#### 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 , 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 .

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 .

Li et al.  found the best possible bounds for the Neuman-Sándor mean in terms of the generalized logarithmic mean . Neuman  and Zhao et al.  proved that the inequalities hold for all with if and only if , , , , , , , and .

In , 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 . Alzer  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  as for all with .

Chu et al.  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
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
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