Sharp Inequalities for Trigonometric Functions

Yang, Zhen-Hang; Jiang, Yun-Liang; Song, Ying-Qing; Chu, Yu-Ming

doi:https://doi.org/10.1155/2014/601839

Abstract

We establish several sharp inequalities for trigonometric functions and present their corresponding inequalities for bivariate means.

1. Introduction

A bivariate real value function is said to be a mean if for all . is said to be homogeneous if for any .

Remark 1 (see [1]). Let be a homogeneous bivariate mean of two positive real numbers and . Then where .

By this remark, almost all of the inequalities for homogeneous symmetric bivariate means can be transformed equivalently into the corresponding inequalities for hyperbolic functions and vice versa. More specifically, let , , and be the logarithmic, identric, and th power means of two distinct positive real numbers and given by respectively. Then, for , we have where . By Remark 1, we can derive some inequalities for hyperbolic functions from certain known inequalities for bivariate means mentioned previously. For example, (see [2, 3]); consider (see [4, 5]); consider that (see [1]) holds for if and only if and ; consider (see [6]); consider (see [7], ( 3.9), and ( 3.10)); if , then the double inequality (see [8]) holds if and only if and ; if , then inequality (11) holds if and only if and ; consider that (see [9]) holds if and only if and .

The main purpose of this paper is to find the sharp bounds for the functions , which include the corresponding trigonometric version of the inequalities listed above. As applications, their corresponding inequalities for bivariate means are presented.

2. Lemmas

Lemma 2 (see [10, Theorem 1.25], [11, Remark 1]). For , let be continuous on and differentiable on ; let on . If is increasing (or decreasing) on , then so are If is one-to-one, then the monotonicity in the conclusion is strict.

Lemma 3 (see [12]). Let and be real numbers and let the power series and be convergent for . If , for , and is increasing (decreasing), for , then the function is also (strictly) increasing (decreasing) on .

Lemma 4 (see [13, pages 227–229]). One has where is the Bernoulli number.

Lemma 5. For every , , the function defined by is increasing if and decreasing if . Consequently, for , one has It is reversed if .

Proof. For , we define and , where . Note that , and can be written as Differentiation and using (14) and (15) yield where Clearly, if the monotonicity of is proved, then by Lemma 3 we can get the monotonicity of , and then the monotonicity of the function easily follows from Lemma 2. For this purpose, since , , for , we only need to show that is decreasing if and increasing if . Indeed, an elementary computation yields It is easy to obtain that, for , which proves the monotonicity of .
Making use of the monotonicity of and the facts that we get inequality (19) and its reverse immediately.

Lemma 6. For every , , the function defined by is increasing if and decreasing if . Consequently, for , one has It is reversed if .

Proof. We define and , where . Note that , and can be written as Differentiating and using (14) and (15) yield where Similarly, we only need to show that is decreasing if and increasing if . In fact, simple computation leads to It is easy to obtain that, for , which proves the monotonicity of .
Making use of the monotonicity of and the facts that we get inequality (27) and its reverse immediately.

Lemma 7 (see [14, 15]). For and , let , , , and be defined by Then, , , and are decreasing with respect to , while is increasing with respect to on .

Proof. It was proved in [14, 15] that the functions and are decreasing with respect to . Now, we prove that has the same property. Logarithmic differentiation gives that, for , Clearly, for and , which yields , and so . This gives and .
Similarly, we get which implies that is decreasing with respect to on . Therefore, which proves the desired result.

3. Main Results

3.1. The First Sharp Bounds for

In this subsection, we present the sharp bounds for in terms of , which give the trigonometric versions of inequalities (6) and (7).

Theorem 8. For , the two-side inequality holds with the best possible constants and , where is the unique root of the equation on . Moreover, one has where the exponents , and coefficients , in (43) are the best possible constants and so is in (44).

Proof. (i) We first prove that the left inequality in (41) for and is the best possible constant. Letting in (19), then we get the first inequality in (41) and the second inequality in (43). If there exists such that for , then Using power series expansion gives Therefore, which derives a contradiction. Hence, is the best possible constant.
(ii) From Lemma 7, we clearly see that the function is decreasing on . Note that Therefore, (42) has a unique root . Numerical calculation gives . Letting in Lemma 5 yields The above inequalities can be rewritten as where the equality is due to the fact that is the unique root of (42). Therefore, we get the right inequality in (41) and the first inequality in (44). We clearly see that is the best possible constant.
(iii) The third inequality in (43) easily follows from which holds due to and . From we clearly see that the coefficients and are the best possible constants.
This completes the proof.

Recently, Yang [16] proved that the inequalities hold for if and only if and , where . Making use of Theorem 8 and Lemma 7, we have the following.

Corollary 9. For , the chain of inequalities hold with the best possible constants , , , and .

3.2. The Second Sharp Bounds for

In this subsection, we give the sharp bounds for in terms of , which give the trigonometric versions of inequalities (8).

Theorem 10. For , the two-side inequality holds with the best possible constants and , where is the unique solution of the equation on . Moreover, the inequalities hold for , where the exponents and the coefficients are the best possible constants. Also, the first member in (57) is decreasing with respect to on , while the third and fourth members are increasing with respect to on . The reverse inequality of (57) holds if .

Proof. For and , we define To prove the desired results, we need two assertions. The first one is which follows by expanding in power series The second one states that the equation , that is, (56), has a unique solution such that for and for . Indeed, Lemma 7 implies that is increasing on , which together with the facts that indicates the second assertion. By using mathematical software, we find .
(i) Now, we prove that the first inequality in (55) holds with the best constant . Letting in Lemma 5 yields the first inequality in (55). Due to the decreasing property of on given by Lemma 7, we assume that there is a with such that the left inequality in (55) holds for ; then we have , which together with the relation (61) leads to . It is clearly impossible. Hence, is the best constant.
(ii) We next show that the second inequality in (55) holds with the best constant . Let us introduce an auxiliary function defined on by Expanding in power series gives where Therefore, we have Differentiation again yields We claim that for . It suffices to show that for . In fact, , and satisfies the recursive relation A direct check leads to due to and satisfies the recursive relation Hence, is decreasing for , and so which yields . From the recursive relation (69), we get for , which proves that for . Note that We also assert that . If not, that is, , then there must be for , which yields and due to being the solution of the equation . This is obviously a contradiction. It follows that there is a such that for and for , which also implies that is decreasing on and increasing on . Therefore, that is, for .
It remains to prove that is the best possible constant. If there is a with such that the right inequality in (55) holds for , then, by the second assertion proved previously, we have , which yields a contradiction.
(iii) The first and second inequalities in (57) and their reverse ones are clearly the direct consequences of Lemma 5. It remains to prove the third one. We have to determine the sign of defined by for and . Arranging leads to As shown previously, for and for , which together with and gives the desired result.
Lemma 7 reveals that the monotonicity of the first, second, and third members in (57) with respect to on due to Finally, we show that is the best possible constant. It easily follows that
Thus, we complete the proof.

Remark 11. Letting and in Theorem 10 and then taking squares, we deduce that the two-side inequality holds for , where .
From the proof of Theorem 10, we clearly see that the constant in (79) is the best possible constant, but is not.

In [15, Theorems 1, 2, and 3], Yang proved that the chain of inequalities holds for with the best constants and . The monotonicity of the function on given in Lemma 7 and Remark 11 lead to the following.

Corollary 12. For , the chain of inequalities holds with the best possible constants , , , and , and .

Using certain known inequalities and the corollary above, we can obtain the following novel inequalities chain for trigonometric functions.

Corollary 13. For , one has

Proof. The first, second, and third inequalities in (82) are due to Neuman [17, Theorem 1].
The fourth one in (82) is equivalent to which holds due to for .
The eighth one is derived from Neuman and Sándor [18, ( 2.5)].
The ninth one easily follows from The tenth, eleventh, and twelfth ones can be obtained by [19, ( 3.9)].
Except the last one, other ones are obviously deduced from Corollary 12.
The last one is equivalent to which follows from the inequality connecting the fourth and sixth members in(82) proved previously.
Thus, the proof is complete.

Remark 14. Sándor [20, page 81, Lemma 2.2] proved that the inequality holds for . Clearly, the sixth and seventh inequalities in (82), that is, for , are a refinement of Sándor's inequality.

Remark 15. Using the decreasing property of the function defined by (83) proved in Corollary 13, we also get for , which can be rewritten as This in conjunction with (43) gives From

Abstract and Applied Analysis