#### Abstract

We improve the Jordan, Adamović-Mitrinović, and Cusa inequalities. As applications, several new Shafer-Fink type inequalities for inverse sine function and bivariate means inequalities are established, and a new estimate for sine integral is given.

#### 1. Introduction

The classical Jordan inequality [1] is given by for .

Some new developments on refinements, generalizations, and applications for the Jordan inequality can be found in [2] and the references therein.

In the recent past, the following two-side inequality
has attracted the attention of many researchers (see, e.g., [2–14]). The left inequality in (2) was obtained by Mitrinović (see [1, p. 238]), while the right one is due to Huygens (see, e.g., [15]) and it is called* Cusa inequality* [3, 5, 6, 8, 14].

In [16], the following open problem was proposed: for each , there are greatest value and least value such that the double inequality holds for all . This was answered by Carver in [17]. In [1, p. 238, 3.4.15], it was listed that for and . Wu [18] proved that the double inequality holds for , , and . In particular, he obtained that for , The first inequality in (6) is equivalent to the Huygen inequality: Jiang [19] showed that for , Li and He [20] gave an improvement of (6) as follows:

The main purpose of this paper is to give sharp bounds for in terms of the functions and , where

The rest of this paper is organized as follows. Several lemmas are given in Section 2. Main results and their proofs are given in Section 3, in which Theorem 7 unifies and generalizes Jordan and Cusa inequalities; Theorem 13 shows that Adamović-Mitrinović and Cusa inequalities (2) can be interpolated by for suitable ; Theorem 18 gives a hyperbolic version of Theorem 7. In Section 4, some new Shafer-Fink type inequalities for inverse sine function and several inequalities for bivariate means are presented, and a simpler but more accurate estimate for sine integral is provided.

#### 2. Lemmas

Lemma 1. *Let and be defined by (10) and (11), respectively. Then and are, respectively, increasing and decreasing with respect to on with the limits
*

*Proof. *From (10) and (11) we have
If , then we clearly see that . If , then , and then .

Simple computations give (12).

Lemma 2. *Let , be defined on by
**
respectively. Then , .*

*Proof. *It is not difficult to see that , for . If , then
and then . It remains to prove that for . Differentiation leads to
Hence, , which implies that is decreasing in on , and therefore,
This completes the proof.

Lemma 3. *Let be defined on by
**
Then*(i)* for all if and only if , where is the unique solution of the equation ;*(ii)* if and only if ;*(iii)*for every , there exists a unique such that for and for .*

*Proof. *In order to prove the desired results, we need to rewrite as
We clearly see that

We claim that there exists unique such that for and for . Indeed, we have
which implies that is increasing on and decreasing on . Since
there exists unique such that and for and for . An easy calculation reveals that .

(i) Now we prove the necessary and sufficient condition for for all . Since , we assume that . Denote the minimum point of by . Then . And then, due to , for all if and only if at least one of the following cases occur.*Case 1.* Consider that , . It is derived that .*Case 2.* Consider that , . It implies that .*Case 3.* Consider that , . It yields .

To sum up, for all if and only if .

(ii) It is clear that if and only if and . Solving the inequalities for leads to .

(iii) In the case of , we clearly see that , , and . This implies that there exists a unique such that for and for .

This completes the proof.

Now let us consider the sign of function defined on by where and are defined by (14) and (15), respectively. We have the following.

Lemma 4. *Let be defined on by (24). Then*(i)* for all if and only if ;*(ii)* for all if and only if , where
*(iii)*in the case of , there exists a unique such that for and for .*

*Proof. *We first give two limit relations as follows:
where
In fact, if , then making use of power series we get
which implies the first relation. Direct computations yield the second one.

Differentiating with respect to leads to
where and are defined by (14) and (15), respectively, and
here is defined by (19) and .

(i) We now prove that for all if and only if . The necessity easily follows from the inequalities and if and together with the relation (26).

Next we prove the sufficiency. If , then by Lemma 3 , and then . This indicates that is decreasing in on , and therefore, we get . If , then and , which yields . This also yields that is decreasing in on , and so .

(ii) Similarly, we can prove that for all if and only if . If for all , then we have and , which together with (26) and lead to .

In order to prove the sufficiency, we distinguish two cases.

In the case of , by Lemma 3 we have , which implies that is increasing in on , and so, .

In the case of , from Lemma 3 there is a unique such that for and for . This in conjunction with (30) and (29) shows that is decreasing in on and increasing on , and consequently, we have
which proves the sufficiency.

(iii) In the case when , we have seen that is decreasing in on and increasing on and for , but
Thus, there is a unique such that for and for .

The whole proof is complete.

We next observe the function defined on by Differentiation yields that where , , and are defined by (14), (15), and (24), respectively. From Lemmas 2 and 4 the following assertion is immediate.

Lemma 5. *Let be the function defined on by (33). Then*(i)* is decreasing in on if and only if ;*(ii)* is increasing in on if and only if , where is given by (25);*(iii)*in the case when , there is a unique such that is increasing in on and decreasing on .*

Lastly, for later use, we also give the following.

Lemma 6. *Let be defined on by (10). Then if and only if , and if and only if .*

*Proof. *For , we define
Then holds for all if and only if .

In fact, if and only if at least one case of the following occurs.*Case 1.* Consider that , . It is impossible.*Case 2.* Consider that , . It indicates .*Case 3.* Consider that , . It is impossible.

In the same way, we can prove that holds for all if and only if .

We now prove that if and only if . Factoring yields
If , then , and then, if and only if , which is equivalent to . If , then , and then, if and only if , which is equivalent to . Consequently, if and only if .

Next we show that if and only if . In fact, if , then if and only if , which yields . It is clearly a contradiction. If , then the statement in question if and only if , which leads to . Thus the proof is complete.

#### 3. Main Results

Theorem 7. *Let . Then for ,
**
holds if and only if . Moreover, we have
**
for , where is the best possible. And the lower and upper bounds in (38) are decreasing and increasing in on , respectively.*

*Proof. *Clearly, the desired result is equivalent to if and only if , where is defined by (33). To this end, we give two limit relations. The first one follows by expanding in power series for . We have
which yields
The second one is derived by a simple computation; that is,
Now we prove that for all if and only if .

The necessity easily follows by solving the simultaneous inequalities:
which implies .

The sufficiency is due to Lemma 5. In fact, If , then by Lemma 5 we see that is decreasing in on . Hence, .

Utilizing the monotonicity of in on gives (38). And from Lemma 1 it is seen that the lower and upper bounds in (38) are decreasing and increasing in on , respectively.

Thus the proof is finished.

By Theorem 7 and Lemma 1, we have the following interesting chain of inequalities.

Corollary 8. *For , one has
*

*Remark 9. *It is clear that our results unify and refine Jordan and Cusa's inequalities and show that the first one in (9) is sharp. Also, Theorem 7 contains other known results, for example, taking in (38) we get
which contain (6). After a simple transformation, (44) can be written as
where the second inequality in (45) is due to Neuman and Sándor [6, ].

Theorem 10. *Let . Then for **
holds if and only if , where .**Moreover, for , one has
**
where , is the best possible. And , are decreasing and increasing in on , respectively.**For one has
**
where is the best possible and is the unique root of the equation
**
on .*

*Proof. *Since the inequality (46) is equivalent to , it suffices to prove that holds for if and only if .

Similarly, solving the simultaneous inequalities and with yields , which proves the necessity.

Conversely, the condition is also sufficient for to be valid. For this end, we divide the proof into two cases.*Case 1.* Consider that . By Lemma 5 it is seen that is increasing in on , which indicates that .*Case 2.* Consider that . By Lemma 5 we see that there is a unique such that is increasing in on and decreasing on . It is acquired that
that is,
which proves the sufficiency.

In the first case, application of the monotonicity of in on leads to (47), and . In the second case, (51) also yields (47), and
Thus we complete the proof.

*Remark 11. *Taking in (46), we get the first inequality in (9).

Letting and solving (49) by mathematical computation software, we find that and . Letting be defined by (25) yields . By Theorem 10 we get the following.

Corollary 12. *For , one has
**
where and are the best possible constants.*

Letting in Lemma 6 and using Theorems 7 and 10, we obtain a chain of inequalities that interpolates Adamović-Mitrinović and Cusa's inequalities (2) by .

Theorem 13. *For , the inequalities
**
hold if and only if , , , and , where .*

Using the monotonicity of in on given by parts one and two of Lemma 5, we see that hold for . And then we have It is clear that the right-hand in (56) is increasing in on , but the monotonicity of left-hand is to be checked. We define where . Logarithmic differentiation leads to where Since we have . Consequently, for .

The result can be stated as a theorem.

Theorem 14. *Let . Then for the inequalities
**
hold if and only if , where is the best constant. And the right-hand and left-hand in (61) are increasing and decreasing in , respectively. Inequality (61) is reversed if and only if , where is defined by (25).*

Putting in Theorem 14 we have the following.

Corollary 15. *For the following inequalities hold:
*

Further, let be defined on by where is defined by (10). We can show that the monotonicity of in for certain fixed . Differentiation again yields It is easy to verify that Consequently, we have It is reversed for . From these we can obtain the following.

Theorem 16. *For the following inequalities hold:
*

Additionally, Lemma 4 implies an optimal two-side inequality.

Theorem 17. *Let and let and be defined by (14) and (15), respectively. Then for the two-side inequality
**
holds if and only if and , where . And, for , the function is decreasing on .*

*Proof. *Since for and by Lemma 2 and defined by (24) can be written as
it follows from Lemma 4 that (71) holds if and only if and . It remains to check the monotonicity of in . Differentiation yields
where . If , then the numerator of the fraction in right-hand above is clearly positive. Consider that . If , then and , which yields that the numerator is nonnegative.

This proves the assertion.

Similarly, we can obtain a hyperbolic version of Theorems 7 and 10

Theorem 18. *Let . Then for **
holds if and only if . It is reversed if and only if .*

*Proof. *Let be the function defined on by
Then the inequalities (74) are equivalent to . Expanding in power series yields
which implies
On the other hand, we have

Now we prove desired results.

(i) We first prove that holds if and only if .

If for all , then we have
Solving the inequalities yields .

We prove the condition is sufficient for to hold for . Differentiation gives
where .

Due to , we see that , which yields

Then ; that is,