot be improved in either
case.
The following two cases will complete the proof for the optimality of parameter −2/λ.
Case I. If λ ∈ 0, 2/3 ∪ 1, 2, then for any ∈ 0, 2/λ − 1, one has
L−2/λ 1, t
lim
t → ∞ Aα 1, tGβ 1, tH 1−α−β 1, t
lim
t → ∞
1 λ/2−λ 1 t 1−λ λ2 /22−λ
2/λ − 1 − 1−
t
t
2t
1 − t1−2/λ
3.14
∞.
Equation 3.14 implies that for any ∈ 0, 2/λ − 1, there exists a sufficiently large
T1 T1 , α, β > 1, such that L−2/λ 1, t > Aα 1, tGβ 1, tH 1−α−β 1, t for t ∈ T1 , ∞.
Case II. If λ ∈ 2/3, 1, then for any > 0, one has
Aα 1, tGβ 1, tH 1−α−β 1, t
t → ∞
L−2/λ− 1, t
⎧
⎫
λ/2λ
⎨
2t 1−λ
⎬
1 − t1−2/λ−
λ2 /22λ
lim
t
⎭
t → ∞⎩ 1 t
2/λ − 11 − 1/t
lim
3.15
∞.
Equation 3.15 implies that for any > 0, there exists a sufficiently large T2 T2 , α, β > 1, such that Aα 1, tGβ 1, tH 1−α−β 1, t > L−2/λ− 1, t for t ∈ T2 , ∞.
10
Abstract and Applied Analysis
The following seven cases will complete the proof for the optimality of parameter
3λ − 5.
Case A. If λ 4/3, then for any > 0 and x > 0, one has
1
L3λ−5− 1, 1 x1 − Aα 1, 1 xGβ 1, 1 xH 1−α−β 1, 1 x
g1 x
,
1 x − 1
3.16
where g1 x x1 x − 1 x/21/3 1 x1/3 1 x − 1.
Upon letting x → 0, the Taylor expansion leads to
g1 x −
1 2
1x3 o x3 .
24
3.17
Equations 3.16 and 3.17 imply that for > 0, there exists a sufficiently small δ1 δ1 > 0, such that L3λ−5− 1, 1 x < Aα 1, 1 xGβ 1, 1 xH 1−α−β 1, 1 x for x ∈ 0, δ1 .
Case B. If λ 5/3, then for any ∈ 0, 1 and x > 0 x → 0, one has
L3λ−5− 1, 1 x − Aα 1, 1 xGβ 1, 1 xH 1−α−β 1, 1 x
x 2/3
− 1 x/6 1 2
1 x − 1
1
1
2 3
− 1 − x o x3 .
1 x1− − 1 24
1 − x
1−
3.18
Equation 3.18 implies that for any ∈ 0, 1, there exists a sufficiently small δ2 δ2 > 0, such that L3λ−5− 1, 1 x < Aα 1, 1 xGβ 1, 1 xH 1−α−β 1, 1 x for x ∈ 0, δ2 .
Case C. If λ ∈ 0, 2/3, then for any > 0 and x > 0 x → 0, one has
5−3λ
L3λ−5− 1, 1 x5−3λ − Aα 1, 1 xGβ 1, 1 xH 1−α−β 1, 1 x
4 − 3λx1 x4−3λ
x λ−1 5−3λ
− 1 x1−λ/2 1 2
1 x4−3λ − 1
1
3
−
o
x3 .
−
3λ4
−
3λx
5
1 x4−3λ − 1 24
3.19
Equation 3.19 implies that for any > 0, there exists a sufficiently small δ3 δ3 , α, β > 0, such that L3λ−5− 1, 1x < Aα 1, 1xGβ 1, 1xH 1−α−β 1, 1x for x ∈ 0, δ3 .
Abstract and Applied Analysis
11
Case D. If λ ∈ 2/3, 1, then for any ∈ 0, 4 − 3λ and x > 0 x → 0, one has
5−3λ−
L3λ−5 1, 1 x5−3λ− − Aα 1, 1 xGβ 1, 1 xH 1−α−β 1, 1 x
4 − 3λ − x1 x4−3λ−
1 x4−3λ− − 1
−
1 x1−λ/2
5−3λ−
1 x/21−λ
3.20
/245 − 3λ − 4 − 3λ − x3 o x3
.
1 x4−3λ− − 1 1 x/21−λ5−3λ−
Equation 3.20 implies that for any ∈ 0, 4 − 3λ, there exists a sufficiently small
δ4 δ4 , α, β > 0, such that L3λ−5 1, 1 x > Aα 1, 1 xGβ 1, 1 xH 1−α−β 1, 1 x for
x ∈ 0, δ4 .
Case E. If λ ∈ 1, 4/3, then for any > 0 and x > 0 x → 0, one has
5−3λ
L3λ−5− 1, 1 x5−3λ − Aα 1, 1 xGβ 1, 1 xH 1−α−β 1, 1 x
4 − 3λx1 x4−3λ
1 x4−3λ − 1
1
1 x4−3λ − 1
x λ−1 5−3λ
− 1 x1−λ/2 1 2
3.21
−/245 − 3λ4 − 3λx3 o x3 .
Equation 3.21 implies that for any > 0, there exists a sufficiently small δ5 δ5 , α, β > 0, such that L3λ−5 1, 1x < Aα 1, 1xGβ 1, 1xH 1−α−β 1, 1x for x ∈ 0, δ5 .
Case F. If λ ∈ 4/3, 5/3, then for any ∈ 0, 3λ − 4 and x > 0 x → 0, one has
5−3λ
L3λ−5− 1, 1 x5−3λ − Aα 1, 1 xGβ 1, 1 xH 1−α−β 1, 1 x
x λ−1 5−3λ
1−λ/2
1
−
x
1
2
1 x3λ−4− − 1
3
−
o
x3 .
−
3λ3λ
−
4
−
x
5
1 x3λ−4− − 1 24
3λ − 4 − x
3.22
1
Equation 3.22 implies that for any ∈ 0, 3λ − 4, there exists a sufficiently small
δ6 δ6 , α, β > 0, such that L3λ−5 1, 1 x < Aα 1, 1 xGβ 1, 1 xH 1−α−β 1, 1 x for
x ∈ 0, δ6 .
12
Abstract and Applied Analysis
Case G. If λ ∈ 5/3, 2, then for any ∈ 0, 3λ − 5 and x > 0 x → 0, one has
3λ−5−
L3λ−5− 1, 1 x3λ−5− − Aα 1, 1 xGβ 1, 1 xH 1−α−β 1, 1 x
x λ−1 3λ−5−
1 x3λ−4− − 1
− 1 x1−λ/2 1 2
3λ − 4 − x
1
− 3λ − 4 − 3λ − 5 − x3 o x3 .
3λ − 4 − x 24
3.23
Equation 3.23 implies that for any ∈ 0, 3λ − 5, there exists a sufficiently small
δ7 δ7 , α, β > 0, such that L3λ−5 1, 1 x < Aα 1, 1 xGβ 1, 1 xH 1−α−β 1, 1 x for
x ∈ 0, δ7 .
Acknowledgments
The authors wish to thank the anonymous referee for his very careful reading of the
manuscript and fruitful comments and suggestions. This research is partly supported by
N S Foundation of China Grant 60850005, N S Foundation of Zhejiang Province Grants
D7080080 and Y607128, and Innovation Team Foundation of the Department of Education
of Zhejiang Province Grant T200924.
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