#### Abstract

We find the least value *α* and the greatest value *β* such that the double inequality holds for all with , where , and are the Neuman-Sándor mean and the first and second Seiffert means of two positive numbers and , respectively.

#### 1. Introduction

For with , the Neuman-Sándor mean [1], the first Seiffert mean [2], and the second Seiffert mean [3] are defined byrespectively. It can be observed that the first Seiffert mean can be rewritten as (see [1])where ), , and are the inverse hyperbolic sine, inverse tangent, and inverse sine functions, respectively.

Recently, the means , , and and other means have been the subject of intensive research. Many remarkable inequalities for means can be found in the literature [4–14].

Let , , , , , , and denote the harmonic, geometric, logarithmic, identric, arithmetic, root-square, and th power means of two positive numbers and with , respectively. Then it is well known that the inequalities hold for with .

Neuman and Sándor [1] established for all with and the following Ky Fan inequalities: for all with , and .

Gao [15] proved the optimal double inequalities hold for all with .

The following bounds for the Seiffert means and in terms of power mean were presented by Jagers in [16]: for all with . Hästö [17] improved the results of [16] and found the sharp lower power mean bound for the Seiffert mean as follows: for all with .

In [18], the authors proved the sharp double inequality holds for all with .

Let denote the Lehmer mean of two positive numbers and with . In [19] the authors presented the following best possible Lehmer mean bounds for the Seiffert means and : for all with .

Let , , and be bivariate means such that for all with . The problems of finding the best possible parameters and such that the inequalities and hold for all with have attracted the interest of many mathematicians.

In [20, 21] the authors proved that the double inequalities hold for all with if and only if , , , and .

In [22], the authors proved that the double inequality holds for all with if and only if , .

The main purpose of this paper is to find the least value and the greatest value such that the double inequality holds for all with .

#### 2. Lemmas

To establish our main result, we need several lemmas which we present in this section.

Lemmas 1 and 2 were proved in [23].

Lemma 1. *Let . Then is strictly decreasing on . Moreover, the inequality**holds for any and the inequality**holds for any .*

Lemma 2. *Let . Then it holds that**for any and**holds for any .*

For , the power series expansion of the function is presented as follows:Therefore, we have the following.

Lemma 3. *If , then it holds that*

*Proof. *The inequality of (22) follows from equality (21) directly. The inequality of (23) follows from for .

Lemma 4. *Let . Then for it holds that*

*Proof. *To prove inequality (25), it suffices to showfor , whereFirstly, we prove inequality (26). From the expression of , we havewhereBecause of for , it follows from (22) and (31) thatfor any . Therefore, inequality (26) follows from (29) and (30) together with (33).

Secondly, we prove inequality (27). From the expression of , we havewhereBecause offor any , it follows from (22) and (35) together with (36) thatfor any . Therefore, inequality (27) follows from (34) and (37).

Lemma 5. *The function is strictly decreasing on , where and , , and are defined as in Lemmas 1, 2, and 4, respectively.*

*Proof. *Direct computations lead to where Then inequality (22) leads tofor any Obviously, it holds thatfor any . Therefore, for any , follows from (39) and (41) together with (42). In other words, is strictly decreasing on ().

Let . It was proved that is strictly decreasing on () in [23] Lemma . Thus, from the monotonicity of and , we have for any . That is to say, is strictly decreasing on (). Considering the monotonicity of in Lemma 1, the proof is completed.

Lemma 6. *Let , where . Then for .*

*Proof. *Let . Then it is easy to verify that is decreasing on , where Considering , we have for . Therefore, for .

#### 3. Main Results

Theorem 7. *The double inequality **holds for any 0 with if and only if and .*

*Proof. *Because , , and are symmetric and homogeneous of degree 1, without loss of generality, we assume that . Let , , and . Then by (1), (3), and (4), direct computations lead toLetThen it follows thatDifferentiating , we havewhere , , and are defined as in Lemmas 1, 2, and 4, respectively.

On one hand, from inequalities in Lemmas 1, 2, and 4, we clearly see thatfor . It leads tofor any . According to (47) and (51), we conclude thatfor all with .

On the other hand, from inequalities in Lemmas 1, 2, and 4, we havefor . According to Lemma 6, we havefor . Lemma 5 shows that is strictly decreasing on . This fact and (54) together with imply that there exists such that is strictly increasing on and strictly decreasing on . Equations (47) and (48) together with the piecewise monotonicity of lead to the conclusion thatfor all with .

At last, by easy computations, (1), (3), and (4) lead toThus, we have the following claims.*Claim 1*. If , then (56) and (57) imply that there exists such that for all with .*Claim 2*. If , then (56) and (58) imply that there exists such that for all with .

Inequalities (52) and (55) in conjunction with the above two claims mean the proof is completed.

#### Conflict of Interests

The authors declare that they have no competing interests.

#### Acknowledgments

This research was supported by the Program of Scientific Research Training for Undergraduate of Anhui University (KYXL2014002) and Doctoral Scientific Research Foundation of Anhui University, China (01001901).