Abstract
We prove that the double inequality holds for all with if and only if and , where , and , , and are the Yang and th one-parameter means of and , respectively.
1. Introduction
Let and with . Then the th one-parameter mean , th power mean , harmonic mean , geometric mean , logarithmic mean , first Seiffert mean , identric mean , arithmetic mean , Yang mean , second Seiffert mean , and quadratic mean are, respectively, defined by
It is well known that both the means and are continuous and strictly increasing with respect to for fixed with . Recently, the one-parameter mean and Yang mean have attracted the attention of many researchers.
Alzer [1] proved that the inequalitieshold for all with and .
In [2, 3], the authors discussed the monotonicity and logarithmic convexity properties of the one-parameter mean .
In [4, 5], the authors proved that the double inequalities hold for all with and if and only if , , , and .
Xia et al. [6] proved that the double inequality holds for all with if , and inequality (4) is reversed if .
Gao and Niu [7] presented the best possible parameters and such that the double inequality holds for all with and .
In [8, 9], the authors proved that the double inequalities hold for all with if and only if , , , and .
Xia et al. [10] found that is the best possible lower power mean bound for the one-parameter mean if and is the best possible upper power mean bound for the one-parameter mean if .
For all with , Yang [11] provided the bounds for the Yang mean in terms of other bivariate means as follows:
In [12, 13], the authors proved that the double inequalities hold for all with if and only if , , , , , and , where is the unique solution of the equation on the interval , and .
Very recently, Zhou et al. [14] proved that and are the best possible parameters such that the double inequality holds for all with .
The aim of this paper is to present the best possible parameters and such that the double inequality holds for all with .
2. Main Result
In order to prove our main result we need a lemma, which we present in this section.
Lemma 1. Let , and Then the following statements are true:(1)if , then for all ;(2)if , then there exists such that for and for .
Proof. For part (1), if , then (9) becomes Therefore, part (1) follows from (10).
For part (2), let , , , , , , , , , , and . Then elaborated computations lead to Note that It follows from (24) and (25) that for .
From (23) and (26) we clearly see that is strictly increasing with respect to on the interval . Then (21) and (22) lead to the conclusion that there exists such that the function is strictly decreasing on and strictly increasing on .
It follows from (19) and (20) together with the piecewise monotonicity of the function that there exists such that the function is strictly decreasing on and strictly increasing on .
Making use of (13)–(18) and the same method as the above we know that there exists such that the function is strictly decreasing on and strictly increasing on .
It follows from (12) and the piecewise monotonicity of the function that there exists such that the function is strictly decreasing on and strictly increasing on .
Therefore, part (2) follows easily from (11) and the piecewise monotonicity of the function .
Theorem 2. The double inequalityholds for all with if and only if and .
Proof. Since and are symmetric and homogeneous of degree one, without loss of generality, we assume that and . Let and . Then (1) lead to where where is defined by (9).
We divide the proof into four cases.
Case 1 (). Then it follows from Lemma 1(2), (29), and (31) that there exists such that the function is strictly decreasing on and strictly increasing on , and Therefore, follows easily from (28), (30), and (32) together with the piecewise monotonicity of the function .
Case 2 (). Then (1) leads to Inequality (34) implies that there exists large enough such that for all with .
Case 3 (). Then from Lemma 1(1) and (31) we know that the function is strictly increasing on the interval . Therefore, follows from (28) and (30) together with the monotonicity of the function .
Case 4 (). Let and ; then making use of Taylor expansion we get Equation (36) implies that there exists small enough such that for all .
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgments
The research was supported by the Natural Science Foundation of China under Grants 61374086, 11371125, and 11401191 and the Natural Science Foundation of Zhejiang Broadcast and TV University under Grant XKT-15G17.