Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2010, Article ID 806825, 10 pages
doi:10.1155/2010/806825
Research Article
Optimal Inequalities for Generalized Logarithmic,
Arithmetic, and Geometric Means
Bo-Yong Long1, 2 and Yu-Ming Chu3
1
College of Mathematics and Econometrics, Hunan University, Changsha 410082, China
School of Mathematical Sciences, Anhui University, Hefei 230039, China
3
Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China
2
Correspondence should be addressed to Yu-Ming Chu, chuyuming2005@yahoo.com.cn
Received 20 October 2009; Revised 16 December 2009; Accepted 1 February 2010
Academic Editor: Alexander I. Domoshnitsky
Copyright q 2010 B.-Y. Long and Y.-M. Chu. 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.
For p ∈ R, the generalized logarithmic mean Lp a, b, arithmetic mean Aa, b, and geometric
mean Ga, b of two positive numbers a and b are defined by Lp a, b a, for a b, Lp a, b 1/b−a
, for
bp1 − ap1 /p 1b − a1/p , for p / 0, p / − 1, and a / b, Lp a, b 1/ebb /aa p 0, and a /
b, Lp a, b b − a/log b − log a, for p −1, and a /
b, Aa, b a b/2, and
√
Ga, b ab, respectively. In this paper, we find the greatest value p or least value q, resp. such
that the inequality Lp a, b < αAa, b 1 − αGa, b or αAa, b 1 − αGa, b < Lq a, b, resp.
holds for α ∈ 0, 1/2or α ∈ 1/2, 1, resp. and all a, b > 0 with a /
b.
1. Introduction
For p ∈ R, the generalized logarithmic mean Lp a, b and power mean Mp a, b with
parameter p of two positive numbers a and b are defined by
⎧
⎪
a,
⎪
⎪
⎪
⎪
⎪
⎪
1/p
⎪
⎪
p1
p1
⎪
⎪
−
a
b
⎪
⎪
,
⎪
⎪
⎪
⎨ p 1b − a
Lp a, b 1/b−a
⎪
⎪
1 bb
⎪
⎪
,
⎪
⎪
⎪
e aa
⎪
⎪
⎪
⎪
⎪
⎪
⎪
b−a
⎪
⎪
,
⎩
log b − log a
a b,
p/
0, p /
− 1, a /
b,
1.1
p 0, a /
b,
p −1, a /
b,
2
Journal of Inequalities and Applications
and
⎧
p
p 1/p
⎪
⎪
⎨ a b
,
2
Mp a, b ⎪
⎪
⎩√ab,
p/
0,
1.2
p 0,
respectively. It is well known that both means are continuous and increasing with respect to
p ∈ R for fixed a and b. Recently, both means have been the subject of intensive research. In
particular, many remarkable inequalities involving Lp a, b and Mp a, b can be found in the
literature 1–9. Let
ab
,
Aa, b 2
⎧ 1/b−a
⎪
b
⎪
⎨1 b
, b/
a,
Ia, b e aa
⎪
⎪
⎩a,
b a,
⎧
⎪
⎨
b−a
, b/
a,
La, b log b − log a
⎪
⎩a,
b a,
1.3
√
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, respectively. Then
min{a, b} < Ha, b M−1 a, b < Ga, b M0 a, b L−2 a, b
< La, b L−1 a, b < Ia, b L0 a, b < Aa, b
1.4
M1 a, b L1 a, b < max{a, b}
for all a /
b.
In 10, Carlson proved that
La, b <
2
1
Aa, b Ga, b
3
3
1.5
for all a, b > 0 with a /
b.
The following inequality is due to Sándor 11, 12:
Ia, b >
2
1
Aa, b Ga, b.
3
3
1.6
In 13, Lin established the following results: 1 p ≥ 1/3 implies that La, b <
Mp a, b for all a, b > 0 with a /
b; 2 p ≤ 0 implies that La, b > Mp a, b for all a, b > 0
with a /
b; 3 p < 1/3 implies that there exist a, b > 0 such that La, b > Mp a, b; 4
p > 0 implies that there exist a, b > 0 such that La, b < Mp a, b. Hence the question
was answered: what are the least value q and the greatest value p such that the inequality
Mp a, b < La, b < Mq a, b holds for all a, b > 0 with a / b.
Pittenger 14 established that
Mp1 a, b ≤ Lp a, b ≤ Mp2 a, b
1.7
Journal of Inequalities and Applications
3
for all a, b > 0, where
⎧
⎪
p log 2
p2
⎪
⎪
,
0,
min
, p > −1, p /
⎪
⎪
⎪
3 log p 1
⎪
⎨
p1 2 ,
p 0,
⎪
⎪
3 ⎪
⎪
⎪
p2
⎪
⎪
,0 ,
p ≤ −1,
⎩min
3
⎧
p log 2
p2
⎪
⎪
⎪
0,
,
max
, p > −1, p /
⎪
⎪
3 log p 1
⎪
⎨
p2 log 2,
p 0,
⎪
⎪
⎪
⎪
p2
⎪
⎪
⎩max
,0 ,
p ≤ −1.
3
1.8
Here, p1 and p2 are sharp and inequality 1.7 becomes equality if and only if a b or
p 1, −2 or −1/2. The case p −1 reduces to Lin’s results 13. Other generalizations of Lin’s
results were given by Imoru 15.
Recently, some monotonicity results of the ratio between generalized logarithmic
means were established in 16–18.
The aim of this paper is to prove the following Theorem 1.1.
Theorem 1.1. Let α ∈ 0, 1 and a, b > 0 with a /
b, then
1 L3α−2 a, b αAa, b 1 − αGa, b for α 1/2;
2 L3α−2 a, b < αAa, b 1 − αGa, b for 0 < α < 1/2, and L3α−2 a, b > αAa, b 1 − αGa, b for 1/2 < α < 1, moreover, in each case, the bound L3α−2 a, b for the sum
αAa, b 1 − αGa, b is optimal.
2. Proof of Theorem 1.1
In order to prove our Theorem 1.1 we need a lemma, which we present in this section.
Lemma 2.1. For α ∈ 0, 1 and ht 6α − 1t6α−4 − 6α − 1t6α−5 − 6α − 5t6α−6 6α − 5t6α−7
one has
1 If α ∈ 1/6, 1, then ht > 0 for t > 1;
2 If α ∈ 0, 1/6, then ht < 0 for t >
5 − 6α/1 − 6α, ht > 0 for 1 < t <
5 − 6α/1 − 6α, and ht 0 for t 5 − 6α/1 − 6α.
Proof. 1 If α 1/6, then we clearly see that
ht 4t−6 t − 1 > 0
for t > 1.
2.1
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Journal of Inequalities and Applications
If α ∈ 1/6, 1, then
ht 6α − 1t − 1 t2 − 1 4
t6α−7 > 0
6α − 1
2.2
for t > 1.
Therefore, Lemma 2.11 follows from 2.1 and 2.2.
2 If α ∈ 0, 1/6, then
⎛
ht 6α − 1t − 1⎝t ⎞⎛ ⎞
5 − 6α ⎠⎝
5 − 6α ⎠ 6α−7
t−
t
.
1 − 6α
1 − 6α
2.3
Therefore, Lemma 2.12 follows from 2.3.
Proof of Theorem 1.1.
Proof. 1 If α 1/2, then 1.1 leads to
ab
L3α−2 a, b L−1/2 a, b 4
√
ab 1
1
Aa, b Ga, b αAa, b 1 − αGa, b.
2
2
2
2.4
2 We divide the proof into two cases.
Case 1. α 1/3 or α 2/3. From inequalities 1.5 and 1.6 we clearly see that
L3α−2 a, b < αAa, b 1 − αGa, b
2.5
L3α−2 a, b > αAa, b 1 − αGa, b
2.6
for α 1/3, and
for α 2/3.
Case2. α ∈ 0, 1 \ {1/3, 1/2, 2/3}. Without loss of generality, we assume that a > b. Let
t a/b > 1, then 1.1 leads to
log L3α−2 a, b − logαAa, b 1 − αGa, b
α
t6α−2 − 1
1
log
1 t2 1 − αt .
− log
2
3α − 2
2
3α − 1t − 1
2.7
Let ft 1/3α − 2 logt6α−2 − 1/3α − 1t2 − 1 − logα/21 t2 1 − αt,
then simple computations yield
Journal of Inequalities and Applications
5
limft 0,
t→1
f t 3α −
2t6α−2
−
1t2
gt
,
− 1α/21 t2 1 − αt
2.8
2.9
where
gt 1 − α3α − 2t6α 3αα − 1t6α−1 − 3α1 − αt6α−2
− α3α − 1t6α−3 α3α − 1t3 3α1 − αt2
2.10
3α1 − αt − 1 − α3α − 2.
Note that
g1 0,
2.11
g t 6α1 − α3α − 2t6α−1 3αα − 16α − 1t6α−2
− 6α1 − α3α − 1t6α−3 − 3α3α − 12α − 1t6α−4
2.12
3α3α − 1t2 6α1 − αt 3α1 − α,
g 1 0,
2.13
g t 6α1 − α3α − 26α − 1t6α−2 6αα − 16α − 1
× 3α − 1t6α−3 − 18α1 − α3α − 12α − 1t6α−4
− 6α3α − 12α − 13α − 2t6α−5 6α3α − 1t
2.14
6α1 − α,
g 1 0,
2.15
g t 12α1 − α3α − 26α − 13α − 1t6α−3
18αα − 16α − 13α − 12α − 1t6α−4
− 36α1 − α3α − 12α − 13α − 2t6α−5
2.16
− 6α3α − 12α − 13α − 26α − 5t6α−6
6α3α − 1,
g 1 0,
g 4 t 36α3α − 13α − 22α − 11 − αht,
where ht is defined as in Lemma 2.1.
2.17
2.18
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Journal of Inequalities and Applications
We divide the proof into five subcases.
Subcase A. α ∈ 0, 1/6. From 2.18 and Lemma 2.12 we clearly see that g 4 t < 0
for t ∈ 1, 5 − 6α/1 − 6α and g 4 t > 0 for t ∈ 5 − 6α/1 − 6α, ∞, then we
know that g t is strictly decreasing in 1, 5 − 6α/1 − 6α and strictly increasing in
5 − 6α/1 − 6α, ∞. Now from the monotonicity of g t and 2.17 together with the
fact that limt → ∞ g t 6α3α − 1 < 0 we clearly see that g t < 0 for t > 1, then from
2.7–2.15 and 3α − 2t6α−2 − 1 > 0 for t > 1 we get L3α−2 a, b < αAa, b 1 − αGa, b
for α ∈ 0, 1/6.
Subcase B. α ∈ 1/6, 1/3. Then 2.18 and Lemma 2.11 lead to
g 4 t < 0
2.19
for t > 1.
From 2.7–2.17 and 2.19 together with the fact that 3α − 2t6α−2 − 1 > 0 for t > 1
we know that L3α−2 a, b < αAa, b 1 − αGa, b for α ∈ 1/6, 1/3.
Subcase C. α ∈ 1/3, 1/2. Then 2.18 and Lemma 2.11 imply that
g 4 t > 0
2.20
for t > 1.
From 2.7–2.17, 2.20 and 3α − 2t6α−2 − 1 < 0 for t > 1 we know that L3α−2 a, b <
αAa, b 1 − αGa, b for α ∈ 1/3, 1/2.
Subcase D. α ∈ 1/2, 2/3. Then 2.19 again yields, and L3α−2 a, b > αAa, b 1 − αGa, b
for α ∈ 1/2, 2/3 follows from 2.7–2.17 and 2.19 together with 3α − 2t6α−2 − 1 < 0.
Subcase E. α ∈ 2/3, 1. Then 2.20 is also true, and L3α−2 a, b > αAa, b 1 − αGa, b for
α ∈ 2/3, 1 follows from 2.7–2.17, 2.20 and the fact that 3α − 2t6α−2 − 1 > 0.
Next, we prove that the bound L3α−2 a, b for the sum αAa, b1−αGa, b is optimal
in each case. The proof is divided into six cases.
Case 1. α 1/3. For any ∈ 0, 1 and x ∈ 0, 1, then 1.1 leads to
L3α−2 1, 1 x1− − αA1, 1 x 1 − αG1, 1 x1−
2
1
A1, 1 x G1, 1 x
3
3
1−
x
1 x 2
1 x1/2
−
3 6 3
1 x − 1
L−1 1, 1 x1− −
f1 x
,
1 x − 1
where f1 x x − 1 x − 11/3 x/6 2/31 x1/2 1− .
1−
2.21
Journal of Inequalities and Applications
7
Let x → 0; making use of Taylor expansion, one has
f1 x 1 2
1 − x3 o x3 .
24
2.22
Equations 2.21 and 2.22 imply that for any ∈ 0, 1, there exists 0 < δ1 δ1 < 1,
such that L3α−2 1, 1 x > αA1, 1 x 1 − αG1, 1 x for any x ∈ 0, δ1 and α 1/3.
Case 2. α 2/3. For any ∈ 0, 1 and x ∈ 0, 1, from 1.1 we have
αA1, 1 x 1 − αG1, 1 x − L3α−2− 1, 1 x
1
2
A1, 1 x G1, 1 x − L− 1, 1 x
3
3
2 x 1
1 − x
1 x1/2 −
3 3 3
1 x1− − 1
f2 x
1 x1− − 1
2.23
,
where f2 x 1 x1− − 12/3 x/3 1/31 x1/2 − 1 − x.
Let x → 0; making use of Taylor expansion, one has
f2 x 1 2
1 − x3 o x3 .
24
2.24
Equations 2.23 and 2.24 imply that for any ∈ 0, 1, there exists 0 < δ2 δ2 < 1,
such that L3α−2− 1, 1 x < αA1, 1 x 1 − αG1, 1 x for x ∈ 0, δ2 and α 2/3.
Case 3. α ∈ 0, 1/3. For ∈ 0, 1 − 3α and x ∈ 0, 1, we get
L3α−2 1, 1 x2−3α− − αA1, 1 x 1 − αG1, 1 x2−3α−
1 − 3α − x1 x1−3α−
1 x1−3α− − 1
f3 x
1 x1−3α− − 1
2−3α−
α
− α x 1 − α1 x1/2
2
2.25
,
where f3 x 1 − 3α − x1 x1−3α− − 1 x1−3α− − 1α α/2x 1 − α1 x1/2 2−3α− .
Let x → 0; making use of Taylor expansion, one has
f3 x 1
1 − 3α − 2 − 3α − x3 o x3 .
24
2.26
8
Journal of Inequalities and Applications
Equations 2.25 and 2.26 imply that for any α ∈ 0, 1/3 and any ∈ 0, 1 − 3α, there
exists 0 < δ3 δ3 , α < 1, such that L3α−2 1, 1 x > αA1, 1 x 1 − αG1, 1 x for
x ∈ 0, δ3 .
Case 4. α ∈ 1/3, 1/2. For any ∈ 0, 2 − 3α and x ∈ 0, 1, we get
L3α−2 1, 1 x2−3α− − αA1, 1 x 1 − αG1, 1 x2−3α−
2−3α−
α
− α x 1 − α1 x1/2
2
−1
3α − 1 x
1 x
3α−1
f4 x
1 x3α−1 − 1
2.27
,
where f4 x 3α − 1 x − 1 x3α−1 − 1α α/2x 1 − α1 x1/2 2−3α− .
Let x → 0; using Taylor expansion we have
f4 x 1
3α − 1 2 − 3α − x3 o x3 .
24
2.28
Equations 2.27 and 2.28 show that for any α ∈ 1/3, 1/2 and any ∈ 0, 2 − 3α,
there exists 0 < δ4 δ4 , α < 1, such that L3α−2 1, 1 x > αA1, 1 x 1 − αG1, 1 x
for x ∈ 0, δ4 .
Case 5. α ∈ 1/2, 2/3. For any ∈ 0, 3α − 1 and x ∈ 0, 1, we have
αA1, 1 x 1 − αG1, 1 x2−3α − L3α−2− 1, 1 x2−3α
2−3α
α
3α − 1 − x
α x 1 − α1 x1/2
−
2
1 x3α−−1 − 1
f5 x
1 x3α−1− − 1
2.29
,
where f5 x 1 x3α−1− − 1α α/2x 1 − α1 x1/2 2−3α − 3α − 1 − x.
Let x → 0; making use of Taylor expansion we get
f5 x 1
3α − 1 − 2 − 3α x3 o x3 .
24
2.30
Equations 2.29 and 2.30 imply that for any α ∈ 1/2, 2/3 and any ∈ 0, 3α − 1,
there exists 0 < δ5 δ5 , α < 1, such that L3α−2− 1, 1 x < αA1, 1 x 1 − αG1, 1 x
for x ∈ 0, δ5 .
Journal of Inequalities and Applications
9
Case 6. α ∈ 2/3, 1. For any ∈ 0, 3α − 2 and x ∈ 0, 1, we get
αA1, 1 x 1 − αG1, 1 x3α−2− − L3α−2− 1, 1 x3α−2−
3α−2− 1 x3α−−1 − 1
α
α x 1 − α1 x1/2
−
2
3α − 1 − x
2.31
f6 x
,
3α − 1 − x
where f6 x 3α − 1 − xα α/2x 1 − α1 x1/2 3α−2− − 1 x3α−1− − 1.
Let x → 0, using Taylor expansion we have
f6 x 1
3α − 2 − 3α − 1 − x3 o x3 .
24
2.32
From 2.31 and 2.32 we know that for any α ∈ 2/3, 1 and any ∈ 0, 3α − 2, there
exists 0 < δ6 δ6 , α < 1, such that L3α−2− 1, 1 x < αA1, 1 x 1 − αG1, 1 x for
x ∈ 0, δ6 .
At last, we propose two open problems as follows.
Open Problem 1
What is the least value p such that the inequality
αAa, b 1 − αGa, b < Lp a, b
2.33
holds for α ∈ 0, 1/2 and all a, b > 0 with a /
b?
Open Problem 2
What is the greatest value q such that the inequality
αAa, b 1 − αGa, b > Lq a, b
2.34
holds for α ∈ 1/2, 1 and all a, b > 0 with a /
b?
Acknowledgments
The authors wish to thank the anonymous referee for their very careful reading of the
manuscript and fruitful comments and suggestions. This research is partly supported by N S
Foundation of China under Grant 60850005, and N S Foundation of Zhejiang Province under
Grants D7080080 and Y607128.
10
Journal of Inequalities and Applications
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