We present the best possible power mean bounds for the product for any p > 0, α ∈ (0,1), and all a, b > 0 with a ≠ b. Here, M_p(a, b) is the pth power mean of two positive numbers a and b.

1. Introduction

For p ∈ ℝ, the pth power mean M_p(a, b) of two positive numbers a and b is defined by

(1.1)

It is well known that M_p(a, b) is continuous and strictly increasing with respect to p ∈ ℝ for fixed a, b > 0 with a ≠ b. Many classical means are special cases 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 [1–22].

Let L(a, b) = (a − b)/(log a − log b),

and

be the logarithmic, Seiffert and identric means of two positive numbers a and b with a ≠ b, respectively. Then it is well known that

(1.2)

for all a, b > 0 with a ≠ b.

In [23–29], 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 a ≠ b.

Alzer and Qiu [12] proved that the inequality

(1.4)

holds for all a, b > 0 with a ≠ b 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 A^α(a, b)L^1−α(a, b) and G^α(a, b)L^1−α(a, b) in terms of power means were proved in [5, 8]:

(1.5)

for any α ∈ (0,1) and all a, b > 0 with a ≠ b.

In [2, 7] the authors answered the questions: 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 a ≠ b?

It is the aim of this paper to present the best possible power mean bounds for the product for any p > 0, α ∈ (0,1) and all a, b > 0 with a ≠ b.

2. Main Result

Theorem 2.1. Let p > 0, α ∈ (0,1) and a, b > 0 with a ≠ b. Then

(1)
for α = 1/2,
(2)
for α > 1/2 and for α < 1/2, and the bounds M_(2α−1)p(a, b) and M₀(a, b) for the product in either case are best possible.

Proof. From (1.1) we clearly see that M_p(a, b) is symmetric and homogenous of degree 1. Without loss of generality, we assume that b = 1, a = x > 1.

(1)
If α = 1/2, then (1.1) leads to

(2.1)

(2)
Firstly, we compare the value of M_(2α−1)p(x, 1) to the value of for α ∈ (0,1/2)∪(1/2,1). From (1.1) we have

(2.2)

Let

(2.3)

then simple computations lead to

(2.4)

(2.5)

where

(2.6)

(2.7)

where

(2.8)

(2.9)

If α ∈ (1/2,1), then (2.9) implies that h(x) is strictly decreasing in [1, +∞). Therefore, follows easily from (2.2)–(2.8) and the monotonicity of h(x).

If α ∈ (0,1/2), then (2.9) leads to the conclusion that h(x) is strictly increasing in [1, +∞). Therefore, follows easily from (2.2)–(2.8) and the monotonicity of h(x).

Secondly, we compare the value of M₀(x, 1) to the value of . It follows from (1.1) that

(2.10)

Let

(2.11)

then simple computations lead to

(2.12)

(2.13)

If α ∈ (1/2,1), then (2.13) implies that F(x) is strictly increasing in [1, +∞). Therefore, follows easily from (2.10)–(2.12) and the monotonicity of F(x).

If α ∈ (0,1/2), then (2.13) leads to the conclusion that F(x) is strictly decreasing in [1, +∞). Therefore, follows easily from (2.10)–(2.12) and the monotonicity of F(x).

Next, we prove that the bound M_(2α−1)p(a, b) for the product in either case is best possible.

If α ∈ (0,1/2), then for any ϵ ∈ (0, (1 − 2α)p) and x > 0 we have

(2.14)

Letting x → 0 and making use of Taylor’s expansion, one has

(2.15)

Equations (2.14) and (2.15) imply that for any α ∈ (0,1/2) and ϵ ∈ (0, (1 − 2α)p) there exists δ₁ = δ₁(ϵ) > 0, such that for x ∈ (0, δ₁).

If α ∈ (1/2,1), then for any ϵ ∈ (0, (2α − 1)p) and x > 0 we have

(2.16)

Letting x → 0 and making use of Taylor’s expansion, one has

(2.17)

Equations (2.16) and (2.17) imply that for any α ∈ (1/2,1) and ϵ ∈ (0, (2α − 1)p) there exists δ₂ = δ₂(ϵ) > 0, such that for x ∈ (0, δ₂).

Finally, we prove that the bound M₀(a, b) for the product in either case is best possible.

If α ∈ (0,1/2), then for any ϵ > 0 we clearly see that

(2.18)

Equation (2.18) implies that for any α ∈ (0,1/2) and ϵ > 0 there exists T₁ = T₁(ϵ) > 1, such that for x ∈ (T₁, +∞).

If α ∈ (1/2,1), then for any ϵ > 0 we have

(2.19)

Equation (2.19) implies that for any α ∈ (1/2,1) and ϵ > 0 there exists T₂ = T₂(ϵ) > 1, such that for x ∈ (T₂, +∞).

Acknowledgments

This paper was supported by the Natural Science Foundation of China under Grants 11071069 and 11171307, the Natural Science Foundation of Hunan Province under Grant 09JJ6003, and the Innovation Team Foundation of the Department of Education of Zhejiang Province under Grant T200924.

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Optimal Inequalities for Power Means

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2. Main Result

Acknowledgments

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