Optimal Bounds for the Neuman-Sándor Mean in terms of the Convex Combination of the First and Second Seiffert Means
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.
For with , the Neuman-Sándor mean , the first Seiffert mean , and the second Seiffert mean  are defined byrespectively. It can be observed that the first Seiffert mean can be rewritten as (see )where ), , and are the inverse hyperbolic sine, inverse tangent, and inverse sine functions, respectively.
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  established for all with and the following Ky Fan inequalities: for all with , and .
Gao  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 : for all with . Hästö  improved the results of  and found the sharp lower power mean bound for the Seiffert mean as follows: for all with .
In , the authors proved the sharp double inequality holds for all with .
Let denote the Lehmer mean of two positive numbers and with . In  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 , 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 .
To establish our main result, we need several lemmas which we present in this section.
Lemma 1. Let . Then is strictly decreasing on . Moreover, the inequalityholds for any and the inequalityholds for any .
Lemma 2. Let . Then it holds thatfor any andholds 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
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).
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  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.
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).
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