#### 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
for .

From (83), we clearly see that and for . This in turn implies that
for .

Equation (81) together with inequalities (82) and (84) lead to the conclusion that
for .

Lemma 10. *Let be defined as in Lemma 2 and
**
Then for . *

*Proof. *Differentiating yields

It follows from (19) and (87) that for .

Therefore, for