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Abstract and Applied Analysis
Volume 2014 (2014), Article ID 592085, 9 pages
http://dx.doi.org/10.1155/2014/592085
Research Article

A Sharp Double Inequality for Trigonometric Functions and Its Applications

1School of Mathematics and Computation Sciences, Hunan City University, Yiyang 413000, China
2Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China

Received 26 April 2014; Accepted 20 June 2014; Published 10 July 2014

Academic Editor: Josip E. Pečarić

Copyright © 2014 Zhen-Hang Yang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We present the best possible parameters and such that the double inequality holds for any . As applications, some new analytic inequalities are established.

1. Introduction

It is well known that the double inequality holds for any . The first inequality in (1) was found by Mitrinović (see [1]), while the second inequality in (1) is due to Huygens (see [2]) and it is called Cusa inequality. Recently, the improvements, refinements, and generalizations for inequality (1) have attracted the attention of many mathematicians [38].

Qi et al. [9] proved that the inequality holds for any . It is easy to verify that and cannot be compared on the interval .

Neuman and Sándor [6] gave an improvement for the first inequality in (1) as follows:

Inequality (3) was also proved by Lv et al. in [10]. In [11, 12], Neuman proved that the inequalities hold for any .

For the second inequality in (1), Klén et al. [13] established for .

Inequality (5) was improved by Yang [14]. In [15], Yang further proved for .

Yang [16] proved that the inequalities hold for .

Zhu [8] and Yang [17] proved that and   are the best possible constants such that the double inequality holds for all .

More results involving inequality (1) can be found in the literature [1822].

Let , , and . Then is defined by

It is well known that is strictly increasing with respect to for fixed and (see [23]). If , then it is easy to check that

It follows from (2) and (3) together with (6) that for .

The main purpose of this paper is to present the best possible parameters and such that the double inequality holds for all . As applications, some new analytic inequalities are found. 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. Let and the function be defined on by
Then the following statements are true:(i) for all if and only if ;(ii) for all if and only if , where   is the unique solution of equation (iii)if , then there exists such that for and for .

Proof. It follows from (13) and (14) that for .
Inequalities (15) lead to the conclusion that the function is strictly decreasing with respect to for fixed and   is the unique solution of (14).
(i) If and , then from the monotonicity of the function we clearly see that
If for all , then (13) leads to
(ii) If and , then the monotonicity of the function leads to the conclusion that .
If and , then (13) and the monotonicity of the function lead to
Inequality (19) implies that the function is concave with respect to on the interval . Therefore, follows from (18) and the concavity of .
If for all , then follows easily from the monotonicity of the function and together with the fact that .
(iii) If and , then from (13) and (19) together with the monotonicity of the function we get and is strictly decreasing on .
It follows from (21) and (22) together with the monotonicity of that there exists such that is strictly increasing on and strictly decreasing on . Therefore, Lemma 1   follows from (20) and the piecewise monotonicity of .
Let and the function be defined on by
Then elaborated computations lead to where is defined by (13).

From Lemma 1 and (24) we get the following Lemma 2 immediately.

Lemma 2. Let and be defined on by (23). Then(i) is strictly decreasing on if and only if ;(ii) is strictly increasing on if and only if , where is the unique solution of (14);(iii)if , then there exists such that is strictly increasing on and strictly decreasing on .

Lemma 3. Let and be defined on by (23). Then(i) for all if and only if ;(ii) for all if and only if ;(iii)if , then there exists such that for and for .

Proof. (i) If and , then from (23) and Lemma 2   we clearly see that
If for all , then (23) leads to (ii) If for all , then from (23) we get
Inequality (27) leads to the conclusion that .
If and , then we divide the proof into two cases.
Case  1. Consider , where is the unique solution of (14). Then from Lemma 2   and (23) we clearly see that
Case  2. Consider . Then (23) and Lemma 2   lead to and there exists such that is strictly increasing on and strictly decreasing on . Therefore, for all follows from (29) and the piecewise monotonicity of .
(iii) If , then . It follows from (23) and Lemma 2   that and there exists such that is strictly increasing on and strictly decreasing on . Therefore, Lemma 3   follows from (30) and the piecewise monotonicity of .
Let and be defined on by
Then elaborated computations give where is defined by (23).

From Lemma 3 and (33) we get Lemma 4 immediately.

Lemma 4. Let and be defined on by (31) and (32). Then(i) is strictly decreasing on if and only if ;(ii) is strictly increasing on if and only if ;(iii)if , then there exists such that is strictly increasing on and strictly decreasing on .

Lemma 5. Let and be defined on by (31) and (32). Then the following statements are true:(i)if for all , then ;(ii)if for all , then , where   is the unique solution of the equation on the interval .

Proof. If for all , then from (31) and (32) we have We first prove that   is the unique solution of (34) on the interval . Let and
Then numerical computations show that
Inequality (38) implies that is strictly decreasing on . Therefore,   is the unique solution of (34) on the interval which follows from (37) and the monotonicity of .
If and for all , then (31) leads to
Therefore, follows from (39) and together with the monotonicity of on the interval .

Lemma 6. Let and , and let be defined by (9). Then the function is strictly decreasing with respect to if .

Proof. Let . Then from (9) we get Inequality (41) and lead to the conclusion that is strictly decreasing with respect to . Therefore, for , and is strictly decreasing with respect to if .

3. Main Results

Theorem 7. Let be defined by (9). Then the double inequality holds for all if and only if , and the double inequality holds for all if and only if , where , and is strictly decreasing with respect to .

Proof. Let and be defined on by (31) and (32). Then
If , then inequality (42) follows from Lemma 4   and (45).
If inequality (42) holds for all , then for all . It follows from Lemma 5   that .
If , then inequality (43) follows from Lemma 4   and (45).
If inequality (43) holds for all , then for all . It follows from Lemma 5   that , where   is the unique solution of (34) on the interval . We claim that ; otherwise , and Lemma 4   leads to the conclusion that there exists such that for .
Note that
It follows from Lemma 6 and (46) that is strictly decreasing with respect to .

From Theorem 7 we get Corollaries 8 and 9 as follows.

Corollary 8. For all one has

Corollary 9. For all one has

Theorem 10. Let be defined by (9). Then the double inequality holds for all if and only if and , where   is the unique solution of (34) on the interval . Moreover, the inequality if and only if where is defined as in Lemma 3  .

Proof. Let and be defined on by (31) and (32). Then Lemma 4   leads to the conclusion that is strictly increasing on and strictly decreasing on . Note that
It follows from the piecewise monotonicity of and (52) that for all . Therefore, for all follows from the first inequality of (53), while for all follows from the second inequality of (42).
Conversely, if the double inequality (49) holds for all , then we clearly see that the inequalities hold for all . Therefore, and follows from Lemma 5 and (54). Moreover, numerical computations show that   and
Therefore, the second conclusion of Theorem 10 follows from (55) and the second inequality of (53).

It follows from Lemma 3 that we get Theorem 11 immediately.

Theorem 11. The double inequalities hold for all if and only if and .

We clearly see that the function is strictly decreasing with respect to for fixed . Let and ; then Theorem 11 leads to the following.

Corollary 12. The inequalities hold for all .

4. Applications

In this section, we give some applications for our main results.

Neuman [24] proved that the Huygens type inequalities