# On New Wilker-Type Inequalities

Authors: Zhengjie Sun; Ling Zhu
Journal: ISRN Mathematical Analysis
 (2011)

---

## Abstract

In this paper, we present and prove a new Wilker-type inequality for hyperbolic functions. We also give a simple proof of the counterpart of the above inequality for the circular functions.


---

## Body

## 1. Introduction

Wilker [1] proposed two open questions as the following statements.Problem 1.
If0<x<π/2, then
(1.1)(sinxx)2+tanxx>2.Problem 2.
There exists a largest constantc such that
(1.2)(sinxx)2+tanxx>2+cx3tanx
for 0<x<π/2.Sumner et al. [2] affirmed the truth of the problems above and obtained a further result as follows.Theorem 1.1.
If0<x<π/2, then
(1.3)16π4x3tanx<(sinxx)2+tanxx-2<845x3tanx.
Furthermore, 16/π4 and 8/45 are the best constants in (1.3).Guo et al. [3] gave new proofs of inequalities (1.1) and (1.2). In recent years, Zhu [4] showed a new simple proof of inequality (1.1); Pinelis [5] got other proof of inequalities (1.3) by using L'Hospital rules for monotonicity; Zhang and Zhu [6] gave a new elementary proof of double inequalities (1.3).Besides, in article [7], Wang offered another type of the inequality and proved it by the power series expansions of circular functions as followsTheorem 1.2.
Let0<x<π/2. Then,
(1.4)245x3sinx<(xsinx)2+xtanx-2<(2π-16π3)x3sinx.
Furthermore, 2/45 and (2/π-16/π3) are the best constants in (1.4).Zhu [8] established a new Wilker-type inequality involving hyperbolic functions as follows: (1.5)(sinh⁡xx)2+tanhxx-2>845x3tanhx,x>0,
where the constant 8/45 is the optimum constant in (1.5).In fact, we can show a new Wilker-type inequality involving hyperbolic functions referring to the result above.Theorem 1.3.
Letx>0. Then
(1.6)(xsinh⁡x)2+xtanhx-2<245x3sinh⁡x.
Furthermore, 2/45 is the best constant in (1.6).The purpose of this paper is to present a concise proof of inequality (1.6) by using the power series expansion of hyperbolic functions and to give an elementary proof of inequality (1.4) in another way.


## 2. The Proof of Theorem1.3

Let((x/sinhx)2+x/tanhx-2)/(x3sinhx)=(x2+xsinhxcoshx-2(sinhx)2)/(xsinhx)3=A(x)/B(x), where A(x)=x2+xsinhxcoshx-2(sinhx)2,  and B(x)=(xsinhx)3. We get that the existence of Theorem 1.3 can be ensured when proving the following two statements: the first is A(x)/B(x)≤2/45 or 2B(x)≥45A(x), and the second is limx→0+(A(x)/B(x))=2/45.Since(2.1)A(x)=x2+12xsinh⁡x-(cosh⁡2x-1)=x2+12x∑n=0∞(2x)2n+1(2n+1)!-∑n=0∞(2x)2n(2n)!+1=x2+12∑n=0∞22n+1(2n+1)!x2n+2-∑n=0∞(2x)2n(2n)!+1=∑n=1∞22n(2n+1)!x2n+2-∑n=1∞22n+2(2n+2)!x2n+2=∑n=1∞(22n(2n+1)!-22n+2(2n+2)!)x2n+2=∑n=2∞(22n(2n+1)!-22n+2(2n+2)!)x2n+2=∑n=2∞anx2n+2,B(x)=14x3(sinh⁡3x-3sinh⁡x)=14x3∑n=0∞32n+1-3(2n+1)!x2n+1=14∑n=0∞32n+1-3(2n+1)!x2n+4=14∑n=1∞32n+1-3(2n+1)!x2n+4=∑n=2∞32n-1-34(2n-1)!x2n+2=∑n=2∞bnx2n+2,
where an=22n/(2n+1)!-22n+2/(2n+2)!, bn=(32n-1-3)/4(2n-1)!, n≥2, and n∈N+, according to the analysis above, the proof of Theorem 1.3 can be completed when proving that (2.2)2bn≥45an,
that is, (2.3)n(n+1)(2n+1)(9n-9)≥135(n-1)4n
for n≥2.Letf(x)=x(x+1)(2x+1)(9x-9)-135(x-1)4x for x∈[2,+∞). We compute(2.4)f′(x)=(6x2+6x+1)9x-9(6x2+6x+1)+(2x3+3x2+x)9xlog⁡9-135⋅4x-135(x-1)4xlog⁡4,f′′(x)=(12x+6)9x+2(6x2+6x+1)9xlog⁡9-9(12x+6)+(2x3+3x2+x)9x(log⁡9)2-270⋅4xlog⁡4-135(x-1)4x(log⁡4)2,f(3)(x)=12⋅9x+(36x+18)9xlog⁡9+3(6x2+6x+1)9x(log⁡9)2-9⋅12+(2x3+3x2+x)9x(log⁡9)3-405⋅4x(log⁡4)2-135(x-1)4x(log⁡4)3,f(4)(x)=(ax3+bx2+cx+d)9x-(ex+m)4x,
where(2.5)a=2(log⁡9)4>0,b=3(log⁡9)4+24(log⁡9)3>0,c=(log⁡9)4+24(log⁡9)3+72(log⁡9)2>0,d=4(log⁡9)3+36(log⁡9)2+48(log⁡9)>0,e=135(log⁡4)4>0,m=540(log⁡4)3-135(log⁡4)4≈940.059>0.
Then let H(x)=(ax3+bx2+cx+d)(9/4)x/(ex+m); we compute H′(x)=(l(x)/(ex+m)2)(9/4)x, where(2.6)l(x)=a4x4+a3x3+a2x2+a1x+a0,a4=aelog⁡94>0,a3=2ae+(eb+am)log⁡94>0,a2=be+3am+(ec+mb)log⁡94>0,a1=2bm+(mc+de)log⁡94>0,a0=m(c+dlog⁡94)-ed≈671800>0.
As we can see, all coefficients of the polynomial l(x) are positive integers. When x>2, we have l(x)>0 and H′(x)>0. In view of that H(2)≈8.4763>1,f(4)(2)≈231740>0,f(3)(2)≈67857>0,f′′(2)≈16922>0,f′(2)≈2849.9>0, we have that H(x)>1 and f(4)(x)>0 for x>2. We can also conclude that f(i)(x)(i=3,2,1,0) is increasing on [2,+∞), respectively. Since f(2)=0, we have f(x)≥0 for x≥2. So the proof of inequality (2.3) is completed. Furthermore, limx→0+(A(x)/B(x))=a2/b2=(24/5!-26/6!)/((33-3)/4·3!)=2/45; the proof of Theorem 1.3 is completed.


## 3. Another Proof of Theorem1.2

We simplify the double inequality (1.4) into another form: (3.1)245<(x/sinx)2+x/tanx-2x3sinx<2π-16π3,0<x<π2.
Here we will discuss the monotonicity of the function g(x)=((x/sinx)2+x/tanx-2)/x3sinx=((1/sinx)(x/sinx)2+xcosx/sin2x-2/sinx)/x3. If the function g(x) is increasing for 0<x<π/2 and both limx→0+g(x)=2/45 and limx→(π/2)-g(x)=2/π-16/π3 are right, the proof of equality (1.4) is completed.From [9] or [10, page 75], we know that for 0<|x|<π the equality (3.2)1sinx=1x+∑n=1∞22n-2(2n)!|B2n|x2n-1
holds. So we can also know that for 0<|x|<π(3.3)cos⁡xsin2x=(-1sinx)′=1x2-∑n=1∞22n-2(2n)!|B2n|(2n-1)x2n-2,1sin3x=(-cos⁡x2sin2x)′+12sinx=1x3+12∑n=2∞22n-2(2n)!|B2n|(2n-1)(2n-2)x2n-3+12x+12∑n=1∞22n-2(2n)!|B2n|x2n-1
hold. Then we compute g(x)=∑n=2∞anx2n-4 by using the results above, where(3.4)an=1222n-2(2n)!|B2n|(2n-1)(2n-2)+1222n-2-2(2n-2)!|B2n-2|-22n-2(2n)!|B2n|(2n-1)-222n-2(2n)!|B2n|=1222n-2-2(2n-2)!|B2n-2|+(2n-1)(2n-2)-2(2n-1)-42(2n)!(22n-2)|B2n|=1222n-2-2(2n-2)!|B2n-2|+2n(2n-5)2(2n)!(22n-2)|B2n|.
When n=2, we can compute a2=(1/2)(2/2!)|B2|+(-4/2·4!)·14·|B4|=(1/2)·(1/6)-(7/6)·(1/30)=2/45>0 by using |B2|=1/6 and |B4|=1/30. Furthermore, there is an obvious fact that an>0 for n≥3. So far, we can demonstrate that an>0 for all n≥2. So the function g(x) is increasing on (0,π/2). Evidently, limx→0+g(x)=2/45 and limx→(π/2)-g(x)=2/π-16/π3 are true. So the proof of the double inequality of (1.4) is completed.Remark 3.1.
Wilker's inequalities (1.1) and (1.2) have been further refined by many scholars in the past few years; the readers can refer to [11–16].



---

*Source: 681702-2011-06-20.xml*