Volume 2012, Issue 1 379785
Research Article
Open Access

A Best Possible Double Inequality for Power Mean

Yong-Min Li

Yong-Min Li

Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China hutc.zj.cn

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Bo-Yong Long

Bo-Yong Long

School of Mathematics Science, Anhui University, Hefei 230039, China ahu.edu.cn

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Yu-Ming Chu

Corresponding Author

Yu-Ming Chu

Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China hutc.zj.cn

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First published: 10 October 2012
Citations: 1
Academic Editor: Huijun Gao

Abstract

We answer the question: for any p, q with pq and p ≠ −q, what are the greatest value λ = λ(p, q) and the least value μ = μ(p, q), such that the double inequality holds for all a, b > 0 with ab? Where Mp(a, b) is the pth power mean of two positive numbers a and b.

1. Introduction

For p, the pth power mean Mp(a, b) of two positive numbers a and b is defined by
(1.1)

It is well known that Mp(a, b) is continuous and strictly increasing with respect to p for fixed a, b > 0 with ab. Many classical means are special case of the power mean, for example, M−1(a, b) = H(a, b) = 2ab/(a + b), , and M1(a, b) = A(a, b) = (a + b)/2 are the harmonic, geometric, and arithmetic means of a and b, respectively. Recently, the power mean has been the subject of intensive research. In particular, many remarkable inequalities and properties for the power mean can be found in literature [115].

Let L(a, b) = (ab)/(log a − log b) and I(a, b) = 1/e(aa/bb) 1/(ab) be the logarithmic and identric means of two positive numbers a and b with ab, respectively. Then it is well known that
(1.2)
for all a, b > 0 with ab.
In [1622], the authors presented the sharp power mean bounds for L, I, (IL) 1/2, and (L + I)/2 as follows:
(1.3)
for all a, b > 0 with ab.
Alzer and Qiu [12] proved that the inequality
(1.4)
holds for all a, b > 0 with ab if and only if p ≤ log 2/(1 + log 2) = 0.40938….
The following sharp bounds for the sum αA(a, b)+(1 − α)L(a, b), and the products Aα(a, b)L1−α(a, b) and Gα(a, b)L1−α(a, b) in terms of power means were proved in [5, 8] as follows:
(1.5)
for any α ∈ (0,1) and all a, b > 0 with ab.
In [2, 7], the authors answered the question: for any α ∈ (0,1), what are the greatest values p1 = p1(α),  p2 = p2(α),   p3 = p3(α), and p4 = p4(α), and the least values q1 = q1(α), q2 = q2(α),  q3 = q3(α), and q4 = q4(α), such that the inequalities
(1.6)
hold for all a, b > 0 with ab?
In [4], the authors presented the greatest value p = p(α, β) and the least value q = q(α, β) such that the double inequality
(1.7)
holds for all a, b > 0 with ab and α, β > 0 with α + β < 1.
It is the aim of this paper to answer the question: for any p, q with pq and p ≠ −q, what are the greatest value λ = λ(p, q) and the least value μ = μ(p, q), such that the double inequality
(1.8)
holds for all a, b > 0 with ab?

2. Main Result

In order to establish our main result, we need a lemma which we present in this section.

Lemma 2.1. Let p, q ≠ 0,  pq and x > 1. Then

(2.1)
for p + q > 0, and
(2.2)
for p + q < 0.

Proof. From (1.1), we have

(2.3)

Let

(2.4)
then simple computations lead to
(2.5)
(2.6)

Equation (2.6) implies that

(2.7)
for p + q > 0, and
(2.8)
for p + q < 0.

Therefore, inequality (2.1) follows from (2.3)–(2.5) and inequality (2.7), and inequality (2.2) follows from (2.3)–(2.5) and inequality (2.8).

Let

(2.9)

Then we clearly see that , and it is not difficult to verify that the identity holds for all a, b > 0 if . Let

(2.10)

Then we have Theorem 2.2 as follows.

Theorem 2.2. The double inequality

(2.11)
holds for all a, b > 0 with ab, and Mλ(a, b) and Mμ(a, b) are the best possible lower and upper power mean bounds for the geometric mean of Mp(a, b) and Mq(a, b).

Proof. From (1.1), we clearly see that Mp(a, b) is symmetric and homogenous of degree 1. Without loss of generality, we assume that b = 1, a = x > 1 and p > q. We divide the proof of inequality (2.11) into three cases.

Case 1. (p, q) ∈ E1⋃ E2. Then from Lemma 2.1, we clearly see that

(2.12)
for (p, q) ∈ E1, and
(2.13)
for (p, q) ∈ E2.

From (1.1), we get

(2.14)

Let

(2.15)
then simple computations lead to
(2.16)
(2.17)
where
(2.18)
(2.19)
(2.20)
where
(2.21)
(2.22)
(2.23)
where
(2.24)
(2.25)
(2.26)
where
(2.27)
(2.28)
(2.29)

If (p, q) ∈ E1, then (2.15), (2.18), (2.21), (2.22), (2.24), (2.25), (2.27), and (2.28) lead to

(2.30)
(2.31)
(2.32)
(2.33)
(2.34)
(2.35)
(2.36)
(2.37)

We divide the discussion into two subcases.

Subcase 1.1. (p, q) ∈ E1. Then (2.26) and (2.29) together with inequalities (2.36) and (2.37) imply that I(x) is strictly decreasing in [1, +). In fact, if (q2p2 + 2pq)/(p + q) ≥ 0, then (2.29) and inequality (2.37) imply that J(x) < 0 for x ∈ [1, +). If (q2p2 + 2pq)/(p + q) < 0, then (2.29) and inequality (2.36) lead to the conclusion that J(x) < 0 for x ∈ [1, +).

From inequalities (2.34) and (2.35) together with the monotonicity of I(x), we know that there exists λ1 > 1 such that I(x) > 0 for x ∈ [1, λ1) and I(x) < 0 for x ∈ (λ1, +). Then (2.23) leads to the conclusion that H(x) is strictly increasing in [1, λ1] and strictly decreasing in [λ1, +).

It follows from (2.32) and (2.33) together with the piecewise monotonicity of H(x) that there exists λ2 > λ1 > 1 such that H(x) > 0 for [1, λ2] and H(x) < 0 for (λ2, +). Then (2.20) leads to the conclusion that G(x) is strictly increasing in [1, λ2] and strictly decreasing in [λ2, +).

From (2.17), (2.19) and (2.31) together with the piecewise monotonicity of G(x), we clearly see that there exists λ3 > λ2 > 1 such that F(x) is strictly increasing in [1, λ3] and strictly decreasing in [λ3, +).

Therefore, follows from (2.14)–(2.16) and (2.30) together with the piecewise monotonicity of F(x).

Subcase 1.2.   (p, q) ∈ E2. Then (2.30) and (2.35) again hold, and (2.18), (2.21), (2.22), and (2.28) lead to

(2.38)
(2.39)
(2.40)
(2.41)

It follows from (2.29) and inequalities (q2p2 + 2pq)/(p + q) < 0 and (2.41) that J(x) > 0 for x ∈ [1, +). Then (2.26) and inequality (2.35) lead to the conclusion that I(x) > 0 for x ∈ [1, +). Therefore, H(x) is strictly increasing in [1, +) follows from (2.23).

It follows from (2.20) and (2.39) together with inequality (2.40) and the monotonicity of H(x) that there exists μ1 > 1 such that G(x) is strictly decreasing in [1, μ1] and strictly increasing in [μ1, +).

From (2.17), (2.19) and (2.38) together with the piecewise monotonicity of G(x), we clearly see that there exists μ2 > μ1 > 1 such that F(x) is strictly decreasing in [1, μ2] and strictly increasing in [μ2, +).

Therefore, follows from (2.14)–(2.16) and (2.30) together with the piecewise monotonicity of F(x).

Case 2. (p, q) ∈ E3⋃ E5. Clearly, we have for (p, q) ∈ E3 and for (p, q) ∈ E5. Therefore, we need only to prove that

(2.42)
for r > 0, and
(2.43)
for r < 0.

From (1.1), one has

(2.44)

Let

(2.45)
then simple computations lead to
(2.46)
(2.47)

If r > 0 (or r < 0, resp.), then (2.47) leads to the conclusion that f(x) is strictly decreasing (or increasing, resp.) in [1, +). Therefore, inequalities (2.42) and (2.43) follow from (2.44)–(2.46) and the monotonicity of f(x).

Case 3.  (p, q) ∈ E4⋃ E6. Then from Lemma 2.1, we clearly see that for (p, q) ∈ E4 and for (p, q) ∈ E6. Therefore, we need only to prove that

(2.48)
for (p, q) ∈ E4, and
(2.49)
for (p, q) ∈ E6.

From (1.1), we get

(2.50)

Let

(2.51)
then simple computations lead to
(2.52)
(2.53)

If (p, q) ∈ E4 (or E6, resp.), then (2.53) implies that f(x) is strictly increasing (or decreasing, resp.) in [1, +). Therefore, inequalities (2.48) and (2.49) follow from (2.50)–(2.52) and the monotonicity of f(x).

Next, we prove that Mλ(a, b) and Mμ(a, b) are the best possible lower and upper power mean bounds for the geometric mean of Mp(a, b) and Mq(a, b). We divide the proof into six cases.

Case A. (p, q) ∈ E1. Then for any ϵ ∈ (0, (p + q)/2) and x > 0, from (1.1), one has

(2.54)
(2.55)

Letting x → 0 and making use of Taylor expansion, we get

(2.56)

Equations (2.54) and (2.56) together with inequality (2.55) imply that for any ϵ ∈ (0, p + q/2), there exist δ1 = δ1(ϵ) > 0 and X1 = X1(p, q, ϵ) > 1 such that for x ∈ (0, δ1) and for x ∈ (X1, +).

Case B.   (p, q) ∈ E2. Then for ϵ ∈ (0, −(p + q)/2) and x > 0, making use of (1.1) and Taylor expansion, we have

(2.57)
(2.58)

Equation (2.57) and inequality (2.58) imply that for any ϵ ∈ (0, −(p + q)/2), there exist δ2 = δ2(ϵ) > 0 and X2 = X2(p, q, ϵ) > 1 such that for x ∈ (0, δ2) and for x ∈ (X2, +).

Case C.   (p, q) ∈ E3. Then for ϵ ∈ (0, p/2) and x > 0, making use of (1.1) and Taylor expansion, we have

(2.59)

Equation (2.59) leads to the conclusion that for any ϵ ∈ (0, p/2), there exist δ3 = δ3(ϵ) > 0 and X3 = X3(p, ϵ) > 1 such that for x ∈ (0, δ3) and for x ∈ (X3, +).

Case D. (p, q) ∈ E4. Then for ϵ ∈ (0, (p + q)/2) and x > 0, making use of (1.1) and Taylor expansion, we have

(2.60)

Equation (2.60) implies that for any ϵ ∈ (0, (p + q)/2), there exist δ4 = δ4(ϵ) > 0 and X4 = X4(p, q, ϵ) > 1 such that for x ∈ (0, δ4) and for x ∈ (X4, +).

Case E. (p, q) ∈ E5. Then for any ϵ ∈ (0, −q/2) and x > 0, making use of (1.1) and Taylor expansion, one has

(2.61)

Equation (2.61) leads to the conclusion that for any ϵ ∈ (0, −q/2), there exist δ5 = δ5(ϵ) > 0 and X5 = X5(q, ϵ) > 1 such that for x ∈ (0, δ5) and for x ∈ (X5, +).

Case F. (p, q) ∈ E6. Then for any ϵ ∈ (0, −(p + q)/2) and x > 0, making use of (1.1) and Taylor expansion, one has

(2.62)

Equation (2.62) shows that for any ϵ ∈ (0, −(p + q)/2), there exist δ6 = δ6(ϵ) > 0 and X6 = X6(p, q, ϵ) > 1 such that for x ∈ (0, δ6) and for x ∈ (X6, +).

Acknowledgments

This research was supported by the Natural Science Foundation of China under Grants 11071069 and 11171307 and the Innovation Team Foundation of the Department of Education of Zhejiang Province under Grant T200924.

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