# Some New Wilker-Type Inequalities for Circular and Hyperbolic Functions

Authors: Ling Zhu
Journal: Abstract and Applied Analysis
 (2009)

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## Abstract

In this paper, we give some new Wilker-type inequalities for circular and hyperbolic functions in exponential form by using generalizations of Cusa-Huygens inequality and Cusa-Huygens-type inequality.


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## Body

## 1. Introduction

Wilker [1] proposed  two  open questions, the first of which was the following statement.Problem 1.
Let0<x<π/2. Then
(1.1)(sinxx)2+tanxx>2
holds.Sumner et al. [2] proved inequality (1.1). Guo et al. [3] gave a new proof of inequality (1.1). Zhu [4, 5] showed two new simple proofs of Wilker’s inequality above, respectively.Recently, Wu and Srivastava [6] obtained Wilker-type inequality as follows:
(1.2)(xsinx)2+xtanx>2,0<x<π2.
Baricz and Sandor [7] found that inequality (1.2) can be proved by using inequality (1.1).On the other hand, in the form of inequality (1.1), Zhu [5] obtained the following Wilker type inequality:
(1.3)(sinhxx)2+tanhxx>2,x>0.In fact, we can obtain further results:(1.4)(sinxx)2+tanxx>(xsinx)2+xtanx>2,0<x<π2,(sinhxx)2+tanhxx>(xsinhx)2+xtanhx>2,x>0.In this note, we establish the following four new Wilker type inequalities in exponential form for circular and hyperbolic functions.Theorem 1.1.
Let0<x<π/2, α∈ℝ and α≠0. Then(i)
whenα>0, the inequality
(1.5)(sinxx)2α+(tanxx)α>(xsinx)2α+(xtanx)α
holds;(ii)
whenα<0, inequality (1.5) is revered.Theorem 1.2.
Let0<x<π/2 and α≥1. Then the inequality
(1.6)(sinxx)2α+(tanxx)α>(xsinx)2α+(xtanx)α>2
holds.Theorem 1.3.
Letx>0, α∈ℝ and α≠0. Then(i)
whenα>0, the inequality
(1.7)(sinhxx)2α+(tanhxx)α>(xsinhx)2α+(xtanhx)α
holds;(ii)
whenα<0, inequality (1.7) is revered.Theorem 1.4.
Letx>0 and α≥1. Then the inequality
(1.8)(sinhxx)2α+(tanhxx)α>(xsinhx)2α+(xtanhx)α>2
holds.


## 2. Lemmas

Lemma 2.1 (see [8–24]).
Letf,g:[a,b]→ℝ be two continuous functions which are differentiable on (a,b). Further, let g′≠0 on (a,b). If f′/g′ is increasing (or decreasing) on (a,b), then the functions (f(x)-f(b))/(g(x)-g(b)) and (f(x)-f(a))/(g(x)-g(a)) are also increasing (or decreasing) on (a,b).Lemma 2.2 (see [25–27]).
Letan and bn(n=0,1,2,…) be real numbers, and let the power series A(t)=∑n=0∞antn and B(t)=∑n=0∞bntn be convergent for |t|<R. If bn>0 for n=0,1,2,…, and if an/bn is strictly increasing (or decreasing) for n=0,1,2,…, then the function A(t)/B(t) is strictly increasing (or decreasing) on (0,R).Lemma 2.3 (see [28, 29]).
Let|x|<π, then the inequality
(2.1)xsinx=1+∑n=1∞22n-2(2n)!|B2n|x2n
holds.Lemma 2.4.
Let|x|<π, then  the inequality
(2.2)1sin2x=csc2x=1x2+∑n=1∞22n(2n)!|B2n|(2n-1)x2n-2
holds.Proof.
The following power series expansion can be found in [30, 1.3.1.4 ( 3)]
(2.3)cotx=1x-∑n=1∞22n(2n)!|B2n|x2n-1,|x|<π.
Then
(2.4)1sin2x=csc2x=-(cotx)′=1x2+∑n=1∞22n(2n)!|B2n|(2n-1)x2n-2,|x|<π.Lemma 2.5 (see [5, 31]).
Let0<x<π/2. Then the inequality
(2.5)(sinxx)3>cosx
holds.Lemma 2.6 (see [5, 31, 32]).
Letx>0. Then the inequality
(2.6)(sinhxx)3>coshx
holds.Lemma 2.7.
Let0<x<π/2. Then the function G(α)=((sinx/x)2α+(tanx/x)α)/((x/sinx)2α+(x/tanx)α) increases as α increases on (-∞,+∞).Lemma 2.8.
Letx>0. Then the function H(α)=((sinhx/x)2α+(tanhx/x)α)/((x/sinhx)2α+(x/tanhx)α) increases as α increases on (-∞,+∞).Lemma 2.9 (a generalization of Cusa-Huygens inequality).
Let0<x<π/2 and α≥1. Then the inequality
(2.7)2(xsinx)α+(xtanx)α>3
or
(2.8)(sinxx)α<23+13(cosx)α
holds.Lemma 2.10 (a generalization of Cusa-Huygens type inequality).
Letx>0 and α≥1. Then the inequality
(2.9)2(xsinhx)α+(xtanhx)α>3
or
(2.10)(sinhxx)α<23+13(coshx)α
holds.


## 3. Proofs of Lemma2.7 and Theorem 1.1

Proof of Lemma2.7.
Direct calculation yieldsG′(α)=J(α)/[(x/sinx)2α+(x/tanx)α]2, where
(3.1)J(α)=[(tanxx)α(xsinx)2α-(xtanx)α(sinxx)2α+2]logtanxx+2[(tanxx)α(xsinx)2α-(xtanx)α(sinxx)2α-2]logxsinx=[(2xsin2x)α-(sin2x2x)α+2]logtanxx+2[(2xsin2x)α-(sin2x2x)α-2]logxsinx=log[(tanxx)(2x/sin2x)α-(sin2x/2x)α+2(x2sin2x)(2x/sin2x)α-(sin2x/2x)α-2]=log[(2xsin2x)(2x/sin2x)α-(sin2x/2x)α((sinxx)31cosx)2].
First, we have [(sinx/x)3(1/cosx)]2>1 by Lemma 2.5. Second, when letting 2x/sin2x=t for 0<x<π/2, we have t>1, and tα-t-α>0 for α>0, so ttα-t-α>1 and (2x/sin2x)(2x/sin2x)α-(sin2x/2x)α[(sinx/x)3(1/cosx)]2>1. Thus J(α)>0 and G′(α)>0. The proof of Lemma 2.7 is complete.Proof of Theorem1.1.
From Lemma2.7 we have G(α)>G(0)=1 for α>0. That is, (1.5) holds. At the same time, we have G(α)<G(0)=1 for α<0. That is, (1.5) is revered.


## 4. Proofs of Lemma2.9 and Theorem 1.2

Proof of Lemma2.9.
LetF(x)=((sinx/x)α-1)/((cosx)α-1)=:f(x)/g(x), where f(x)=(sinx/x)α-1, and g(x)=(cosx)α-1. Then
(4.1)k(x)=f′(x)g′(x)=(sinxxcosx)α-1sinx-xcosxx2sinx,k′(x)=(sinxxcosx)α-2u(x)x4sinxcos2x,
where
(4.2)u(x)=(α-1)(x-sinxcosx)(sinx-xcosx)+cosx(x2-2sin2x+xsinxcosx)=(xsinx-sin2xcosx-x2cosx+xcos2xsinx)α-(xsinx+sin2xcosx-2x2cosx)=(xsinx-sin2xcosx-x2cosx+xcos2xsinx)(α-G(x)),
where G(x)=(xsinx+sin2xcosx-2x2cosx)/(xsinx-sin2xcosx-x2cosx+xcos2xsinx). Then
(4.3)G(x)=2x/sin2x+1-2x2/sin2x2x/sin2x-1-(x/sinx)2+xcotx:=A(x)B(x),
where A(x)=2x/sin2x+1-(2x2/sin2x), and B(x)=2x/sin2x-1-(x/sinx)2+xcotx. By (2.1), (2.2), and (2.3), we have
(4.4)A(x)=1+∑n=1∞22n-2(2n)!|B2n|(2x)2n+1-2(1+∑n=1∞22n(2n)!|B2n|(2n-1)x2n)=∑n=1∞(22n-4n)22n(2n)!|B2n|x2n=∑n=2∞(22n-4n)22n(2n)!|B2n|x2n=:∑n=2∞anx2n,B(x)=1+∑n=1∞22n-2(2n)!|B2n|(2x)2n-1-(1+∑n=1∞22n(2n)!|B2n|(2n-1)x2n)+1-∑n=1∞22n(2n)!|B2n|x2n=∑n=1∞(22n-2n-2)22n(2n)!|B2n|x2n=∑n=2∞(22n-2n-2)22n(2n)!|B2n|x2n=:∑n=2∞bnx2n,
where an=((22n-4n)22n/(2n)!)|B2n| and bn=((22n-2n-2)22n/(2n)!)|B2n|>0.
When settingcn=an/bn, we have that cn=(22n-4n)/(22n-2n-2) is increasing for n=2,3,…,A(x)/B(x) is increasing from (0,π/2) onto (4/5,1) by Lemma 2.2. When α≥1 , we have u(x)≥0. So k(x) is increasing on (0,π/2). This leads to that f′(x)/g′(x) is increasing on (0,π/2). Thus F(x)=f(x)/g(x)=(f(x)-f(0+))/(g(x)-g(0+)) is increasing on (0,π/2) by Lemma 2.1. At the same time, limx→0+F(x)=1/3. So the proof of Lemma 2.9 is complete.Proof of Theorem1.2.
From Theorem1.1, when α≥1 we have
(4.5)(sinxx)2α+(tanxx)α>(xsinx)2α+(xtanx)α.
On the other hand, when α≥1 we can obtain
(4.6)1+(xsinx)2α+(xtanx)α≥2(xsinx)α+(xtanx)α>3
by the arithmetic mean-geometric mean inequality and Lemma 2.9. So
(4.7)(xsinx)2α+(xtanx)α>2
holds.
Combining (4.5) and (4.7) gives (1.6).


## 5. Proofs of Lemma2.8 and Theorem 1.3

Proof of Lemma2.8.
Direct calculation yieldsH′(α)=I(α)/[(x/sinhx)2α+(x/tanhx)α]2, where
(5.1)I(α)=[(tanhxx)α(xsinhx)2α+(xtanhx)α(sinhxx)2α+2]logtanhxx+2[(tanhxx)α(xsinhx)2α+(xtanhx)α(sinhxx)2α+2]logsinhxx=[(tanhxx)α(xsinhx)2α+(xtanhx)α(sinhxx)2α+2]log[(sinhxx)31coshx].
First, (tanhx/x)α(x/sinhx)2α+(x/tanhx)α(sinhx/x)2α+2>0 for x>0. Second, we have log[(sinhx/x)3(1/coshx)]>0 by Lemma 2.6. Thus I(α)>0 and H′(α)>0. The  proof of Lemma 2.8 is complete.Proof of Theorem1.3.
From Lemma2.8 we have H(α)>H(0)=1 for α>0. That is, (1.7) holds. At the same time, we have H(α)<H(0)=1 for α<0. That is, (1.7) is revered.


## 6. Proofs of Lemma2.10 and Theorem 1.4

Proof of Lemma2.10.
LetQ(x)=((sinhx/x)α-1)/((coshx)α-1)=:f(x)/g(x), where f(x)=(sinhx/x)α-1, and g(x)=(coshx)α-1. Then
(6.1)k(x)=:f′(x)g′(x)=(sinhxtcoshx)α-1xcoshx-sinhxx2sinhx=:(sinhxtcoshx)α-1A(x)B(x),
where A(x)=xcoshx-sinhx and B(x)=x2sinhx. Since
(6.2)A(x)=x∑n=0∞x2n(2n)!-∑n=0∞x2n+1(2n+1)!=∑n=1∞(2n)x2n+1(2n+1)!=∑n=0∞(2n+2)x2n+3(2n+3)!=:∑n=0∞anx2n+3,B(x)=∑n=0∞x2n+3(2n+1)!=:∑n=0∞bnx2n+3,
where an=(2n+2)/(2n+3)! and bn=1/(2n+1)!.
When settingcn=an/bn, we have cn=1/(2n+3) is decreasing for n=0,1,2,…,A(x)/B(x) is decreasing on (0,+∞) by Lemma 2.2. At the same time, the function (tanhx/x)α-1 is decreasing on (0,+∞) when α≥1. By (6.1), we obtain that k(x)=f′(x)/g′(x) is decreasing on (0,+∞). Thus Q(x)=f(x)/g(x)=(f(x)-f(0+))/(g(x)-g(0+)) is decreasing on (0,+∞) by Lemma 2.1. At the same time, limx→0+Q(x)=1/3. So the proof of Lemma 2.10 is complete.Proof of Theorem1.4.
By the same way as Theorem1.2, we can prove Theorem 1.4.


## 7. Open Problem

In this  section, we pose the following open problem: find the respective largest range ofα such that the inequalities (1.6) and (1.8) hold.



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*Source: 485842-2009-07-09.xml*