Sharp Becker Stark’s type inequalities with power exponential functions

R E S E A R C H

Open Access

Sharp Becker-Stark’s type inequalities with

power exponential functions

Yusuke Nishizawa

*_{Correspondence:}

[email protected]

General Education, Ube National College of Technology, Tokiwadai 2-14-1, Ube, Yamaguchi 755-8555, Japan

Abstract

In this paper, we give some inequalities with power exponential functions derived from the left hand side of Becker-Stark’s inequality:

8

π

tanx x <

π

π

for 0 <x<

π/2.

MSC: Primary 26D05

Keywords: Becker-Stark’s inequality; Bernoulli’s inequality; monotonically decreasing function; monotonically increasing function

1 Introduction

Becker-Stark’s inequality is well known: 

π_{– x} <

tanx x <

π

π_{– x} (.)

for  <x<π/. The research of Becker-Stark’s inequality is one of the active areas in math-ematical analysis [–]. Recently, Zhu [] gave the following refinement of Becker-Stark’s inequality: For  <x<π/, the inequalities



π_{– x} +



π –

π_{– }

π

π– x<tanx

x (.)

and

tanx x <



π_{– x}+



π –

 –π π

π– x (.)

hold, where the constants –(π_{– )/(π}_{) and –( –π}_)/π_{are the best possible.}

More-over, from the right hand side of the inequality (.), Chen and Cheung [] gave the fol-lowing inequality: For  <x<π/, the inequality

π π_{– x}

θ

<tanx

x <

π π_{– x}

ϑ

(.)

holds, where the constantsθ=π_{/ andϑ=  are the best possible. In [], Sun and Zhu}

gave a simple proof of the results. The above inequality (.) is created based on the right hand side of Becker-Stark’s inequality (.). However, in this paper we establish some in-equalities created based on the left hand side of the inequality (.).

2 Results and discussion

Motivated by (.), in this paper, we give some inequalities with power exponential func-tions derived from the left hand side of Becker-Stark’s inequality (.). Since we note that /(π– x_{) <  for  <x< (}√_π_{– )/ and /(π}_{– x}_{) >  for (}√_π_{– )/ <x<π/, we}

obtain the two inequalities as follows.

Theorem . For <x< (√π_{– )/,we have}



π_{– x}

θ

<tanx

x <



π_{– x}

ϑ(x)

with the best possible constantθ= and the function

ϑ(x) =  x–√π_{– }+



√

π_{– }.

Theorem . For(√π_{– )/ <x<π/,we have}



π_{– x}

θ

<tanx

x <



π_{– x}

ϑ(x)

with the best possible constantθ= and the function

ϑ(x) =  x–√π_{– }–

√

π_{–  –π+ }

π–√π_{– } .

From Theorems . and ., we have the best possible constantθsuch that



π_{– x}

θ

<tanx

x .

If  <x< (√π_{– )/, the constantθmust beθ<  in order to satisfy ≤tanx/x< (/(π}_–

x₎₎θ_{. On the other hands, if (}√_π_{– )/ <x<π/, the constantθmust be  <θin order}

to satisfy /(π– x)≤tanx/x< (/(π– x))θ_{. Here, we obtain the two inequalities as}

follows.

Theorem . For/ <x< (√π_{– )/,we have}



π_{– x}

ϑ(x) 

<tanx

x ,

Corollary . For <x<π/,we do not have the best possible constantϑsuch that

tanx x <



π_{– x}

ϑ

.

3 Proofs of main theorems 3.1 Proof of Theorem 2.1

Proof of Theorem. We set

f(x) =



π_{– x}



x–√π–+ 

√

π–

–tanx

x .

From  x–√π_{– }+



√

π_{– }< 

for  <x< (√π_{– )/, by Bernoulli’s inequality, we have}



π_{– x}



x–√π–+ 

√

π–

>  +



π_{– x}– 

 x–√π_{– }+



√

π_{– }

.

By the right hand side of the inequality (.), for  <x< (√π_{– )/,}

f(x) >  +



π_{– x}– 

 x–√π_{– }+



√

π_{– }

– π



π_{– x}

=x(

√

π_{– x}_{– x}_–π_{x+ x+π}_{– )} √

π_{– (π– x)(}√_π_{–  – x)(x+π)}

=√ xg(x)

π_{– (π– x)(}√_π_{–  – x)(x+π)},

where

g(x) = √π_{– x}_{– x}_–π_{x+ x+π}_{– .}

From√π_{–  – x>  for  <x< (}√_π_{– )/, it suffices to show that}

g(x) > .

Here, the derivative ofg(x) is

g(x) =  –π+ √π_{–  – x.}

By  –π_{<  and}√_π_{–  –  < , we haveg(x) <  for any  <x< (}√_π_{– )/. Sinceg(x)}

is strictly decreasing for  <x< (√π_{– )/, we have}

g(x) >g

√

π_{– }



Therefore, we can get

tanx x <



π_{– x}

ϑ(x)

,

where

ϑ(x) =  x–√π_{– }+



√

π_{– }.

Sincetanx/xis strictly increasing for  <x<π/, we have 

π_{– x} <  <

tanx x

for any  <x< (√π_{– )/. Hence, for  <x< (}√_π_{– )/, we obtain}



π_{– x}

θ

<tanx

x <



π_{– x}

ϑ(x)

,

where the constantθ= . Sinceϑ(x) is strictly decreasing for  <x< (√π_{– )/ and}

ϑ(x) <ϑ() = ,

the constant θ =  is the best possible. Therefore, the proof of Theorem . is

com-plete.

3.2 Proof of Theorem 2.2

Proof of Theorem. We set

f(x) =



π_{– x}



x–√π––

√

π––π+

π–√π–

–tanx

x .

From  x–√π_{– }–

√

π_{–  –π+ }

π–√π_{– } > 

for (√π_{– )/ <x<π/, by Bernoulli’s inequality, we have}



π_{– x}



x–√π––

√

π––π+

π–√π–

>  +



π_{– x} – 



x–√π_{– }– √

π_{–  –π+ }

π–√π_{– }

By the inequality (.), for (√π_{– )/ <x<π/,}

f(x) >  +



π_{– x}– 

 x–√π_{– }–

√

π_{–  –π+ }

π–√π_{– }

–



π_{– x}+



π –

 –π π

π– x

= g(x)

π₍√_π_{–  –π)(}√_π_{–  – x)(x+π)},

where

g(x) = πx– π√π_{– x}_{+ }√_π_{– x}_{– πx}

+ πx– πx– π√π_{– x}_{+ π}√_π_{– x}_{+ x}

+ πx– πx+ π√π_{– x}_{+ πx}

– πx+ πx– πx+ π√π_{– x}

– π√π_{– x–π}_{– π}_{+ π}_{+ π}

– π+π√π_{–  – π}√_π_{– .}

From (√π_{–  –π)(}√_π_{–  – x) >  for (}√_π_{– )/ <x<π/, it suffices to show that}

g(x) > .

We have the derivatives

g(x) = πx– π√π_{– x}_{+ }√_π_{– x}_{– πx}

+ πx– πx– π√π_{– x}_{+ π}√_π_{– x}_{+ x}

+ πx– πx+ π√π_{– x+ πx}

–π+ π– π+π√π_{–  – π}√_π_{– }

= h(x)

and

h(x) = πx_{– π}√_π_{– x}_{+ }√_π_{– x}_{– πx}

+ πx– πx– π√π_{– x+ π}√_π_{– x+ x}

+π– π+π√π_{–  + π}

= k(x).

From

and

–π– π√π_{–  –π}_{+ }∼

= –. < , we have

k(x) = –π– √π_{–  –πx}_{– π}_{– π}√_π_{–  –π}_{+ x}

+π√π_{–  +π}_{– π}_{+ π}

> –π– √π_{–  –π}π





– π– π√π_{–  –π}_{+ }π



+π√π_{–  +π}_{– π}_{+ π} ∼_{= ..}

Sinceh(x) is strictly increasing for (√π_{– )/ <x<π/, we have}

h(x) >h

√

π_{– }



∼_{= ..}

Thus,g(x) is strictly increasing for (√π_{– )/ <x<π/ and we have}

g(x) >g

√

π_{– }



= .

Therefore, we can get

tanx x <



π_{– x}

ϑ(x)

,

where

ϑ(x) =  x–√π_{– }–

√

π_{–  –π+ }

π–√π_{– } .

Since we have  < 

π_{– x} <

tanx x

for any (√π_{– )/ <x<π/, we obtain}



π_{– x}

θ

<tanx

x <



π_{– x}

ϑ(x)

,

where the constantθ= . Sinceϑ(x) is strictly decreasing for (√π_{– )/ <x<π/ and}

ϑ(x) >ϑ

π



= ,

3.3 Proof of Theorem 2.3 and Corollary 2.4

We need two lemmas to prove Theorem ..

Lemma . For–/ <t< ,we have

ln(t+ ) > t.

Proof We set

f(x) =ln(t+ ) – t, then

f(t) = – t+  (t+ ).

Fromf(t) >  for –/ <t< –/ andf(t) <  for –/ <t< ,f(t) is strictly increasing for –/ <t< –/ andf(t) is strictly decreasing for –/ <t< . Since

f

– 

=  –ln

 

∼_{= .}

and

f() = ,

we can getf(t) >  for –/ <t< .

Lemma . For <s< /,we have

ln(s+ ) > s.

Proof We set

f(s) =ln(s+ ) – s, then

f(s) = – s–  (s+ ).

Fromf(s) >  for  <s< / andf(s) <  for / <s< /,f(s) is strictly increasing for  <s< / andf(s) is strictly decreasing for / <s< /. Since

f

 

=ln

 

– 

∼= . and

f() = ,

Proof of Theorem. We set

f(x) =lntanx

x –

ϑ(x)



ln 

π_{– x}

=lntanx

x –



x– √π_{– }+

 √π_{– }

ln 

π_{– x}.

If

t= – + 

π_{– x},

then –/ <t<  for / <x< (√π_{– )/, by Lemma ., we can get}

ln 

π_{– x} >

 

– + 

π_{– x}

.

If

s= – + 

π_{– x} +



π –

π–  π

π– x,

then  <s< / for / <x< (√π_{– )/, by Lemma . and the inequality (.), we can}

get

lntanx

x >ln



π_{– x} +



π–

π–  π

π– x

> 

– + 

π_{– x}+



π –

π_{– }

π

π– x.

Since 

x– √π_{– }+

 √π_{– }< 

and  

– + 

π_{– x}

<ln 

π_{– x}< 

for / <x< (√π_{– )/, we obtain}

f(x) > 

– + 

π_{– x} +



π –

π–  π

π– x

–



x– √π_{– }+

 √π_{– }

×



– + 

π_{– x}

= g(x)

where

g(x) = –,√π_{– x}_{+ ,π}√_π_{– x}

– ,x+ ,πx– ,πx

+ πx+ ,π√π_{– x}_{– ,π}√_π_{– x}

+ ,πx– ,πx+ ,πx

+ ,πx– πx– ,π√π_{– x+ π}√_π_{– x}

– ,π+ ,π– π.

It suffices to show thatg(x) >  for / <x< (√π_{– )/. We have derivatives}

g(x) = –,√π_{– x}_{+ ,π}√_π_{– x}

– ,x+ ,πx– ,πx

+ ,πx+ ,π√π_{– x}_{– ,π}√_π_{– x}

+ ,πx– ,πx+ ,πx

+ ,π– π– ,π√π_{–  + π}√_π_{– ,}

g(x) = –,√π_{– x}_{+ ,π}√_π_{– x}

– ,x+ ,πx– ,πx

+ πx+ ,π√π_{– x– ,π}√_π_{– x}

+ ,π– ,π+ π

= h(x), and

h(x)

 = –,

√

π_{– x}_{+ ,π}√_π_{– x}

– ,x+ ,πx– ,πx

+ π+ ,π√π_{–  – ,π}√_π_{– }

< –,√π_{– }

 



+ ,π√π_{– }

 

√

π_{– }



– ,

 

+ ,π

 

√

π_{– }

– ,π

 

+ π+ ,π√π_{–  – ,π}√_π_{– }

= –, + ,π– π+π

, + 

√

π_{– }

∼_{= –,..}

Thus,h(x) is strictly decreasing for / <x< (√π_{– )/. From}

h

 

we haveg(x) <  for / <x< (√π_{– )/. Therefore,g(x) is strictly decreasing forx} <

x< (√π_{– )/. From}

g

 

∼_{= ,.}

and

g

√

π_{– }



∼_{= –,,}

there exists uniquely a real numberx with / <x< ( √

π_{– )/ such thatg(x} ) = .

Hence,g(x) is strictly increasing for / <x<xandg(x) is strictly decreasing forx<x<

(√π_{– )/. From}

g

 

∼_{= ,}

and

g

√

π_{– }



= ,

we can getg(x) >  for / <x< (√π_{– )/. Hence, the proof of Theorem . is}

com-plete.

Proof of Corollary. By Theorem ., for / <x< (√π_{– )/, we have the following:}

lntan_xx ln 

π–x

< 

x– √π_{– }+

 √π_{– }

=

– 

x

√

π_{– }



√

π_–

 –x

.

Therefore

lim

x→(√π_{–)/–}

lntan_xx ln_π_–_x

= –∞.

The proof of Corollary . is complete.

4 Conclusions

In this paper, we gave four inequalities derived from the left hand side of Becker-Stark’s inequality (.), which are natural generalizations of the inequality (.). Since the value of /(π_{– x}_{) is less than  for  <x< (}√_π_{– )/ and the value of /(π}_{– x}_{) is larger}

than  for (√π_{– )/ <x<π/, we established the inequalities in Theorems . and ..}

By Theorem ., we obtained Corollary . immediately.

Competing interests

Acknowledgements

The author would like to thank Professor Mitsuhiro Miyagi and the referees for their helpful suggestions and good advice.

Received: 29 June 2015 Accepted: 6 December 2015 References

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