On New Wilker-Type Inequalities

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.

1. Introduction

Wilker [1] proposed two open questions as the following statements.

Problem 1. If 0<𝑥<𝜋/2, then sin𝑥𝑥²+ta𝑥n𝑥>2.(1.1)

Problem 2. There exists a largest constant 𝑐 such that sin𝑥𝑥²+ta𝑥n𝑥>2+𝑐𝑥³tan𝑥(1.2) 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. If 0<𝑥<𝜋/2, then 16𝜋⁴𝑥³tan𝑥<sin𝑥𝑥²+tan𝑥𝑥−2<485𝑥³tan𝑥.(1.3) 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 follows

Theorem 1.2. Let 0<𝑥<𝜋/2. Then, 245𝑥³sin𝑥<𝑥sin𝑥²+𝑥tan𝑥−2<𝜋2−1𝜋6³𝑥³sin𝑥.(1.4) 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: sin𝑥h𝑥²+tan𝑥h𝑥−2>485𝑥³tanh𝑥,𝑥>0,(1.5) 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. Let 𝑥>0. Then sin𝑥h𝑥²+tan𝑥h𝑥−2<425𝑥³sinh𝑥.(1.6) 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 Theorem 1.3

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 lim_𝑥→0⁺(𝐴(𝑥)/𝐵(𝑥))=2/45.

Since𝐴(𝑥)=𝑥²+12𝑥sinh𝑥−(cosh2𝑥−1)=𝑥²+12𝑥^∞_𝑛=0(2𝑥)^2𝑛+1(2𝑛+1)!−^∞_𝑛=0(2𝑥)^2𝑛=𝑥₂(2𝑛)!+1+12^∞_𝑛=02^2𝑛+1(2𝑛+1)!𝑥^2𝑛+2−^∞_𝑛=0(2𝑥)^2𝑛=(2𝑛)!+1_∞_𝑛=12^2𝑛(2𝑛+1)!𝑥^2𝑛+2−^∞_𝑛=12^2𝑛+2(2𝑛+2)!𝑥^2𝑛+2=^∞_𝑛=1(2𝑛2+^2𝑛1)!−(2𝑛2+2^2𝑛+2)!𝑥^2𝑛+2=^∞_𝑛=2(2𝑛2+^2𝑛1)!−(2𝑛2+2^2𝑛+2)!𝑥^2𝑛+2=^∞_𝑛=2𝐵(𝑥)=1𝑎_𝑛𝑥^2𝑛+2,4𝑥³(sinh3𝑥−3sinh𝑥)=14𝑥^3∞_𝑛=03^2𝑛+1−3(2𝑛+1)!𝑥^2𝑛+1=14^∞_𝑛=03^2𝑛+1−3(2𝑛+1)!𝑥^2𝑛+4=14^∞_𝑛=13^2𝑛+1−3(2𝑛+1)!𝑥^2𝑛+4=^∞_𝑛=23^2𝑛−1−34(2𝑛−1)!𝑥^2𝑛+2=^∞_𝑛=2𝑏_𝑛𝑥^2𝑛+2,(2.1) where 𝑎_𝑛=2^2𝑛/(2𝑛+1)!−2^2𝑛+2/(2𝑛+2)!, 𝑏_𝑛=(3^2𝑛−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 2𝑏_𝑛≥45𝑎_𝑛,(2.2) that is, 𝑛(𝑛+1)(2𝑛+1)(9^𝑛−9)≥135(𝑛−1)4^𝑛(2.3) for 𝑛≥2.

Let 𝑓(𝑥)=𝑥(𝑥+1)(2𝑥+1)(9^𝑥−9)−135(𝑥−1)4^𝑥 for 𝑥∈[2,+∞). We compute𝑓^(𝑥)=+6𝑥2²𝑥+₃6𝑥+19^𝑥−96𝑥²+6𝑥+1𝑓+3𝑥²+𝑥9^𝑥log9−135⋅4^𝑥−135(𝑥−1)4^𝑥log4,_(𝑥)=+(122𝑥𝑥+₃6)9^𝑥+26𝑥²+6𝑥+19^𝑥log9−9(12𝑥+6)𝑓₍₃₎+3𝑥²+𝑥9^𝑥(log9)²−270⋅4^𝑥log4−135(𝑥−1)4^𝑥(log4)²,(𝑥)=+122⋅𝑥9^𝑥₃+(36𝑥+18)9^𝑥log9+36𝑥²+6𝑥+19^𝑥(log9)²−9⋅12𝑓₍₄₎+3𝑥²+𝑥9^𝑥(log9)³−405⋅4^𝑥(log4)²−135(𝑥−1)4^𝑥(log4)³,(𝑥)=𝑎𝑥³+𝑏𝑥²+𝑐𝑥+𝑑9^𝑥−(𝑒𝑥+𝑚)4^𝑥,(2.4) where𝑏=3(l𝑎og=92)(₄log9)⁴>0,𝑐=(log9)₄+24(log9)³>0,𝑑=4(log9)₃+24(log9)³+72(log9)²>0,𝑒=+13365(l(ologg94)²)₄+48(log9)>0,𝑚=540(log4)₃>0,−135(log4)⁴≈940.059>0.(2.5) 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, where𝑙(𝑥)=𝑎₄𝑥𝑎⁴₄+𝑎₃𝑥³+𝑎₂𝑥²+𝑎₁𝑥+𝑎₀,=𝑎𝑒log9𝑎4>0,₃=2𝑎𝑒+(𝑒𝑏+𝑎𝑚)log9𝑎4>0,₂=𝑏𝑒+3𝑎𝑚+(𝑒𝑐+𝑚𝑏)log9𝑎4>0,₁=2𝑏𝑚+(𝑚𝑐+𝑑𝑒)log9𝑎4>0,₀=𝑚𝑐+𝑑log94−𝑒𝑑≈671800>0.(2.6) 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, lim_𝑥→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 Theorem 1.2

We simplify the double inequality (1.4) into another form: 245<(𝑥/sin𝑥)²+𝑥/tan𝑥−2𝑥³sin𝑥<𝜋2−1𝜋6³,0<𝑥<𝜋2.(3.1) 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 lim_𝑥→0⁺𝑔(𝑥)=2/45 and lim_{𝑥→(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 1sin𝑥=𝑥1+^∞_𝑛=12^2𝑛−2(2𝑛)!||𝐵^2𝑛||𝑥^2𝑛−1(3.2) holds. So we can also know that for 0<|𝑥|<𝜋cos𝑥sin²𝑥=−sin1𝑥^=1𝑥²−^∞_𝑛=12^2𝑛−2(2𝑛)!||𝐵^2𝑛1||(2𝑛−1)𝑥^2𝑛−2,sin³𝑥=−2csoins²𝑥𝑥^+1=12sin𝑥𝑥³+12^∞_𝑛=22^2𝑛−2(2𝑛)!||𝐵^2𝑛||(2𝑛−1)(2𝑛−2)𝑥^2𝑛−3+12𝑥+12^∞_𝑛=12^2𝑛−2(2𝑛)!||𝐵^2𝑛||𝑥^2𝑛−1(3.3) hold. Then we compute 𝑔(𝑥)=∑^∞𝑛=2𝑎_𝑛𝑥^2𝑛−4 by using the results above, where𝑎_𝑛=122^2𝑛−2(2𝑛)!||𝐵^2𝑛||(2𝑛−1)(2𝑛−2)+122^2𝑛−2−2(2𝑛−2)!||𝐵^2𝑛−2−2_2𝑛||−2(2𝑛)!||𝐵^2𝑛||(2𝑛−1)−22^2𝑛−2(2𝑛)!||𝐵^2𝑛=1||22^2𝑛−2−2(2𝑛−2)!||𝐵^2𝑛−2||+(2𝑛−1)(2𝑛−2)−2(2𝑛−1)−42(2𝑛)!2^2𝑛=1−2||𝐵_2𝑛||22^2𝑛−2−2(2𝑛−2)!||𝐵^2𝑛−2||+2𝑛(2𝑛−5)2(2𝑛)!2^2𝑛−2||𝐵_2𝑛||.(3.4) 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 𝑎_𝑛>0 for 𝑛≥3. So far, we can demonstrate that 𝑎_𝑛>0 for all 𝑛≥2. So the function 𝑔(𝑥) is increasing on (0,𝜋/2). Evidently, lim_𝑥→0⁺𝑔(𝑥)=2/45 and lim_{𝑥→(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].