Volume 2013, Issue 1 832591
Research Article
Open Access

Bounds of the Neuman-Sándor Mean Using Power and Identric Means

Yu-Ming Chu

Corresponding Author

Yu-Ming Chu

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

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Bo-Yong Long

Bo-Yong Long

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

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First published: 14 February 2013
Citations: 18
Academic Editor: Wenchang Sun

Abstract

In this paper we find the best possible lower power mean bounds for the Neuman-Sándor mean and present the sharp bounds for the ratio of the Neuman-Sándor and identric means.

1. Introduction

For p the pth power mean Mp(a, b), Neuman-Sándor Mean M(a, b) [1], and identric mean I(a, b) of two positive numbers a and b are defined by
()
()
()
respectively, where is the inverse hyperbolic sine function.

The main properties for Mp(a, b) and I(a, b) are given in [2]. It is well known that Mp(a, b) is continuously and strictly increasing with respect to p for fixed a, b > 0 with ab. Recently, the power, Neuman-Sándor, and identric means have been a subject of intensive research. In particular, many remarkable inequalities for these means can be found in the literature [326].

Let H(a, b) = 2ab/(a + b), , L(a, b) = (ba)/(log b − log a), , A(a, b) = (a + b)/2, T(a, b) = (ab)/[2arctan((ab)/(a + b))], , and C(a, b) = (a2 + b2)/(a + b) be the harmonic, geometric, logarithmic, first Seiffert, arithmetic, second Seiffert, quadratic, and contraharmonic means of two positive numbers a and b with ab, respectively. Then, it is well known that the inequalities
()
hold for all a, b > 0 with ab.
The following sharp bounds for L, I, (IL) 1/2, and (I + L)/2 in terms of power means are presented in [2732]:
()
for all a, b > 0 with ab.
Pittenger [31] found the greatest value r1 and the least value r2 such that the double inequality
()
holds for all a, b > 0, where Lr(a, b) is the rth generalized logarithmic means which is defined by
()
The following sharp power mean bounds for the first Seiffert mean P(a, b) are given in [10, 33]:
()
for all a, b > 0 with ab.
In [17], the authors answered the question: for α ∈ (0,1), what are the greatest value p and the least value q such that the double inequality
()
holds for all a, b > 0 with ab?
Neuman and Sándor [1] established that
()
for all a, b > 0 with ab.
Let 0 < a, b ≤ 1/2 with ab, a = 1 − a and b = 1 − b. Then, the Ky Fan inequalities
()
were presented in [1].
In [24], Li et al. found the best possible bounds for the Neuman-Sándor mean M(a, b) in terms of the generalized logarithmic mean Lr(a, b). Neuman [25] and Zhao et al. [26] proved that the inequalities
()
hold for all a, b > 0 with ab if and only if , β ≥ 1/3, , μ ≥ 1/6, α1 ≥ 2/9, , α2 ≥ 1/3, and .
In [7], Sándor and Trif proved that the inequalities
()
hold for all a, b > 0 with ab.

Neuman and Sándor [15] and Gao [20] proved that α1 = 1, β1 = e/2, α2 = 1, , α3 = 1, β3 = 3/e, α4 = e/π, β4 = 1, α5 = 1, and β5 = 2e/π are the best possible constants such that the double inequalities α1 < A(a, b)/I(a, b) < β1,  α2 < I(a, b)/M2/3(a, b) < β2, α3 < I(a, b)/He(a, b) < β3, α4 < P(a, b)/I(a, b) < β4, and α5 < T(a, b)/I(a, b) < β5 hold for all a, b > 0 with ab, where is the Heronian mean of a and b.

In [34], Sándor established that
()
for all a, b > 0 with ab.
It is not difficult to verify that the inequality
()
holds for all a, b > 0 with ab.
From inequalities (10), (14), and (15), one has
()
for all a, b > 0 with ab.
It is the aim of this paper to find the best possible lower power mean bound for the Neuman-Sándor mean M(a, b) and to present the sharp constants α and β such that the double inequality
()
holds for all a, b > 0 with ab.

2. Main Results

Theorem 1. is the greatest value such that the inequality

()
holds for all a, b > 0 with ab.

Proof. From (1) and (2), we clearly see that both M(a, b) and Mp(a, b) are symmetric and homogenous of degree one. Without loss of generality, we assume that b = 1 and a = x > 1.

Let , then from (1) and (2) one has

()
Let
()
Then, simple computations lead to
()
()
()
where
()
()
()
where
()
()
()
()
()
()
()
where
()
for x > 1.

Equation (33) and inequality (34) imply that is strictly decreasing on [1, +). Then, the inequality (31) and (32) lead to the conclusion that there exists x1 > 1, such that is strictly increasing on [1, x1] and strictly decreasing on [x1, +).

From (29) and (30) together with the piecewise monotonicity of , we clearly see that there exists x2 > x1 > 1, such that f2(x) is strictly increasing on [1, x2] and strictly decreasing on [x2, +).

It follows from (26)–(28) and the piecewise monotonicity of f2(x) that there exists x3 > x2 > 1, such that f1(x), is strictly increasing on [1, x3] and strictly decreasing on [x3, +).

From (23)–(25) and the piecewise monotonicity of f1(x) we see that there exists x4 > x3 > 1, such that f(x) is strictly increasing on (1, x4] and strictly decreasing on [x4, +).

Therefore, for x > 1 follows easily from (19)–(22) and the piecewise monotonicity of f(x).

Next, we prove that is the greatest value such that for all x > 1.

For any ε > 0 and x > 1, from (1) and (2), one has

()

Inequality (35) implies that for any ε > 0, there exists X = X(ε) > 1, such that for x ∈ (X, +).

Remark 2. 4/3 is the least value such that inequality (16) holds for all a, b > 0 with ab, namely, M4/3(a, b) is the best possible upper power mean bound for the Neuman-Sándor mean M(a, b).

In fact, for any ε ∈ (0,4/3) and x > 0, one has

()

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

()

Equations (36) and (37) imply that for any ε ∈ (0,4/3) there exists δ = δ(ε) > 0, such that M(1 + x, 1) > M(4/3)−ε(1 + x, 1) for x ∈ (0, δ).

Theorem 3. For all a, b > 0 with ab, one has

()
with the best possible constants 1 and .

Proof. From (2) and (3), we clearly see that both M(a, b) and I(a, b) are symmetric and homogenous of degree one. Without loss of generality, we assume that b = 1 and a = x > 1. Let

()
Then, simple computations lead to
()
where
()
()
where
()
()
()
()
()
for x > 1.

From (46) and (47), we clearly see that is strictly increasing on [1, +). Then, (45) leads to the conclusion that is strictly increasing on [1, +).

Equations (43) and (44) together with the monotonicity of impliy that f2(x) > 0 for x > 1. Then, (42) leads to the conclusion that f1(x) is strictly increasing on [1, +).

It follows from equations (40) and (41) together with the monotonicity of f1(x) that f(x) is strictly increasing on (1, +).

Therefore, Theorem 3 follows from (39) and the monotonicity of f(x) together with the facts that

()

Acknowledgments

This research was supported by the Natural Science Foundation of China under Grants nos. 11071069 and 11171307, and the Innovation Team Foundation of the Department of Education of Zhejiang Province under Grant no. T200924.

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