Sharp Power Mean Bounds for the Combination of Seiffert and

Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2010, Article ID 108920, 12 pages doi:10.1155/2010/108920

Research Article Sharp Power Mean Bounds for the Combination of Seiffert and Geometric Means Yu-Ming Chu,1 Ye-Fang Qiu,2 and Miao-Kun Wang2 1 2

Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China Department of Mathematics, Zhejiang Sci-Tech University, Hangzhou 310018, China

Correspondence should be addressed to Yu-Ming Chu, [email protected] Received 29 March 2010; Revised 23 July 2010; Accepted 29 July 2010 Academic Editor: Lance Littlejohn Copyright q 2010 Yu-Ming Chu et al. 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 ∈ 0, 1, what are the greatest value p and the least value q such that the double inequality Mp a, b 0 with a / b. Here, Mp a, b, P a, b, and Ga, b denote the power of order p, Seiffert, and geometric means of two positive numbers a and b, respectively.

1. Introduction For p ∈ R, the power mean Mp a, b of order p and the Seiffert mean P a, b of two positive numbers a and b are defined by ⎧ ⎪ ap bp 1/p ⎪ ⎨ , p/ 0, 2 Mp a, b ⎪ ⎪ ⎩√ab, p 0, ⎧ a−b ⎪ ⎪ , a/ b,

⎨ a/b − π P a, b 4 arctan ⎪ ⎪ ⎩a, a b.

1.1

The main properties of the power mean are given in 1. It is well known that Mp a, b is strictly increasing with respect to p for fixed a, b > 0 with a / b. Recently, the power mean has been the subject of intensive research. In particular, many remarkable inequalities for the power mean can be found in the literature 2–16.

2

Abstract and Applied Analysis

The Seiffert mean P was introduced by Seiffert in 17, it can be rewritten in the following symmetric form see 18, 2.4:

P a, b

⎧ ⎪ ⎨

a−b , 2 arcsina − b/a b ⎪ ⎩a,

a / b, a b.

1.2

√ b−log a, b / a, Let Aa, b 1/2a b, Ga, b ab, La, b b−a/log Ha, b a, ba, b a 1/b−a , b/ a, be the arithmetic, geometric, logarithmic, 2ab/a b and Ia, b 1/eb /aa, ba harmonic, and identric means of two positive numbers a and b, respectively. Then it is well known that min{a, b} < Ha, b M−1 a, b < Ga, b M0 a, b < La, b < Ia, b < Aa, b M1 a, b < max{a, b}

1.3

for all a, b > 0 with a / b. In 9, Alzer and Janous presented the sharp power mean bounds for the sum 2/3Aa, b 1/3Ga, b as follows: Mlog 2/ log 3 a, b
0 with a / b. In 17, Seiffert proved that La, b 0 with a / b. The following power mean bounds for the Seiffert mean was given by Jagers 19: M1/2 a, b 0 with a / b. In 20, 21, the authors presented the bounds for the Seiffert mean P in terms of A and G as follows: P a, b >

3Aa, bGa, b , Aa, b 2Ga, b

1 1 2 1 Aa, b Ga, b A1/3 a, bG2/3 a, b for all a, b > 0 with a / b.

1.7


3

The following sharp lower power mean bounds for 1/3Ga, b 2/3Ha, b, 2/3Ga, b 1/3Ha, b, and P a, b can be found in 4, 6: 2 1 Ga, b Ha, b > M−2/3 a, b, 3 3 1 2 Ga, b Ha, b > M−1/3 a, b, 3 3

1.8

P a, b > Mlog 2/ log π a, b for all a, b > 0 with a / b. The purpose of this paper is to answer the question: for α ∈ 0, 1, what are the greatest value p and the least value q such that the double inequality Mp a, b 0 with a /

2. Lemmas In order to prove our main result, we need several lemmas which we present in this section. Lemma 2.1. Let λ ∈ 0, 1/3, x ∈ 1, ∞ and hx 1−3λx2λ1 −13λx2λ −13λx1−3λ. Then there exists x0 > 1 such that hx < 0 for x ∈ 1, x0 , hx > 0 for x ∈ x0 , ∞ and hx0 0. Proof. Simple computations lead to h1 −12λ < 0,

2.1

lim hx ∞,

2.2

h x 1 − 3λ2λ 1x2λ − 2λ1 3λx2λ−1 − 1 3λ, h 1 −6λ1 2λ < 0,

2.3

lim h x ∞,

2.4

h x 2λx2λ−2 1 2λ1 − 3λx 1 − 2λ1 3λ > 0.

2.5

x → ∞

x → ∞

Inequality 2.5 implies that h x is strictly increasing in 1, ∞. Then 2.3 and 2.4 lead to that there exists x1 > 1 such that h x < 0 for x ∈ 1, x1 and h x > 0 for x ∈ x1 , ∞. Hence, hx is strictly decreasing in 1, x1 and strictly increasing in x1 , ∞. Therefore, Lemma 2.1 follows from 2.1 and 2.2 together with the monotonicity of hx. √ Lemma 2.2. If λ ∈ 1/ 10, 1/3, then the following statements are true: 1 300λ4 324λ3 29λ2 − 45λ − 8 < 0; 2 −1176λ5 − 24λ4 114λ3 10λ2 − 3λ − 1 < 0; 3 −24λ5 72λ4 178λ3 6λ2 − 25λ − 3 < 0.

4


Proof. Simple computations lead to √ 3 29λ2 − 45λ − 8 < 300 × 1/34 324 × 1/33 29 × 1/32 − 45/ 10 − 8 1 300λ4 324λ√ 590 − 243 10/54 < 0; √ 5 √ 4 2 −1176λ5 − 24λ4 114λ3 √10λ2 − 3λ − 1 < −1176 √ × 1/ 10 − 24 × 1/ 10 114 × 1/33 10 × 1/32 − 3/ 10 − 1 3070 − 1107 10/750 < 0; √ 5 10 72 × 1/34 3 −24λ5 72λ4 178λ3 6λ2 √− 25λ − 3 < −24 × 1/ √ 3 2 178 × 1/3 6 × 1/3 − 25/ 10 − 3 34750 − 17037 10/6750 < 0. √ Lemma 2.3. If α ∈ 0, 3/ 10, then I α a, bG1−α a, b < M2α/3 a, b

2.6

holds for all a, b > 0 with a / b. Proof.√Without loss of generality, we assume that a > b. Let t a/b > 1 and β α/3 ∈ 0, 1/ 10. Then

logM2α/3 a, b − log I α a, bG1−α a, b

log M2β a, b − log I 3β a, bG1−3β a, b

3βt 1 − 3β 1 t2β 1 log − log t − log t 3β. 2β 2 t−1 2

ft

3βt 1 − 3β 1 t2β 1 log − log t − log t 3β. 2β 2 t−1 2

2.7

Let 2.8

Then simple computations lead to limft 0,

2.9

t→1

f t

gt t − 12

,

2.10

where gt t2β1 − 2t2β t2β−1 / 1 t2β − 3βt − 1 3β log t − 1 − 3βt − 2 1/t/2, g1 0, g t

g1 t 2 , 2 2t 1 t2β

2.11


5

where g1 t 1−3βt4β2 6βt4β1 −13βt4β −2βt2β2 4βt2β1 −2βt2β −13βt2 6βt1−3β, g1 1 0,

2.12

g1 1 0,

2.13

g1 1 0.

2.14

Let g2 t g1 t/8βt2β−3 and g3 t g2 t/2βt2β−3 . Then g2 t 1 − 3β 2β 1 4β 1 t2β2 3β 4β 1 4β − 1 t2β1 − 1 3β 4β − 1 2β − 1 t2β − β β 1 2β 1 t2 β 2β 1 2β − 1 t − β β − 1 2β − 1 ,

2.15

g2 1 0,

√ √ g2 1 2 1 − 10β 1 10β ≥ 0,

2.17

√ √ g2 1 2 6β 1 1 − 10β 1 10β ≥ 0,

2.18

2 g3 t 2 1 − 3β 2β 1 4β 1 β 1 t2 3β 4β 1 4β − 1 2 × 2β 1 2β − 1 t − 2 1 3β 4β − 1 2β − 1 β − 1 ,

g3 1 9β 3 − 28β2 > 0, g3 t 2β 1 4β 1 4 1 − 3β 2β 1 β 1 t 3β 4β − 1 2β − 1 ,

g3 1 2β 1 4β 1 4 3β − 46β2 , min

β∈0,1/10

4 3β − 46β

2

√ 3 10 − 6 > 0. 10

2.16

2.19 2.20 2.21 2.22 2.23

√ From 2.21−2.23 we clearly see that g3 t > g3 1 ≥ 3 10 − 6/102β 14β 1 > 0 for t ∈ 1, ∞, hence g3 t is strictly increasing in 1, ∞. Then 2.20 implies that g3 t > 0 for t ∈ 1, ∞, hence g2 t is strictly increasing in 1, ∞. It follows from 2.18 and the monotonicity of g2 t that g2 t > 0 for t ∈ 1, ∞, hence g2 t is strictly increasing in 1, ∞. Then 2.17 implies that g2 t > 0 for t ∈ 1, ∞, therefore g2 t is strictly increasing in 1, ∞. Equation 2.16 and the monotonicity of g2 t lead to that g2 t > 0 for t ∈ 1, ∞, so g1 t is strictly increasing in 1, ∞. From 2.9−2.14 and the monotonicity of g1 t we can deduce that ft > 0 for t ∈ 1, ∞.

2.24

6

Abstract and Applied Analysis Therefore, Lemma 2.3 follows from 2.7 and 2.8 together with 2.24.

Remark 2.4. In 22, Theorem 3.1, the authors proved that I 1/2 a, bG1/2 a, b < M1/3 a, b,

2.25

I 2/3 a, bG1/3 a, b < M4/9 a, b,

2.26

I 1/3 a, bG2/3 a, b < M2/9 a, b

2.27

for all a, b > 0 with a / b. Obviously, 2.6 is the generalization of 2.25−2.27. √ √ Remark 2.5. If α ∈ 3/ 10, 1, then β α/3 ∈ 1/ 10, 1/3 and 2.17 leads to

√ √ g2 1 2 1 − 10β 1 10β < 0.

2.28

Inequality 2.28 and the continuity of g2 t imply that there exists δ > 0 such that g 2 t < 0

2.29

for t ∈ 1, 1 δ. From 2.29 and 2.9−2.16 we can deduce that ft < 0

2.30

for t ∈ 1, 1 δ. Equations 2.7 and 2.8 together with 2.30 lead to I α a, bG1−α a, b > M2α/3 a, b

2.31

for all a, b > 0 with a/b ∈√ 1, 1 δ ∪ 1/1 δ, 1. Therefore, α0 3/ 10 is the largest value in 0, 1 such that inequality 2.6 holds for α ∈ 0, α0 .

3. Main Result Theorem 3.1. If α ∈ 0, 1, then M0 a, b 0 with a / b, and M0 a, b and M2α/3 a, b are the best possible lower and upper power mean bounds for the product Pα a, bG1−α a, b.


7

Proof. For all a, b > 0 with a / b, from 1.3, 1.5 and Lemma 2.1 we clearly see that P α a, bG1−α a, b > M0 a, b

3.2

P α a, bG1−α a, b < M2α/3 a, b

3.3

for all α ∈ 0, 1, and

√ for α ∈ 0, 3/ 10. √ b. Next, we prove that 3.3 is also true for α ∈ 3/ 10, 1 and all a, b > 0 with a / Without loss of generality, we assume that a > b. Let t a/b > 1 and β α/3 ∈ √ 1/ 10, 1/3. Then 1.1 leads to

logM2α/3 a, b − log P α a, bG1−α a, b

log M2β a, b − log P 3β a, bG1−3β a, b

1 1 t4β t2 − 1 log − 3β log − 1 − 3β log t. 2β 2 4 arctan t − π

Ft

1 1 t4β t2 − 1 log − 3β log − 1 − 3β log t. 2β 2 4 arctan t − π

3.4

Let 3.5

Then simple computations lead to limFt 0,

F t

t→1

1 − 3β t4β2 − 1 3β t4β − 1 3β t2 1 − 3β 12β . 2 t 14 arctan t − π tt2 − 1 t4β 1

3.6 3.7

From Lemma 2.1 we know that there exists λ0 > 1 such that 1 − 3β λ0 4β2 − 1 3β λ0 4β − 1 3β λ0 2 1 − 3β 0, 1 − 3β t4β2 − 1 3β t4β − 1 3β t2 1 − 3β < 0

3.9

1 − 3β t4β2 − 1 3β t4β − 1 3β t2 1 − 3β > 0

3.10

3.8

for t ∈ 1, λ0 , and

for t ∈ λ0 , ∞.

8

Abstract and Applied Analysis We divide two cases to prove that F t > 0

3.11

for t > 1. Case 1. t ∈ λ0 , ∞. Then 3.11 follows from 3.7 and 3.8 together with 3.10. Case 2. t ∈ 1, λ0 . Then 3.7 can be written as 1 − 3β t4β2 − 1 3β t4β − 1 3β t2 1 − 3β F1 t, F t tt2 − 1 t4β 1 4 arctan t − π

3.12

where 12βt t2 − 1 t4β 1 F1 t 4 arctan t 2 − π. t 1 1 − 3β t4β2 − 1 3β t4β − 1 3β t2 1 − 3β

3.13

Let x t2 ∈ 1, λ0 2 , then 3.13 leads to F1 1 0, F1 t

t2 1

4F2 x 2 , 4β2 − 1 3β t4β − 1 3β t2 1 − 3β 1 − 3β t

2

3.14

where

F2 x 18β2 − 9β 1 x4β3 − 18β2 − 3β 1 x4β2

− 18β2 3β 1 x4β1 18β2 9β 1 x4β

2 6β2 − 1 x2β3 − 2 6β2 − 1 x2β2

− 2 6β2 − 1 x2β1 2 6β2 − 1 x2β

18β2 9β 1 x3 − 18β2 3β 1 x2

− 18β2 − 3β 1 x 18β2 − 9β 1 .

3.15


9

Let F3 x x4−2β /8βF2 4 x, then 3.15 leads to F2 1 0, F2 1 0,

3.16

F2 1 0, F2 1 0,

F3 x 18β2 − 9β 1 4β 1 4β 3 2β 1 x2β3

− 18β2 − 3β 1 2β 1 4β 1 4β − 1 x2β2

− 18β2 3β 1 4β 1 4β − 1 2β − 1 x2β1

18β2 9β 1 4β − 1 2β − 1 4β − 3 x2β

6β − 1 2β 3 β 1 2β 1 x3

3.17

2

− 6β2 − 1 β 1 2β 1 2β − 1 x2

− 6β2 − 1 2β 1 2β − 1 β − 1 x

6β2 − 1 2β − 1 β − 1 2β − 3 . √ From β ∈ 1/ 10, 1/3 and x > 1, we clearly see that 18β2 − 9β 14β 14β 32β 1x2β3 < 18β2 − 9β 14β 14β 32β 1x3 , −18β2 − 3β 12β 14β 14β − 1x2β2 < −18β2 − 3β 12β 14β 14β − 1x2 , −18β2 3β 14β 14β − 12β − 1x2β1 < −18β2 3β 14β 14β − 12β − 1x2 and 18β2 9β 14β − 12β − 14β − 3x2β < 18β2 9β 14β − 12β − 14β − 3x. These inequalities and Lemma 2.2 lead to

F3 x < 18β2 − 9β 1 4β 1 4β 3 2β 1 x3

− 18β2 − 3β 1 2β 1 4β 1 4β − 1 x2

− 18β2 3β 1 4β 1 4β − 1 2β − 1 x2

18β2 9β 1 4β − 1 2β − 1 4β − 3 x

6β2 − 1 2β 3 β 1 2β 1 x3

− 6β2 − 1 β 1 2β 1 2β − 1 x2

− 6β2 − 1 2β 1 2β − 1 β − 1 x

6β2 − 1 2β − 1 β − 1 2β − 3

10


2β 300β4 324β3 29β2 − 45β − 8 x3

−1176β5 − 24β4 114β3 10β2 − 3β − 1 x2

2 276β5 − 276β4 3β3 43β2 − 3β − 1 x

24β5 − 72β4 62β3 − 6β2 − 11β 3

< 2β 300β4 324β3 29β2 − 45β − 8 x

−1176β5 − 24β4 114β3 10β2 − 3β − 1 x

2 276β5 − 276β4 3β3 43β2 − 3β − 1 x 24β5 − 72β4 62β3 − 6β2 − 11β 3

−24β5 72β4 178β3 6β2 − 25β − 3 x 24β5 − 72β4 62β3 − 6β2 − 11β 3 < −24β5 72β4 178β3 6β2 − 25β − 3 24β5 − 72β4 62β3 − 6β2 − 11β 3

12β 20β2 − 3 < 0. 3.18

From 3.16−3.18 we can deduce that F2 x < 0

3.19

for x ∈ 1, λ0 2 . Equation 3.14 together with 3.19 imply that F1 t < 0

3.20

for t ∈ 1, λ0 . Therefore, 3.11 follows from 3.9 and 3.12 together with 3.20. It follows from 3.4−3.6 and 3.11 that P α a, bG1−α a, b < M2α/3 a, b

3.21

√ for α ∈ 3/ 10, 1 and all a, b > 0 with a / b. At last, we prove that M0 a, b and M2α/3 a, b are the best possible lower- and upperpower mean bounds for the product P α a, bG1−α a, b, respectively.


11

For any 0 < ε < 2/3α and x > 0, from 1.1 one has Mε 1, x lim x → ∞ P α 1, xG1−α 1, x

lim

x → ∞

1 x−ε /21/ε 1−α/2 ∞, α x √ 1 − x−1 / 4 arctan x − π

3.22

2α/3−ε 2α/3−ε P α 1, 1 xG1−α 1, 1 x − M2α/3−ε 1, 1 x

Jx

√ 4 arctan 1 x − π

α2α−3ε/3 ,

3.23

√ where Jx xα2α−3ε/3 1 x1−α2α−3ε/6 − 1/2 4 arctan 1 x − πα2α−3ε/3 1 1 x2α−3ε/3 . Let x → 0, making use of the Taylor expansion we get Jx

1 ε2α − 3εx1/3α2α−3ε2 o x1/3α2α−3ε2 . 24

3.24

Equation 3.22 implies that for any 0 < ε < 2/3α there exists X Xε, α > 1 such that Mε 1, x > P α 1, xG1−α 1, x for x ∈ X, ∞. Equations 3.23 and 3.24 imply that for any 0 < ε < 2/3α there exists δ δε, α > 0 such that P α 1, 1 xG1−α 1, 1 x > M2α/3−ε 1, 1 x for x ∈ 0, δ.

Acknowledgments The authors wish to thank the anonymous referees for their very careful reading of the paper and fruitful comments and suggestions. This paper is partly supported by N S Foundation of China under Grant 60850005, N S Foundation of Zhejiang Province under Grants Y7080106 and Y607128, and the Innovation Team Foundation of the Department of Education of Zhejiang Province under Grant T200924.

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12


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