PDF
Research Article | Open Access
Best Possible Bounds for Neuman-Sándor Mean by the Identric, Quadratic and Contraharmonic Means
Tie-Hong Zhao,1 Yu-Ming Chu
,2 Yun-Liang Jiang,3 and Yong-Min Li41Department of Mathematics, Hangzhou Normal University, Hangzhou 310036, China
2School of Mathematics and Computation Sciences, Hunan City University, Yiyang 413000, China
3School of Information & Engineering, Huzhou Teachers College, Huzhou 313000, China
4School of Automation, Southeast University, Nanjing 210096, China
Abstract
We prove that the double inequalities hold for all with if and only if , , , and , where , , , and are the identric, Neuman-Sándor, quadratic, and contraharmonic means of and , respectively.
1. Introduction
For and with , the identric mean , Neuman-Sándor mean [1], quadratic mean , contraharmonic mean , and th power mean are defined by respectively, where is the inverse hyperbolic sine function.
Recently, the identric, Neuman-Sándor, quadratic, and contraharmonic means have attracted the interest of numerous eminent mathematicians. In particular, many remarkable inequalities for these means can be found in the literature [1–18].
Let , , , , , and be the harmonic, geometric, logarithmic, first Seiffert, arithmetic, and second Seiffert means of two distinct positive numbers and , respectively. Then it is well known that the inequalities hold for all with .
Neuman and Sándor [1, 8] established that for all with .
Let with , , and . Then the Ky Fan inequalities were presented in [1].
Li et al. [19] found the best possible bounds for the Neuman-Sándor mean in terms of the generalized logarithmic mean . Neuman [20] and Zhao et al. [21] proved that the inequalities hold for all with if and only if , , , , , , , and .
In [22], Chu and Long gave the best possible constants , and such that the double inequalities and hold for all with .
The ratio of identric means leads to the weighted geometric mean which has been investigated in [23–25]. Alzer [26] proved that the inequalities hold for all with .
The following sharp bounds for , , and in terms of the power mean and the convex combination of arithmetic and geometric means are given in [27] as for all with .
Chu et al. [28] presented the optimal constants , and such that the double inequalities hold for all with .
The aim of this paper is to find the best possible constants and such that the double inequalities hold for all with . All numerical computations are carried out using MATHEMATICA software.
2. Lemmas
In order to prove our main results, we need several lemmas, which we present in this section.
Lemma 1. The double inequality holds for .
Proof. To prove Lemma 1, it suffices to prove that
for .
From the expressions of and , we get
where
for .
Therefore, inequality (12) follows from (14)–(16), and inequality (13) follows from (14)–(17).
Lemma 2. Let Then for , and for .
Proof. To prove inequalities (19) and (20), it suffices to show that
for , and
for .
From (21) and (22), one has
for , and
for .
Therefore, inequality (21) follows from (23) and (24), and inequality (22) follows from (23) and (25).
Lemma 3. Let
Then the double inequality
holds for .
Proof. To prove inequality (27), it suffices to show that
for .
First, we prove inequality (28). From the expression of , we have
where
Equation (32) leads to
where
Note that
for , and
for .
It follows from (32) and (34)–(36) together with Lemma 1 that
for , and
for .
From (33), (37), and (38), we clearly see that there exists such that for and for . Then (31) leads to the conclusion that is strictly increasing on and strictly decreasing on .
Therefore, inequality (28) follows from (30) and the piecewise monotonicity of .
Next, we prove inequality (29). From the expression of , we get
where
It follows from Lemma 1 and (40) that
for .
Therefore, inequality (29) follows from (39) together with (41).
Lemma 4. Let Then the double inequality holds for .
Proof. To prove Lemma 4, it suffices to prove that
for .
We first prove inequality (44). From the expression of , we obtain
where
Equation (48) leads to
where
Note that
for , and
for .
It follows from Lemma 1, (48), and (51)–(53) that
for , and
for .
From (50) and (55), we know that is strictly decreasing on , and this in conjunction with (49) and (54) leads to the conclusion that there exists such that for and for . Then (47) implies that is strictly increasing on and strictly decreasing on . Therefore, inequality (44) follows from (46) and the piecewise monotonicity of .
Next, we prove inequality (45). From the expression of one has
where
It follows from Lemma 1 and (52) that
for .
Therefore, inequality (45) follows from (56) together with (58).
Lemma 5. Let be defined as in Lemma 2 and
Then the double inequality
holds for .
Proof. From Lemma 2, one has
for .
Therefore, Lemma 5 follows easily from (61).
Lemma 6. Let be defined as in Lemma 2 and
Then the double inequality
holds for .
Proof. It follows from Lemma 2 that
for .
Therefore, Lemma 6 follows from (64).
Lemma 7. The inequality holds for .
Proof. Let
Then
where
It follows from Lemma 1 and (68) that
for .
Therefore, Lemma 7 follows from (67) together with (69).
Lemma 8. Let
Then for .
Proof. Let
Then
Lemma 7 and give and
for . This in turn implies that
for .
On the other hand, from the expression of , we get
where
for .
From (75)–(76), we clearly see that and for . This in turn implies that
for .
Equation (72) together with inequalities (74) and (77) lead to the conclusion that
for .
Lemma 9. Let Then for .
Proof. Let
then
From (74), we clearly see that
for .
On the other hand, from the expression of together with Lemma 1, we get