Bounds of the Neuman-Sándor Mean Using Power and Identric Means
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
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 ∈ ℝ for fixed a, b > 0 with a ≠ b. 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 [3–26].
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 a ≠ b, where is the Heronian mean of a and b.
2. Main Results
Theorem 1. is the greatest value such that the inequality
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
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 a ≠ b, 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 a ≠ b, one has
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
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.