Volume 2009, Article ID 741923,6pages doi:10.1155/2009/741923
Research Article
Two Sharp Inequalities for Power Mean,
Geometric Mean, and Harmonic Mean
Yu-Ming Chu
1and Wei-Feng Xia
21Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China 2School 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
Mpa, b ap bp/21/p, p / 0, and Mpa, b ab, p 0. In this paper, we establish two
sharp inequalities as follows: 2/3Ga, b 1/3Ha, b M−1/3a, b and 1/3Ga, b
2/3Ha, b M−2/3a, 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
Mpa, b ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ap bp 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 forMpa, b can be found in literature 1–12. It is well known
thatMpa, 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
14, page 350:
Mlog 2/ log 3a, b 23Aa, b 13Ga, b M2/3a, b 1.3
for alla, b > 0.
In15, Mao proved
M1/3a, b 13Aa, 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 Mpa, b 2/3Ga, b 1/3Ha, b Mra, b
andMqa, b 1/3Ga, b 2/3Ha, b Msa, b for all a, b > 0?
2. Main Results
Theorem 2.1. 2/3Ga, b 1/3Ha, b M−1/3a, b for all a, b > 0, equality holds if and
only ifa b, and M−1/3a, b is the best possible lower power mean bound for the sum 2/3Ga, b
1/3Ha, b.
Proof. Ifa b, then we clearly see that 2/3Ga, b 1/3Ha, b M−1/3a, b a.
Ifa / b and a/b t6, then simple computation leads to
2 3Ga, b 1 3Ha, b − M−1/3a, b b 2t3 3 2t6 31 t6− 8t6 1 t23 2bt3 31 t23 t4− t2 1 × t2 13t4− t2 1 t3t2 12− 12t3t4− t2 1 2bt3 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
For any 0< ε <1
3 and 0< x < 1, one has M−1/3ε1 x2, 1 1/3−ε − 2 3G1 x 2, 1 1 3H1 x 2, 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 fx 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 x 2 ox2 × 11− 3ε 3 x 1 − 3ε2 18 x 2 ox2 − 2 11− 3ε 3 x − 1 − 3ε1 6ε 18 x 2 ox2 × 12− 6ε 3 x − 2ε1 − 3ε 3 x 2 ox2 2 1 1 − 3εx 1 − 3ε1 − 9ε 6 x 2 ox2 − 2 1 1 − 3εx 1 − 3ε1 − 10ε 6 x 2 ox2 ε1 − 3ε 3 x 2 ox2. 2.3
Equations2.2 and 2.3 imply that for any 0 < ε < 1/3 there exists 0 < δ δε < 1,
such thatM−1/3ε1 x2, 1 > 2/3G1 x2, 1 1/3H1 x2, 1 for x ∈ 0, δ.
Remark 2.2. For anyε > 0, one has
sum2/3Ga, b 1/3Ha, b.
Theorem 2.3. 1/3Ga, b 2/3Ha, b M−2/3a, b for all a, b > 0, equality holds if and
only ifa b, and M−2/3a, b is the best possible lower power mean bound for the sum 1/3Ga, b
2/3Ha, b.
Proof. Ifa b, then we clearly see that 1/3Ga, b 2/3Ha, b M−2/3a, b a.
Ifa / b and a/b t6, then elementary calculation yields
1 3Ga, b 2 3Ha, b 2 − M−2/3a, 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 3 t24 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 3 t24 8t21 3t20− 54t18 24t17 3t16 8t15 54t14 24t13− 142t12 24t11 54t10 8t9 3t8 24t7− 54t6 3t4 8t3 1 b2t6t − 14 91 t62 1 t4 3 t20 4t19 10t18 28t17 70t16 148t15 220t14 268t13 277t12 240t11 240t10 240t9 277t8 268t7 220t6 148t5 70t4 28t3 10t2 4t 1> 0. 2.5
For any 0< ε < 2/3 and 0 < x < 1, one has M−2/3ε1, 1 x2 2/3−ε − 1 3G1, 1 x 2 2 3H1, 1 x 2 2/3−ε 21 x4−6ε/3 1 1 x4−6ε/3 − 1 2x 7/6x2 1/6x3 2−3ε/3 1 x 1/2x22−3ε/3 fx 1 1 x4−6ε/31 x 1/2x22−3ε/3, 2.6 where fx 21 x4−6ε/31 x x2/22−3ε/3 − 1 2x 7/6x2 1/6x32−3ε/31 1 x4−6ε/3.
Letx → 0, then the Taylor expansion leads to
fx 2 14− 6ε 3 x 2 − 3ε1 − 6ε 9 x 2 ox2 × 12− 3ε 3 x 2 − 3ε2 18 x 2 ox2 − 2 14− 6ε 3 x 2 − 3ε1 − 4ε 6 x 2 ox2 × 12− 3ε 3 x 2 − 3ε1 − 6ε 18 x 2 ox2 2 1 2 − 3εx 2 − 3ε4 − 9ε 6 x 2 ox2 − 2 1 2 − 3εx 2 − 3ε4 − 10ε 6 x 2 ox2 ε2 − 3ε 3 x 2 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 anyε > 0, one has
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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