Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2009, Article ID 741923, 6 pages
doi:10.1155/2009/741923
Research Article
Two Sharp Inequalities for Power Mean,
Geometric Mean, and Harmonic Mean
Yu-Ming Chu1 and Wei-Feng Xia2
1
2
Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China
School of Teacher Education, Huzhou Teachers College, Huzhou 313000, China
Correspondence should be addressed to Yu-Ming Chu, chuyuming2005@yahoo.com.cn
Received 23 July 2009; Accepted 30 October 2009
Recommended by Wing-Sum Cheung
For p ∈ R, the power mean of order p of twopositive numbers a and b is defined by
Mp a, b ap bp /21/p , p /
0, and Mp a, b ab, p 0. In this paper, we establish two
sharp inequalities as follows: 2/3Ga, b 1/3Ha, b
√ M−1/3 a, b and 1/3Ga, b 2/3Ha, b M−2/3 a, b for all a, b > 0. Here Ga, b ab and Ha, b 2ab/a b denote
the geometric mean and harmonic mean of a and b, respectively.
Copyright q 2009 Y.-M. Chu and W.-F. Xia. 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.
1. Introduction
For p ∈ R, the power mean of order p of two positive numbers a and b is defined by
⎧
⎪
ap bp 1/p
⎪
⎨
,
2
Mp a, b ⎪
⎪
⎩√
ab,
p
/ 0,
1.1
p 0.
Recently, the power mean has been the subject of intensive research. In particular,
many remarkable inequalities for Mp a, b can be found in literature 1–12. It is well known
that Mp a, b is continuous and increasing with respect to p ∈ R for fixed a and b. If we
√
denote by Aa, b a b/2, Ga, b ab, and Ha, b 2ab/a b the arithmetic mean,
geometric mean and harmonic mean of a and b, respectively, then
min{a, b} Ha, b M−1 a, b Ga, b M0 a, b Aa, b M1 a, b max{a, b}.
1.2
2
Journal of Inequalities and Applications
In 13, Alzer and Janous established the following sharp double-inequality see also
14, page 350:
Mlog 2/ log 3 a, b 1
2
Aa, b Ga, b M2/3 a, b
3
3
1.3
for all a, b > 0.
In 15, Mao proved
M1/3 a, b 2
1
Aa, b Ga, b M1/2 a, b
3
3
1.4
for all a, b > 0, and M1/3 a, b is the best possible lower power mean bound for the sum
1/3Aa, b 2/3Ga, b.
The purpose of this paper is to answer the questions: what are the greatest values p and
q, and the least values r and s, such that Mp a, b 2/3Ga, b 1/3Ha, b Mr a, b
and Mq a, b 1/3Ga, b 2/3Ha, b Ms a, b for all a, b > 0?
2. Main Results
Theorem 2.1. 2/3Ga, b 1/3Ha, b M−1/3 a, b for all a, b > 0, equality holds if and
only if a b, and M−1/3 a, b is the best possible lower power mean bound for the sum 2/3Ga, b
1/3Ha, b.
Proof. If a b, then we clearly see that 2/3Ga, b 1/3Ha, b M−1/3 a, b a.
If a /
b and a/b t6 , then simple computation leads to
2
Ga, b 3
2t3
b
3
1
Ha, b − M−1/3 a, b
3
2t6
8t6
−
31 t6 1 t2 3
3 2
2bt3
2
4
2
3 2
3 4
2
t − t 1 t t 1 − 12t t − t 1
× t 1
31 t2 3 t4 − t2 1
2bt
10
8
7
6
5
4
3
2
×
t
2t
−
11t
t
14t
t
−
11t
2t
1
31 t2 3 t4 − t2 1
2.1
3
2bt3 t − 14
6
5
4
3
2
×
t
4t
12t
17t
12t
4t
1
31 t2 3 t4 − t2 1
> 0.
Next, we prove that M−1/3 a, b is the best possible lower power mean bound for the
sum 2/3Ga, b 1/3Ha, b.
Journal of Inequalities and Applications
For any 0 < ε <
3
1
and 0 < x < 1, one has
3
1/3−ε
1/3−ε 2
1
2
2
2
M−1/3ε 1 x , 1
− G1 x , 1 H1 x , 1
3
3
1/3−ε
−1 1 1 x−2/32ε
2
21 x2
−
1 x 2
3
3x2 2x 2
21 x2/3−2ε
1 1 x2/3−2ε
−
1 2x 4/3x2 x3 /3
1 x x2 /2
1/3−ε
2.2
fx
,
1/3−ε
2/3−2ε
1 1 x
1 x x2 /2
1/3−ε
where fx 21 x2/3−2ε 1 x x2 /2
− 1 1 x2/3 −2ε 1 2x 4/3x2 x3 /31/3−ε .
Let x → 0, then the Taylor expansion leads to
2 − 6ε
1 − 3ε1 6ε 2
fx 2 1 x−
x o x2
3
9
1 − 3ε
1 − 3ε2 2
2
× 1
x
x o x
3
18
1 − 3ε
1 − 3ε1 6ε 2
−2 1
x−
x o x2
3
18
2 − 6ε
2ε1 − 3ε 2
× 1
x−
x o x2
3
3
1 − 3ε1 − 9ε 2
2 1 1 − 3εx x o x2
6
1 − 3ε1 − 10ε 2
− 2 1 1 − 3εx x o x2
6
ε1 − 3ε 2
x o x2 .
3
2.3
Equations 2.2 and 2.3 imply that for any 0 < ε < 1/3 there exists 0 < δ δε < 1,
such that M−1/3ε 1 x2 , 1 > 2/3G1 x2 , 1 1/3H1 x2 , 1 for x ∈ 0, δ.
Remark 2.2. For any ε > 0, one has
lim
t → ∞
ε 1/ε 2√
1
2t
2
2t
∞.
G1, t H1, t − M−ε 1, t lim
−
t
t → ∞ 3
3
3
31 t
1 tε
2.4
4
Journal of Inequalities and Applications
Therefore, M0 a, b Ga, b is the best possible upper power mean bound for the
sum 2/3Ga, b 1/3Ha, b.
Theorem 2.3. 1/3Ga, b 2/3Ha, b M−2/3 a, b for all a, b > 0, equality holds if and
only if a b, and M−2/3 a, b is the best possible lower power mean bound for the sum 1/3Ga, b
2/3Ha, b.
Proof. If a b, then we clearly see that 1/3Ga, b 2/3Ha, b M−2/3 a, b a.
If a /
b and a/b t6 , then elementary calculation yields
2
2
1
Ga, b Ha, b − M−2/3 a, b2
3
3
⎡
3 ⎤
2 3
6
4
4t
2t
t
⎦
b2 ⎣
−
3 31 t6 1 t4
3 2
2 b 2 t6
4
6
3
6 6
t 4t 1 − 72t t 1
3 t 1
91 t6 2 1 t4
b 2 t6
24
21
20
18
17
16
15
14
13
3 t 8t 3t 18t 24t 3t 8t 54t 24t
91 t6 2 1 t4
2t12 24t11 54t10 8t9 3t8 24t7 18t6 3t4 8t3 1
− 72t18 144t12 72t6
b 2 t6
24
21
20
18
17
16
15
14
13
12
3 t 8t 3t − 54t 24t 3t 8t 54t 24t − 142t
91 t6 2 1 t4
24t11 54t10 8t9 3t8 24t7 − 54t6 3t4 8t3 1
b2 t6 t − 14
t20 4t19 10t18 28t17 70t16 148t15 220t14 268t13
3
91 t6 2 1 t4
277t12 240t11 240t10 240t9 277t8 268t7 220t6
148t5 70t4 28t3 10t2 4t 1 > 0.
2.5
Next, we prove that M−2/3 a, b is the best possible lower power mean bound for the
sum 1/3Ga, b 2/3Ha, b.
Journal of Inequalities and Applications
5
For any 0 < ε < 2/3 and 0 < x < 1, one has
2/3−ε
2/3−ε 1
2
M−2/3ε 1, 1 x2 − G1, 1 x2 H1, 1 x2 3
3
2−3ε/3
1 2x 7/6x2 1/6x3
21 x4−6ε/3
−
2−3ε/3
1 1 x4−6ε/3
1 x 1/2x2 2.6
fx
,
1 1 x4−6ε/3 1 x 1/2x2 2−3ε/3
where fx 21 x4−6ε/3 1 x x2 /2
1 x
2−3ε/3
− 1 2x 7/6x2 1/6x3 2−3ε/3
1 4−6ε/3
.
Let x → 0, then the Taylor expansion leads to
4 − 6ε
2 − 3ε1 − 6ε 2
x
x o x2
fx 2 1 3
9
2 − 3ε
2 − 3ε2 2
2
× 1
x
x o x
3
18
4 − 6ε
2 − 3ε1 − 4ε 2
−2 1
x
x o x2
3
6
2 − 3ε
2 − 3ε1 − 6ε 2
× 1
x
x o x2
3
18
2 − 3ε4 − 9ε 2
2 1 2 − 3εx x o x2
6
2 − 3ε4 − 10ε 2
− 2 1 2 − 3εx x o x2
6
ε2 − 3ε 2
x o x2 .
3
2.7
Equations 2.6 and 2.7 imply that for any 0 < ε < 2/3 there exists 0 < δ δε < 1,
such that
M−2/3ε 1, 1 x2 > 1/3G 1, 1 x2 2/3H 1, 1 x2
2.8
for x ∈ 0, δ.
Remark 2.4. For any ε > 0, one has
lim
t → ∞
ε 1/ε 1√
2
2t
1
4t
∞.
G1, t H1, t − M−ε 1, t lim
−
t
t → ∞ 3
3
3
31 t
1 tε
2.9
6
Journal of Inequalities and Applications
Therefore, M0 a, b Ga, b is the best possible upper power mean bound for the
sum 1/3Ga, b 2/3Ha, b.
Acknowledgments
This research is partly supported by N S Foundation of China under Grant 60850005 and the
N S Foundation of Zhejiang Province under Grants Y7080185 and Y607128.
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