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A best-possible double inequality between Seiffert and harmonic means

  • Yu-Ming Chu
  • Miao-Kun Wang
  • Zi-Kui Wang
Open Access
Research

Abstract

In this paper, we establish a new double inequality between the Seiffert and harmonic means.

The achieved results is inspired by the papers of Sándor (Arch. Math., 76, 34-40, 2001) and Hästö (Math. Inequal. Appl., 7, 47-53, 2004), and the methods from Wang et al. (J. Math. Inequal., 4, 581-586, 2010). The inequalities we obtained improve the existing corresponding results and, in some sense, are optimal.

2010 Mathematics Subject Classification: 26E60.

Keywords

harmonic mean Seiffert mean inequality 

1 Introduction

For a, b > 0 with ab, the Seiffert mean P(a, b) was introduced by Seiffert [1] as follows:
P ( a , b ) = a - b 4 arctan a b - π . Open image in new window
(1.1)

Recently, the bivariate mean values have been the subject of intensive research. In particular, many remarkable inequalities for the Seiffert mean can be found in the literature [1, 2, 3, 4, 5, 6, 7, 8, 9].

Let H(a, b) = 2ab/(a+b), G ( a , b ) = a b Open image in new window, L(a, b) = (a - b)/(log a - log b), I(a, b) = 1/e(b b /a a )1/(b-a), A(a, b) = (a+b)/2, C(a, b) = (a2+b2)/(a+b), and M p (a, b) = ((a p + b p )/2)1/p(p ≠ 0) and M 0 ( a , b ) = a b Open image in new window be the harmonic, geometric, logarithmic, identric, arithmetic, contraharmonic, and p-th power means of two different positive numbers a and b, respectively. Then, it is well known that
min { a , b } < H ( a , b ) = M - 1 ( a , b ) < G ( a , b ) = M 0 ( a , b ) < L ( a , b ) < I ( a , b ) < A ( a , b ) = M 1 ( a , b ) < C ( a , b ) < max { a , b } . Open image in new window

For all a, b > 0 with ab, Seiffert [1] established that L(a, b) < P(a, b) < I(a, b); Jagers [4] proved that M1/2(a, b) < P (a, b) < M2/3(a, b) and M2/3(a, b) is the best-possible upper power mean bound for the Seiffert mean P(a, b); Seiffert [7] established that P(a, b) > A(a, b)G(a, b)/L(a, b) and P(a, b) > 2 A(a, b)/π; Sándor [6] presented that ( A ( a , b ) + G ( a , b ) ) 2 < P ( a , b ) < A ( a , b ) ( A ( a , b ) + G ( a , b ) ) 2 Open image in new window and A 2 ( a , b ) G ( a , b ) 3 < P ( a , b ) < ( G ( a , b ) + 2 A ( a , b ) ) 3 Open image in new window; Hästö [3] proved that P(a, b) > Mlog 2/ log π(a, b) and Mlog 2/ log π(a, b) is the best-possible lower power mean bound for the Seiffert mean P(a, b).

Very recently, Wang and Chu [8] found the greatest value α and the least value β such that the double inequality A α (a, b)H1-α(a, b) < P(a, b) < A β (a, b)H1-β(a, b) holds for a, b > 0 with ab; For any α ∈ (0, 1), Chu et al. [10] presented the best-possible bounds for P α (a, b)G1(a, b) in terms of the power mean; In [2], the authors proved that the double inequality αA(a, b) + (1 - α)H(a, b) < P(a, b) < βA(a, b) + (1 - β)H(a, b) holds for all a, b > 0 with ab if and only if α ≤ 2/π and β ≥ 5/6; Liu and Meng [5] proved that the inequalities
α 1 C ( a , b ) + ( 1 - α 1 ) G ( a , b ) < P ( a , b ) < β 1 C ( a , b ) + ( 1 - β 1 ) G ( a , b ) Open image in new window
and
α 2 C ( a , b ) + ( 1 - α 2 ) H ( a , b ) < P ( a , b ) < β 2 C ( a , b ) + ( 1 - β 2 ) H ( a , b ) Open image in new window

hold for all a, b > 0 with ab if and only if α1 ≤ 2/9, β1 ≥ 1/π, α2 ≤ 1/π and β2 ≥ 5/12.

For fixed a, b > 0 with ab and x ∈ [0, 1/2], let
h ( x ) = H ( x a + ( 1 - x ) b , x b + ( 1 - x ) a ) . Open image in new window

Then, it is not difficult to verify that h(x) is continuous and strictly increasing in [0, 1/2]. Note that h(0) = H(a, b) < P(a, b) and h(1/2) = A(a, b) > P(a, b). Therefore, it is natural to ask what are the greatest value α and least value β in (0, 1/2) such that the double inequality H(αa + (1 - α)b, αb + (1 - α)a) < P(a, b) < H(βa + (1 - β)b, βb + (1 - β)a) holds for all a, b > 0 with ab. The main purpose of this paper is to answer these questions. Our main result is the following Theorem 1.1.

Theorem 1.1. If α, β ∈ (0, 1/2), then the double inequality
H ( α a + ( 1 - α ) b , α b + ( 1 - α ) a ) < P ( a , b ) < H ( β a + ( 1 - β ) b , β b + ( 1 - β ) a ) Open image in new window

holds for all a, b > 0 with ab if and only if α ( 1 - 1 - 2 π ) 2 Open image in new window and β ( 6 - 6 ) 1 2 Open image in new window.

2 Proof of Theorem 1.1

Proof of Theorem 1.1. Let λ = ( 1 - 1 - 2 π ) 2 Open image in new window and μ = ( 6 - 6 ) 1 2 Open image in new window. We first prove that inequalities
P ( a , b ) > H ( λ a + ( 1 - λ ) b , λ b + ( 1 - λ ) a ) Open image in new window
(2.1)
and
P ( a , b ) < H ( μ a + ( 1 - μ ) b , μ b + ( 1 - μ ) a ) Open image in new window
(2.2)

hold for all a, b > 0 with ab.

Without loss of generality, we assume that a > b. Let t = a b > 1 Open image in new window and p ∈ (0, 1/2); then, from (1.1), one has
H ( p a + ( 1 - p ) b , p b + ( 1 - p ) a ) - P ( a , b ) = 2 [ p t 2 + ( 1 - p ) ] [ ( 1 - p ) t 2 + p ] t 2 + 1 - t 2 - 1 4 arctan t - π = 2 [ p t 2 + ( 1 - p ) ] [ ( 1 - p ) t 2 + p ] ( t 2 + 1 ) ( 4 arctan t - π ) × 4 arctan t - t 4 - 1 2 [ p t 2 + ( 1 - p ) ] [ ( 1 - p ) t 2 + p ] - π . Open image in new window
(2.3)
Let
f ( t ) = 4 arctan t - t 4 - 1 2 [ p t 2 + ( 1 - p ) ] [ ( 1 - p ) t 2 + p ] - π , Open image in new window
(2.4)
then, simple computations lead to
(2.5)
lim t + f ( t ) = π - 1 2 p ( 1 - p ) Open image in new window
(2.6)
and
f ( t ) = f 1 ( t ) ( t 2 + 1 ) [ p ( 1 - p ) t 4 + ( 2 p 2 - 2 p + 1 ) t 2 + p ( 1 - p ) ] 2 , Open image in new window
(2.7)
where
f 1 ( t ) = 4 p 2 ( 1 - p ) 2 t 8 - ( 2 p 2 - 2 p + 1 ) t 7 + 8 p ( 1 - p ) ( 2 p 2 - 2 p + 1 ) t 6 + ( 2 p 2 - 2 p - 1 ) t 5 + 4 ( 6 p 4 - 1 2 p 3 + 1 0 p 2 - 4 p + 1 ) t 4 + ( 2 p 2 - 2 p - 1 ) t 3 + 8 p ( 1 - p ) ( 2 p 2 - 2 p + 1 ) t 2 - ( 2 p 2 - 2 p + 1 ) t + 4 p 2 ( 1 - p ) 2 . Open image in new window
(2.8)
Note that
f 1 ( 1 ) = 0 , Open image in new window
(2.9)
lim t + f 1 ( t ) = + , Open image in new window
(2.10)
f 1 ( t ) = 3 2 p 2 ( 1 - p ) 2 t 7 - 7 ( 2 p 2 - 2 p + 1 ) t 6 + 4 8 p ( 1 - p ) ( 2 p 2 - 2 p + 1 ) t 5 + 5 ( 2 p 2 - 2 p - 1 ) t 4 + 1 6 ( 6 p 4 - 1 2 p 3 + 1 0 p 2 - 4 p + 1 ) t 3 + 3 ( 2 p 2 - 2 p - 1 ) t 2 + 1 6 p ( 1 - p ) ( 2 p 2 - 2 p + 1 ) t - ( 2 p 2 - 2 p + 1 ) , Open image in new window
f 1 ( 1 ) = 0 , Open image in new window
(2.11)
lim t + f 1 ( t ) = + . Open image in new window
(2.12)
Let f 2 ( t ) = f 1 ( t ) 2 Open image in new window, f 3 ( t ) = f 2 ( t ) 3 Open image in new window, f 4 ( t ) = f 3 ( t ) 4 Open image in new window, f 5 ( t ) = f 4 ( t ) 5 Open image in new window, f 6 ( t ) = f 5 ( t ) 6 Open image in new window and f 7 ( t ) = f 6 ( t ) 7 Open image in new window. Then, simple computations lead to
f 2 ( t ) = 1 1 2 p 2 ( 1 - p ) 2 t 6 - 2 1 ( 2 p 2 - 2 p + 1 ) t 5 + 1 2 0 p ( 1 - p ) ( 2 p 2 - 2 p + 1 ) t 4 + 1 0 ( 2 p 2 - 2 p - 1 ) t 3 + 2 4 ( 6 p 4 - 1 2 p 3 + 1 0 p 2 - 4 p + 1 ) t 2 + 3 ( 2 p 2 - 2 p - 1 ) t + 8 p ( 1 - p ) ( 2 p 2 - 2 p + 1 ) , Open image in new window
f 2 ( 1 ) = - 2 ( 2 4 p 2 - 2 4 p + 5 ) , Open image in new window
(2.13)
lim t + f 2 ( t ) = + , Open image in new window
(2.14)
f 3 ( t ) = 2 2 4 p 2 ( 1 - p ) 2 t 5 - 3 5 ( 2 p 2 - 2 p + 1 ) t 4 + 1 6 0 p ( 1 - p ) ( 2 p 2 - 2 p + 1 ) t 3 + 1 0 ( 2 p 2 - 2 p - 1 ) t 2 + 1 6 ( 6 p 4 - 1 2 p 3 + 1 0 p 2 - 4 p + 1 ) t + ( 2 p 2 - 2 p - 1 ) , Open image in new window
f 3 ( 1 ) = - 6 ( 2 4 p 2 - 2 4 p + 5 ) , Open image in new window
(2.15)
lim t + f 3 ( t ) = + , Open image in new window
(2.16)
f 4 ( t ) = 2 8 0 p 2 ( 1 - p ) 2 t 4 - 3 5 ( 2 p 2 - 2 p + 1 ) t 3 + 1 2 0 p ( 1 - p ) ( 2 p 2 - 2 p + 1 ) t 2 + 5 ( 2 p 2 - 2 p - 1 ) t + 4 ( 6 p 4 - 1 2 p 3 + 1 0 p 2 - 4 p + 1 ) , Open image in new window
f 4 ( 1 ) = 4 ( 1 6 p 4 - 3 2 p 3 - 2 5 p 2 + 4 1 p - 9 ) , Open image in new window
(2.17)
lim t + f 4 ( t ) = + , Open image in new window
(2.18)
f 5 ( t ) = 2 2 4 p 2 ( 1 - p ) 2 t 3 - 2 1 ( 2 p 2 - 2 p + 1 ) t 2 + 4 8 p ( 1 - p ) ( 2 p 2 - 2 p + 1 ) t + ( 2 p 2 - 2 p - 1 ) , Open image in new window
f 5 ( 1 ) = 2 ( 6 4 p 4 - 1 2 8 p 3 + 2 0 p 2 + 4 4 p - 1 1 ) , Open image in new window
(2.19)
lim t + f 5 ( t ) = + , Open image in new window
(2.20)
f 6 ( t ) = 1 1 2 p 2 ( 1 - p ) 2 t 2 - 7 ( 2 p 2 - 2 p + 1 ) t + 8 p ( 1 - p ) ( 2 p 2 - 2 p + 1 ) , Open image in new window
(2.21)
f 6 ( 1 ) = 9 6 p 4 - 1 9 2 p 3 + 7 4 p 2 + 2 2 p - 7 , Open image in new window
(2.22)
lim t + f 6 ( t ) = + , Open image in new window
(2.23)
f 7 ( t ) = 3 2 p 2 ( 1 - p ) 2 t - ( 2 p 2 - 2 p + 1 ) Open image in new window
(2.24)
and
f 7 ( 1 ) = 3 2 p 4 - 6 4 p 3 + 3 0 p 2 + 2 p - 1 . Open image in new window
(2.25)

We divide the proof into two cases.

Case 1. p = λ = ( 1 - 1 - 2 π ) 2 Open image in new window. Then equations (2.6), (2.13), (2.15), (2.17), (2.19), (2.22) and (2.25) become
lim t + f ( t ) = 0 , Open image in new window
(2.26)
f 2 ( 1 ) = - 2 ( 5 π - 1 2 ) π < 0 , Open image in new window
(2.27)
f 3 ( 1 ) = - 6 ( 5 π - 1 2 ) π < 0 , Open image in new window
(2.28)
f 4 ( 1 ) = - 2 ( 1 8 π 2 - 4 1 π - 8 ) π 2 < 0 , Open image in new window
(2.29)
f 5 ( 1 ) = - 2 ( 1 1 π 2 - 2 2 π - 1 6 ) π 2 < 0 , Open image in new window
(2.30)
f 6 ( 1 ) = - 7 π 2 - 1 1 π - 2 4 π 2 < 0 Open image in new window
(2.31)
and
f 7 ( 1 ) = π + 8 - π 2 π 2 > 0 . Open image in new window
(2.32)

From (2.24), we clearly see that f7(t) is strictly increasing in [1, +∞), and then (2.32) leads to the conclusion that f7(t) > 0 for t ∈ [1, +∞). Thus, f6(t) is strictly increasing in [1, +∞).

It follows from (2.23) and (2.31) together with the monotonicity of f6(t) that there exists t1> 1 such that f6(t) < 0 for t ∈ (1, t1) and f6(t) > 0 for t ∈ (t1, +∞). Thus, f5(t) is strictly decreasing in [1, t1] and strictly increasing in [t1, +∞).

From (2.20) and (2.30), together with the piecewise monotonicity of f5(t), we clearly see that there exists t2> t1> 1 such that f4(t) is strictly decreasing in [1, t2] and strictly increasing in [t2, +∞). Then, equation (2.18) and inequality (2.29) lead to the conclusion that there exists t3> t2> 1 such that f3(t) is strictly decreasing in [1, t3] and strictly increasing in [t3, +∞).

It follows from (2.16) and (2.28) together with the piecewise monotonicity of f3(t) we conclude that there exists t4> t3> 1 such that f2(t) is strictly decreasing in [1, t4] and strictly increasing in [t4, +∞). Then, equation (2.14) and inequality (2.27) lead to the conclusion that there exists t5> t4> 1 such that f 1 ( t ) Open image in new window is strictly decreasing in [1, t5] and strictly increasing in [t5, +∞).

From equations (2.11) and (2.12), together with the piecewise monotonicity of f 1 ( t ) Open image in new window, we know that there exists t6> t5> 1 such that f1(t) is strictly decreasing in [1, t6] and strictly increasing in [t6, +∞). Then, equations (2.7)-(2.10) lead to the conclusion that there exists t7> t6> 1 such that f(t) is strictly decreasing in [1, t7] and strictly increasing in [t7, +∞).

Therefore, inequality (2.1) follows from equations (2.3)-(2.5) and (2.26) together with the piecewise monotonicity of f(t).

Case 2. p = μ = ( 6 - 6 ) 1 2 Open image in new window. Then, equations (2.13), (2.15), (2.17), (2.19) and (2.21) become
f 2 (1) = 0, Open image in new window
(2.33)
f 3 (1) = 0, Open image in new window
(2.34)
f 4 ( 1 ) = 1 7 1 8 > 0 , Open image in new window
(2.35)
f 5 ( 1 ) = 1 7 9 > 0 Open image in new window
(2.36)
and
f 6 ( t ) = 1 3 6 ( 1 7 5 t 2 - 1 4 7 t + 3 5 ) > 0 Open image in new window
(2.37)

for t > 1.

From inequality (2.37), we know that f5(t) is strictly increasing in [1, +∞), and then inequality (2.36) leads to the conclusion that f5(t) > 0 for t ∈ [1, +∞). Thus, f4(t) is strictly increasing in [1, +∞).

It follows from inequality (2.35) and the monotonicity of f4(t) that f3(t) is strictly increasing in [1, +∞).

Therefore, inequality (2.2) follows easily from equations (2.3)-(2.5), (2.7), (2.9), (2.11), (2.33), and (2.34) together with the monotonicity of f3(t).

Next, we prove that λ = ( 1 - 1 - 2 π ) 2 Open image in new window is the best-possible parameter such that inequality (2.1) holds for all a, b > 0 with ab. In fact, if ( 1 - 1 - 2 π ) 2 = λ < p < 1 2 Open image in new window, then equation (2.6) leads to
lim t + f ( t ) = π - 1 2 p ( 1 - p ) > 0 . Open image in new window
(2.38)
Inequality (2.38) implies that there exists T = T(p) > 1 such that
(2.39)

for t ∈ (T, +∞).

From equations (2.3) and (2.4), together with inequality (2.39), we clearly see that P(a, b) < H(pa + (1 - p)b, pb + (1 - p)a) for a/b ∈ (T2, +∞).

Finally, we prove that μ = ( 6 - 6 ) 1 2 Open image in new window is the best-possible parameter such that inequality (2.2) holds for all a, b > 0 with ab. In fact, if 0 < p < μ = ( 6 - 6 ) 1 2 Open image in new window, then equation (2.13) leads to
f 2 ( 1 ) = - 2 ( 2 4 p 2 - 2 4 p + 5 ) < 0 . Open image in new window
(2.40)
Inequality (2.40) implies that there exists δ = δ (p) > 0 such that
(2.41)

for t ∈ (1, 1 + δ).

Therefore, P(a, b) > H(pa + (1 - p)b, pb + (1 - p)a) for a/b ∈ (1, (1 + δ)2) follows from equations (2.3)-(2.5), (2.7), (2.9), and (2.11) together with inequality (2.41).

Notes

Acknowledgements

This study is partly supported by the Natural Science Foundation of China (Grant no. 11071069), the Natural Science Foundation of Hunan Province (Grant no. 09JJ6003), and the Innovation Team Foundation of the Department of Education of Zhejiang Province(Grant no. T200924).

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© Chu et al; licensee Springer. 2011

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Authors and Affiliations

  1. 1.Department of MathematicsHuzhou Teachers CollegeHuzhouChina
  2. 2.Department of MathematicsHangzhou Normal UniversityHangzhouChina

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