Wu, “Generalization and sharpness of the power means inequality and their applications,” Journal of Mathematical Analysis and Applications, vol.. Richards, “Sharp power mean bounds for t
Trang 1Volume 2009, Article ID 741923, 6 pages
doi:10.1155/2009/741923
Research Article
Two Sharp Inequalities for Power Mean,
Geometric Mean, and Harmonic Mean
1 Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China
2 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 two positive numbers a and b is defined by
M p a, b a p b p /21/p , p / 0, and M p 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 ofa and b, respectively.
Copyrightq 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
Forp ∈ R, the power mean of order p of two positive numbers a and b is defined by
M p a, b
⎧
⎪
⎪
a p b p 2
1/p
, p / 0,
√
ab, p 0.
1.1
Recently, the power mean has been the subject of intensive research In particular, many remarkable inequalities forM p a, b can be found in literature 1 12 It is well known thatM p a, b is continuous and increasing with respect to p ∈ R for fixed a and b If we
denote byAa, b a b/2, Ga, b √ab, and Ha, b 2ab/a b the arithmetic mean,
geometric mean and harmonic mean ofa and b, respectively, then
Trang 2In13, Alzer and Janous established the following sharp double-inequality see also
14, page 350:
Mlog 2/ log 3a, b 2
3Aa, b 1
3Ga, b M2/3a, b 1.3
for alla, b > 0.
In15, Mao proved
M1/3a, b 1
3Aa, b 23Ga, b M1/2a, b 1.4
for alla, b > 0, and M1/3a, 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 valuesp and
q, and the least values r and s, such that M p a, b 2/3Ga, b 1/3Ha, b M r a, b
andM q a, b 1/3Ga, b 2/3Ha, b M s 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.
Ifa / b and a/b t6, then simple computation leads to
2
3Ga, b 13Ha, b − M −1/3 a, b
b
2t3
3 2t6
31 t6−
8t6
1 t23
31 t23
t4− t2 1 × t
2 13t4− t2 1 t3
t2 12− 12t3
t4− t2 1
31 t23
t4− t2 1 ×
t10 2t8− 11t7 t6 14t5 t4− 11t3 2t2 1
2bt3t − 14
31 t23
t4− t2 1 ×
t6 4t5 12t4 17t3 12t2 4t 1
> 0.
2.1
Next, we prove thatM −1/3 a, b is the best possible lower power mean bound for the
Trang 3For any 0< ε <1
3 and 0< x < 1, one has
M −1/3ε 1 x2, 11/3−ε− 2
3G1 x2, 1 1
3H1 x2, 1
1/3−ε
1 1 x −2/32ε
2
−1
−
2
31 x 21 x2
3x2 2x 2
1/3−ε
21 x2/3−2ε
1 1 x2/3−2ε −
1 2x 4/3x2 x3/3
1 x x2/2
1/3−ε
1 1 x2/3−2ε1 x x2/21/3−ε,
2.2
wherefx 21 x2/3−2ε 1 x x2/21/3−ε − 1 1 x2/3 −2ε1 2x 4/3x2
x3/31/3−ε
Letx → 0, then the Taylor expansion leads to
fx 2 12− 6ε
3 x − 1 − 3ε1 6ε
9 x2 ox2
×
11− 3ε
3 x 1 − 3ε2
18 x2 ox2
− 2 11− 3ε
3 x − 1 − 3ε1 6ε
18 x2 ox2
× 12− 6ε
3 x − 2ε1 − 3ε
3 x2 ox2
2 1 1 − 3εx 1 − 3ε1 − 9ε
6 x2 ox2
− 2 1 1 − 3εx 1 − 3ε1 − 10ε6 x2 ox2
ε1 − 3ε
3 x2 ox2
.
2.3
Equations2.2 and 2.3 imply that for any 0 < ε < 1/3 there exists 0 < δ δε < 1,
such thatM −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
Trang 4Therefore, M0a, b Ga, b is the best possible upper power mean bound for the
sum2/3Ga, 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.
Ifa / b and a/b t6, then elementary calculation yields
1
3Ga, b 23Ha, b
2
− M −2/3 a, b2
b2
⎡
⎣
t3
3 4t6
31 t6
2
−
2t4
1 t4
3⎤
⎦
b2t6
91 t62
1 t4 3 t4 13t6 4t3 12− 72t6t6 12
b2t6
91 t62
1 t4 3t24 8t21 3t20 18t18 24t17 3t16 8t15 54t14 24t13
2t12 24t11 54t10 8t9 3t8 24t7 18t6 3t4 8t3 1
−72t18 144t12 72t6
b2t6
91 t62
1 t4 3t24 8t21 3t20− 54t18 24t17 3t16 8t15 54t14 24t13− 142t12
24t11 54t10 8t9 3t8 24t7− 54t6 3t4 8t3 1
b2t6t − 14
91 t62
1 t4 3t20 4t19 10t18 28t17 70t16 148t15 220t14 268t13
277t12 240t11 240t10 240t9 277t8 268t7 220t6
148t5 70t4 28t3 10t2 4t 1> 0.
2.5
Next, we prove thatM −2/3 a, b is the best possible lower power mean bound for the
Trang 5For any 0< ε < 2/3 and 0 < x < 1, one has
M −2/3ε 1, 1 x22/3−ε− 1
3G1, 1 x2 2
3H1, 1 x2
2/3−ε
21 x4−6ε/3
1 1 x 4−6ε/3 −
1 2x 7/6x2 1/6x3 2−3ε/3
1 x 1/2x22−3ε/3
1 1 x 4−6ε/31 x 1/2x22−3ε/3 ,
2.6
where fx 21 x 4−6ε/3 1 x x2/2 2−3ε/3 − 1 2x 7/6x2 1/6x32−3ε/31
1 x 4−6ε/3
Letx → 0, then the Taylor expansion leads to
fx 2 14− 6ε
3 x 2 − 3ε1 − 6ε
9 x2 ox2
×
12− 3ε
3 x 2 − 3ε2
18 x2 ox2
− 2 14− 6ε
3 x 2 − 3ε1 − 4ε
6 x2 ox2
× 12− 3ε
3 x 2 − 3ε1 − 6ε
18 x2 ox2
2 1 2 − 3εx 2 − 3ε4 − 9ε
6 x2 ox2
− 2 1 2 − 3εx 2 − 3ε4 − 10ε6 x2 ox2
ε2 − 3ε
3 x2 ox2
.
2.7
Equations2.6 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
> 1/3G1, 1 x2
2/3H1, 1 x2
2.8 forx ∈ 0, δ.
Remark 2.4 For any ε > 0, one has
Trang 6Therefore, M0a, b Ga, b is the best possible upper power mean bound for the
sum1/3Ga, 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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... lower power mean bound for the Trang 3For any 0< ε <1
3 and. .. best possible lower power mean bound for the
Trang 5For any 0< ε < 2/3 and < x < 1,... x2, 1 for x ∈ 0, δ Remark 2.2 For any ε > 0, one has
Trang 4Therefore, M0a,