On approximating the modified Bessel function of the first kind and Toader-Qi mean

Abstract

(ProQuest: ... denotes formulae and/or non-USASCII text omitted; see image)

In the article, we present several sharp bounds for the modified Bessel function of the first kind ...... and the Toader-Qi mean ...... for all ...... and ...... with .......

Yang and Chu Journal of Inequalities and Applications (2016) 2016:40 DOI 10.1186/s13660-016-0988-1

R E S E A R C H Open Access

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Web End = On approximating the modied Bessel function of the rst kind and Toader-Qi mean

Zhen-Hang Yang and Yu-Ming Chu*

*Correspondence: mailto:[email protected]

Web End [email protected] School of Mathematics and Computation Sciences, Hunan City University, Yiyang, 413000, China

Abstract

In the article, we present several sharp bounds for the modied Bessel function of the rst kind I0(t) = [summationtext]

n=0

t2n22n(n!)2 and the Toader-Qi mean TQ(a, b) = 2 [integraltext]

/20 acos2 bsin2 d

for all t > 0 and a, b > 0 with a = b.

MSC: 33C10; 26E60

Keywords: modied Bessel function; Toader-Qi mean; logarithmic mean; identric mean

1 Introduction

Let a, b > , p : (, )

R+ be a strictly monotone real function, (, ) and

rn() =

[braceleftBigg](an cos + bn sin )/n, n = ,acos bsin , n = . (.)

Then the mean Mp,n(a, b) was rst introduced by Toader in [] as follows:

Mp,n(a, b) = p

[parenleftbigg]

[integraldisplay]

p

[parenleftbig]rn()[parenrightbig] d

[parenrightbigg]

= p

[parenleftbigg]

[integraldisplay] /

p

[parenleftbig]rn()[parenrightbig] d

[parenrightbigg]

, (.)

where p is the inverse function of p.From (.) and (.) we clearly see that

M/x,(a, b) =

[integraltext]

= AGM(a, b)

is the classical arithmetic-geometric mean, which is related to the complete elliptic integral of the rst kind K(r) = [integraltext]

/

/

d

a cos +b sin

( r sin )/ d. The Toader mean Mx,(a, b) =

[integraldisplay] /

[radicalbig]a cos + b sin d = T(a, b)

is related to the complete elliptic integral of the second kind E(r) = [integraltext]

/

( r sin )/ d.

We have

Mxq,(a, b) =

[parenrightbigg]/q (q = ).

2016 Yang and Chu. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License

(http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, pro

vided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and

indicate if changes were made.

[parenleftbigg]

[integraldisplay] /

aqcos bqsin d

Yang and Chu Journal of Inequalities and Applications (2016) 2016:40 Page 2 of 21

In particular,

Mx,(a, b) =

[integraldisplay] /

acos bsin d

= TQ(a, b) (.)

is the Toader-Qi mean.

Recently, the arithmetic-geometric mean AGM(a, b) and the Toader mean T(a, b) have attracted the attention of many researchers. In particular, many remarkable inequalities for AGM(a, b) and T(a, b) can be found in the literature [].

For q = , the mean Mx

q,(a, b) seems to be mysterious, Toader [] said that he did not know how to determine any sense for this mean.

Let z

C,

R

\{, , , . . .} and (z) = limn n!nz/[ k=(z + k)] be the classical gamma function. Then the modied Bessel function of the rst kind I(z) [] is given by

I (z) =

n=

zn+

[summationdisplay] n!n+ ( + n + ). (.)

Very recently, Qi et al. [] proved the identity

Mxq,(a, b) =

[parenleftbigg]

[integraldisplay] /

aqcos bqsin d

[parenrightbigg]/q = abI/q

[parenleftbigg]q

[parenrightbigg]

log a b

(.)

and inequalities

L(a, b) < TQ(a, b) < A(a, b) + G(a, b)

<

A(a, b) + G(a, b) < I(a, b) (.)

for all q = and a, b > with a = b, where L(a, b) = (b a)/(log b log a), A(a, b) = (a + b)/, G(a, b) = ab, and I(a, b) = (bb/aa)/(ba)/e are, respectively, the logarithmic, arithmetic, geometric, and identric means of a and b.

Let b > a > , p

R, t = (log b log a)/ > , and the pth power mean Ap(a, b) be dened

by

(p = ), A(a, b) = ab = G(a, b).

Then the logarithmic mean L(a, b), the identric mean I(a, b), and the pth power mean Ap(a, b) can be expressed as

L(a, b) = ab

sinh t

t , I(a, b) =

/p

Ap(a, b) =

[parenleftbigg]ap + bp

abet/tanht,

(.)

Ap(a, b) = ab cosh/p(pt) (p = )

and (.)-(.) lead to

TQ(a, b)

ab =

Mx,(a, b)

ab =

[integraldisplay] /

etcos() d

= I(t)

=

[integraldisplay] /

cosh(t cos ) d =

[integraldisplay] /

cosh(t sin ) d. (.)

Yang and Chu Journal of Inequalities and Applications (2016) 2016:40 Page 3 of 21

The main purpose of this paper is to present several sharp bounds for the modied Bessel function of the rst kind I(t) and the Toader-Qi mean TQ(a, b).

2 Lemmas

In order to establish our main results we need several lemmas, which we present in this section.

Lemma . (See []) Let [parenleftbig]n

k[parenrightbig] be the number of combinations of n objects taken k at a time,

that is,

[parenleftbigg]n k

[parenrightbigg] = n!

k!(n k)!.

Then

[summationdisplay]

k=

[parenleftbigg]n k

=

[parenleftbigg]n n

Lemma . (See []) Let {an}n= and {bn}n= be two real sequences with bn > and

limn an/bn = s. Then the power series [summationtext]n= antn is convergent for all t

R and

n= antn [summationtext]

n= bntn = s if the power series [summationtext]

n= bntn is convergent for all t

[summationtext]

lim

t

R.

Lemma . The Wallis ratio

Wn =

(n + ) ( ) (n + )

(.)

is strictly decreasing with respect to all integers n and strictly log-convex with respect to all real numbers n .

Proof It follows from (.) that

Wn+

Wn =

(n + ) < (.)

for all integers n .

Therefore, Wn is strictly decreasing with respect to all integers n follows from (.). Let f (x) = (x + /)/ (x + ) and (x) = (x)/ (x) be the psi function. Then it follows from the monotonicity of (x) that

[bracketleftbig]log f (x)[bracketrightbig] =

x +

[parenleftbigg]

[parenrightbigg]

(x + ) > (.)

for all x .

Therefore, Wn is strictly log-convex with respect to all real numbers n follows from (.) and (.).

Yang and Chu Journal of Inequalities and Applications (2016) 2016:40 Page 4 of 21

Lemma . (See []) The double inequality

(x + a)a <

(x + a) (x + ) <

xa

holds for all x > and a (, ).

Lemma . Let sn = (n)!(n+)!/[n(n!)]. Then the sequence {sn}n= is strictly decreasing and

lim

n sn =

. (.)

Proof The monotonicity of the sequence {sn}n= follows from

sn+

sn =

(n + )(n + )(n + ) < .

To prove (.), we rewrite sn as

sn = (n + )

[bracketleftbigg](n )!! nn!

= n +

( )

(n + ) (n + )

. (.)

It follows from Lemma . and (.) that

=

= (n + )

(n + ) (n + )

n +

n +

< sn <

n +

n . (.)

Therefore, equation (.) follows from (.).

Lemma . (See []) Let A(t) = [summationtext]

k= aktk and B(t) = [summationtext]k= bktk be two real power series converging on (r, r) (r > ) with bk > for all k. If the non-constant sequence {ak/bk} is increasing (decreasing) for all k, then the function A(t)/B(t) is strictly increasing (decreasing)

on (, r).

Lemma . (See []) Let A(t) = [summationtext]

k= aktk and B(t) = [summationtext]k= bktk be two real power series converging on R with bk > for all k. If there exists m

N such that the non-constant sequence {ak/bk} is increasing (decreasing) for k m and decreasing (increasing) for k m, then there exists t (, ) such that the function A(t)/B(t) is strictly increasing (decreasing) on (, t) and strictly decreasing (increasing) on (t, ).

Lemma . The identity

I(t) =

(n)! n(n!) tn

holds for all t

R.

Yang and Chu Journal of Inequalities and Applications (2016) 2016:40 Page 5 of 21

Proof From (.) and Lemma . together with the Cauchy product we have

I(t) =

[summationdisplay]

n=

[parenleftBigg] n

[summationdisplay]

k=

k(k!)

[parenrightBigg]

(nk)[(n k)!]

tn

[parenleftBigg]

[parenrightBigg]

(n!) (k!)[(n k)!]

(n)!n(n!) tn.

Lemma . (See []) Let < a < b < and f , g : [a, b]

R. Then

=

n(n!)

n

[summationdisplay]

k=

tn =

[integraldisplay] ba g(x) dx

if both f and g are increasing or decreasing on (a, b).

Lemma . (See []) Let < a < b < and f , g : (a, b)

R. Then

[integraldisplay] ba f (x)g(x) dx

b a

[integraldisplay] ba f (x) dx

[integraldisplay] ba f (x)g(x) dx

b a

[integraldisplay] ba f (x) dx

[integraldisplay] ba g(x) dx

[parenrightbigg]

(b a)

[integraldisplay] b a

[parenleftbigg] x a + b

[parenrightbigg] f (x) dx [integraldisplay]

b a

[parenleftbigg] x a + b

g(x) dx (.)

if both f and g are convex on the interval (a, b), and inequality (.) becomes an equality if and only if f or g is a linear function on (a, b).

3 Main results

Theorem . The double inequalities

et + t < I(t) <

et + t (.)

and

b + log(b/a) < TQ(a, b) <

b[radicalbig] + log(b/a) (.)

hold for all t > and b > a > .

Proof From (.) we have

I(t) =

[integraldisplay] /

cosh(t sin ) d =

[integraldisplay]

cosh(tx)

x dx (.)

and

etI(t) =

[integraldisplay] /

et[cos()] d

=

[integraldisplay] /

d

etsin

<

[integraldisplay] /

d + t sin =

+ t . (.)

Yang and Chu Journal of Inequalities and Applications (2016) 2016:40 Page 6 of 21

We clearly see that both cosh(tx) and / x are increasing with respect to x on (, ). Then Lemma . and (.) lead to

I(t)

[integraldisplay]

dx x

[integraldisplay]

cosh(tx) dx =

sinh t

t

= et

t

[parenleftbigg]

et

> et

t

[parenrightbigg]

[parenleftbigg]

+ t

[parenrightbigg] = et

+ t . (.)

Therefore, inequality (.) follows from (.) and (.).

Let t = log(b/a)/. Then it follows from (.) and (.) that

b/a + log(b/a) <

TQ(a, b)

ab <

b/a[radicalbig] + log(b/a). (.)

Therefore, inequality (.) follows from (.).

Remark . From Theorem . we clearly see that

lim

t etI(t) =

lim

x

+ TQ(x, ) = .

Theorem . The double inequalities

[radicalbigg]sinh(t)t < I(t) <

[radicalbigg]sinh(t)t (.)

and

[radicalbig]L(a, b)A(a, b) < TQ(a, b) < [radicalbig]L(a, b)A(a, b) (.)

hold for all t > and a, b > with a = b if and only if /, /, / and .

Proof Let

R(t) = I(t)

sinh(t)/(t), (.)

an = (n)!

n(n!) , bn =

n

(n + )!. (.)

Then simple computation leads to

an

bn =

(n)!(n + )!

n(n!) . (.)

It follows from Lemma . and (.) that the sequence {an/bn}n= is strictly decreasing and

lim

n

an

bn =

. (.)

Yang and Chu Journal of Inequalities and Applications (2016) 2016:40 Page 7 of 21

From Lemma . we have

R(t) =

n= antn [summationtext]

n= bntn . (.)

Lemma . and (.) together with the monotonicity of the sequence {an/bn}n= lead to the conclusion that R(t) is strictly decreasing on the interval (, ). Therefore, we have

lim

t R(t) < R(t) <

lim

t

+ R(t) =

[summationtext]

a

b = . (.)

From Lemma ., (.), and (.) we know that

lim

t R(t) =

. (.)

Therefore, inequality (.) holds for all t > if and only if / and / follows easily from (.), (.), and (.) together with the monotonicity of R(t).

Let b > a > and t = log(b/a)/. Then inequality (.) holds for a, b > with a = b if and only if /, and follows from (.) and (.) together with inequality (.)

for all t > if and only if / and /.

Remark . Equations (.) and (.) imply that

lim

t et

tI(t) =

or we have the asymptotic formula

I(t)

ett (t ).

Theorem . Let , > , t = . . . . be the unique solution of the equation

d dt

[bracketleftbigg] tI(t) sinh t (cosh t ) sinh t

[bracketrightbigg]

= (.)

on (, ) and

= tI(t)

sinh t(cosh t ) sinh t = . . . . . (.)

Then the double inequality

[radicalbigg]

( cosh t + )

sinh t

t < I(t) <

[radicalbigg]

( cosh t + )

sinh t

t (.)

or

[radicalBig][bracketleftbig]A(a, b) + ( )G(a, b)[bracketrightbig]L(a, b) < TQ(a, b) < [radicalBig][bracketleftbig]A(a, b) + ( )G(a, b)[bracketrightbig]L(a, b)

holds for all t > or a, b > with a = b if and only if /, > .

Yang and Chu Journal of Inequalities and Applications (2016) 2016:40 Page 8 of 21

Proof Let

R(t) = I(t) sinhtt

(cosh t) sinh t

t

, (.)

cn = (n)!

n(n!)

(n + )!, dn =

n

(n + )!, sn =

(n)!(n + )!n(n!) , (.)

and

s n = [parenleftbig]n + n + n + [parenrightbig]sn [parenleftbig]n + n + [parenrightbig]. (.)

Then it follows from Lemma ., Lemma ., Lemma ., and (.)-(.) that

R(t) =

n= cntn [summationtext]

n= dntn , (.)

lim

[summationtext]

t R(t) =

lim

n

cn

dn =

lim

n

nsn

n =

, (.)

c

d =

<

c

d =

>

c

d =

, (.)

cn+

dn+

cn

dn =

ns n

(n + )(n+ )(n ), (.)

and we have the inequality

s n >

[parenleftbig]n + n + n + [parenrightbig] [parenleftbig]n + n + [parenrightbig]

>

[parenleftbig]n + n + n + [parenrightbig] [parenleftbig]n + n + [parenrightbig] =

[bracketleftbig]n [parenleftbig]n + n + [parenrightbig][bracketrightbig] > (.)

for all n .

From (.)-(.) we know that the sequence {cn/dn}n= is strictly increasing for

n and strictly decreasing for n . Then Lemma . and (.) lead to the conclusion

that there exists t (, ) such that R(t) is strictly increasing on (, t) and decreasing

on (t, ). Therefore, we have

min[braceleftBig]R[parenleftbig]+[parenrightbig], lim

t R(t)

[bracerightBig] < R(t) R(t) (.)

for all t > , and t is the unique solution of equation (.) on (, ).

Note that

R[parenleftbig]+[parenrightbig] = cd =

. (.)

From (.), (.), (.), (.), and (.) we get

< R(t) R(t) = . (.)

Therefore, inequality (.) holds for all t > if and only if /, follows from

(.) and (.) together with the piecewise monotonicity of R(t) on (, ). Numerical

computations show that t = . . . . and = . . . . .

Yang and Chu Journal of Inequalities and Applications (2016) 2016:40 Page 9 of 21

Theorem . Let p, q

R. Then the double inequality

p

< I(t) < q

coshp t

sinh t t

sinh t

t + ( q)

cosh t (.)

or

Lp(a, b)Ap(a, b) < TQ(a, b) < qL(a, b) + ( q)A(a, b)

holds for all t > or a, b > with a = b if and only if p / and q /.

Proof If the rst inequality of (.) holds for all t > , then

lim

t

+

I(t) coshp t(sinhtt )pt =

[parenleftbigg]

[parenrightbigg]

p

,

which implies that p /.

It is not dicult to verify that the function coshp t(sinh t/t)p is strictly decreasing with respect to p

R for any xed t > , hence we only need to prove the rst inequality of (.) for all t > and p = /, that is,

I(t) >

sinh t t

cosh t. (.)

Making use of the power series and Cauchy product formulas together with Lemma . we have

I(t)

sinh t t

cosh t

[bracketrightBigg]

=

[summationdisplay]

n=

[bracketleftBigg] n

[summationdisplay]

k=

[parenleftbigg] (k)!

k(k!)

tn. (.)

Let Wn and sn be, respectively, dened by Lemma . and Lemma ., and

un =

n

[summationdisplay]

k=

((n k))! (nk)((n k)!)

[parenrightbigg] n+ n+

(n + )!

[parenleftbigg] (k)!

k(k!)

((n k))! (nk)((n k)!)

[parenrightbigg] n+ n+

(n + )! . (.)

Then simple computations lead to

u = u = , u =

, u =

. (.)

It follows from Lemma . and Lemmas .-. together with (.) that

un =

n

[summationdisplay]

k=

[parenleftbigg]

(k!)((n k)!)

(k)! k(k!)

((n k))! (nk)((n k)!)

[parenrightbigg] n+ n+

(n + )!

n

k=

[parenleftbigg]

= [summationdisplay] (k!)((n k)!) WkWnk

[parenrightbigg] n+ n+

(n + )!

Yang and Chu Journal of Inequalities and Applications (2016) 2016:40 Page 10 of 21

n

k=

[parenleftbigg]

> [summationdisplay] (k!)((n k)!) Wn/

[parenrightbigg] n+ n+

(n + )!

=

n

[summationdisplay]

k=

[bracketleftbigg] (k!)((n k)!)

[parenleftbigg] (n/ + /) (/) (n/ + )

n+ n+

(n + )!

[bracketrightbigg]

>

n

[summationdisplay]

k=

[parenleftbigg] (n/ + /)(k!)((n k)!)

[parenrightbigg] n+ n+

(n + )!

=

(n + )(n!)

n

[summationdisplay]

k=

(n!) (k!)((n k)!)

n+ n+ (n + )!

=

(n + )(n!)

(n)!

(n!)

n+ n+ (n + )!

= n+(n+ )

(n + )!

[bracketleftbigg] n+

n+

n +

[parenleftbigg]

[parenrightbigg]

[bracketrightbigg]

sn

> n+(n+ )

(n + )!

+

[parenrightbigg]

[bracketleftbigg][parenleftbigg]

[bracketrightbigg]

> (.)

for all n .Therefore, inequality (.) follows from (.)-(.).

If the second inequality of (.) holds for all t > , then we have

lim

t

+

I(t) qsinhtt ( q) cosh t

t =

q

[parenleftbigg]

[parenrightbigg]

,

which implies that q /.

Since cosh t > sinh t/t, we only need to prove that the second inequality of (.) holds for all t > and q = /, that is,

cosh t I(t) cosh t sinh t/t >

. (.)

Let

n = nn! (n )!!

nn!(n)! ,

n = n

(n + )!,

n = (n + )(n + )

(n + ) Wn,

and Wn be dened by (.).

Then simple computations lead to

cosh t I(t) cosh t sinh t/t =

n= ntn [summationtext]

n= ntn , (.)

n+

n+

[summationtext]

n

n =

n + n + ( Wn+)

n +

n ( Wn) =

n n(n + ), (.)

n+

n = +

n + (n + ) > ,

=

> . (.)

From (.) and (.) we clearly see that the sequence {n/n}n= is strictly increasing, then Lemma . and (.) lead to the conclusion that the function (cosh t I(t))/[cosh t

Yang and Chu Journal of Inequalities and Applications (2016) 2016:40 Page 11 of 21

sinh t/t] is strictly increasing on the interval (, ). Therefore, inequality (.) follows from the monotonicity of (cosh t I(t))/[cosh t sinh t/t] and the fact that

lim

t

+

.

Theorem . Let p, q > , t be the unique solution of the equation

d[p

(I(t))

cosh(pt) ]

dt = (.)

and

= p(I(t) )

cosh(pt) . (.)

Then the following statements are true:(i) The double inequality

p +

p

cosh t I(t) cosh t sinh t/t =

=

cosh(pt) < I(t) <

q +

q

cosh(qt) (.)

or

[parenleftbigg]

p

[parenrightbigg] G(a, b) +

p App(a, b)Gp(a, b)

< TQ(a, b) <

q Aqq(a, b)Gq(a, b)

holds for all t > or a, b > with a = b if and only if p / and q .

(ii) The inequality

I(t)

p +

[parenleftbigg]

q

[parenrightbigg] G(a, b) +

p

cosh(pt) (.)

or

[parenleftbigg]

[parenrightbigg]

p App(a, b)Gp(a, b)

holds for all t > or a, b > with a = b if p (/, ).

Proof (i) Let

R(t) = p(I(t) )

cosh(pt) , (.)

un =

n(n!) , vn =

pn

(n)! .

TQ(a, b)

p

G(a, b) +

Then simple computations lead to

R(t) =

n= untn [summationtext]

n= vntn , (.)

[summationtext]

Yang and Chu Journal of Inequalities and Applications (2016) 2016:40 Page 12 of 21

un+

vn+

[parenrightbigg]

un

vn =

(n)! (p)n(n!)

[parenleftbigg] p n + n +

. (.)

From (.) we clearly see that the sequence {un/vn}n= is strictly decreasing if p and strictly increasing if p /. Then Lemma . and (.) lead to the conclusion that the function R(t) is strictly decreasing if p and strictly increasing if p /. Hence, we have

R(t) < lim

t

+ R(t) =

u

v =

(.)

for all t > if p and

R(t) > lim

t

+ R(t) =

u

v =

(.)

for all t > if p /.

Therefore, inequality (.) holds for all t > if p / and q follows easily from (.) and (.) together with (.).

If the rst inequality (.) holds for all t > , then we have

lim

t

+

I(t) (

p +

p cosh(pt)) t =

[parenleftbigg]

p

[parenrightbigg]

,

which implies that p /.

If there exists q (/, ) such that the second inequality of (.) holds for all t > , then we have

lim

t

I(t) (

q

cosh(qt))eqt . (.)

But the rst inequality of (.) leads to

I(t) (

q

+

+

cosh(qt)) eqt

> e(q)t

+ t

[parenleftbigg]

q

[parenrightbigg] eqt + eqt

q (t ),

which contradicts inequality (.).(ii) If p (/, ), then from (.) we know that there exists n

N such that the sequence {un/vn}n= is strictly decreasing for n n and strictly increasing for n n.

Then (.) and Lemma . lead to the conclusion that there exists t (, ) such that the function R(t) is strictly decreasing on (, t] and strictly increasing on [t, ). We clearly see that t satises equation (.). It follows from (.) and (.) together with the piecewise monotonicity of R(t) that

R(t) R(t) = . (.)

Therefore, inequality (.) holds for all t > follows from (.) and (.).

Yang and Chu Journal of Inequalities and Applications (2016) 2016:40 Page 13 of 21

It is not dicult to verify that the function

p +

p

cosh(pt)

is strictly increasing with respect to p on the interval (, ) and

cosh

[parenleftbigg] t

[parenrightbigg]

> cosh/ t

for t > .

Letting p = /, /, /, /, / and q = in Theorem .(i), then we get Corollary . immediately.

Corollary . The inequalities

cosh/(t) < cosh

[parenleftbigg] t

[parenrightbigg]

[parenrightbigg] <

cosh[parenleftbigg]t

<

cosh[parenleftbigg]t

[parenrightbigg]

[parenrightbigg]

<

cosh[parenleftbigg]t

+

<

cosh[parenleftbigg]t

+

[parenrightbigg] < I(t) <

+ cosh t

or

G/(a, b)A/(a, b) < A//(a, b)G/(a, b) G(a, b)

<

A//(a, b)G/(a, b)

G(a, b)

< A

/

/(a, b)G

/(a, b)

<

A//(a, b)G/(a, b) +

G(a, b)

<

A

/

/(a, b)G

/(a, b) +

G(a, b)

< TQ(a, b) < A(a, b) + G(a, b)

hold for all t > or all a, b > with a = b.

Theorem . Let p > . Then the following statements are true:(i) The inequality

I(t) > [bracketleftbig]cosh(pt)[bracketrightbig]

p (.)

or

TQ(a, b) > G

p (a, b)A

pp (a, b) (.)

holds for all t > or a, b > with a = b if and only if p /.(ii) The inequality (.) or (.) is reversed if and only if p /.

Yang and Chu Journal of Inequalities and Applications (2016) 2016:40 Page 14 of 21

(iii) The inequalities

cosh/ t < cosh

[parenleftbigg]t

[parenrightbigg][bracketrightbigg]/

[parenrightbigg]

<

[parenrightbigg] [bracketleftbigg]cosh[parenleftbigg]t

< I(t) < cosh

[parenleftbigg] t

< et/ (.)

or

G/(a, b)A/(a, b) < A

/

/(a, b)G

/(a, b)

< A

/(a, b)< TQ(a, b) < A(a, b) + G(a, b)

< G(a, b)e(A(a,b)G(a,b))/(L(a,b))

hold for all t > or all a, b > with a = b.

Proof (i) If inequality (.) holds for all t > , then we have

lim

t

+

I(t) [cosh(pt)]

/

/(a, b)G

p

[parenleftbigg] p

[parenrightbigg]

t =

,

which implies that p /.

It follows from Lemma of [] that the function [cosh(pt)]/(p) is strictly decreasing

with respect to p (, ) for any xed t > , hence we only need to prove that inequality (.) holds for all t > and p = /. From the sixth inequality of Corollary . we clearly see that it suces to prove that

cosh[parenleftbigg]t

[parenrightbigg]

+

>

[bracketleftbigg]cosh[parenleftbigg]t

[parenrightbigg][bracketrightbigg]/

for all t > , which is equivalent to

log[bracketleftbigg]

cosh(x) +

>

log(cosh x) (.)

for all x > , where x = t/.

Let

f(x) = log

[bracketleftbigg]

[bracketrightbigg]

[bracketrightbigg]

cosh(x) +

log(cosh x), (.)

f(x) = sinh(x) cosh x cosh(x) sinh x sinh x,

n = ( )( + )n + ( + )( )n ,

n = ( + )n.

Then simple computations lead to

f() = , (.)

Yang and Chu Journal of Inequalities and Applications (2016) 2016:40 Page 15 of 21

f (x) = f(x)

cosh x[ cosh(x) + ], (.)

f(x) =

[summationdisplay]

n=

n(n )!xn, (.)

= = , (.)

nn = ( )(n )(n ). (.)

From (.)-(.) and n > > > for n we know that

f(x) > (.)

for all x > .Therefore, inequality (.) follows easily from (.)-(.) and (.).(ii) The suciency follows easily from the monotonicity of the function p [cosh(pt)]/(p) and the last inequality in Corollary . together with the identity ( + cosh t)/ = cosh(t/).

Next, we prove the necessity. If there exists p (/, /) such that I(t) < [cosh(pt)]/(p) for all t > , then we have

lim

t

I(t) [cosh(pt)]/(p)

et/(p) . (.)

But the rst inequality of (.) leads to

I(t) [cosh(pt)]/(p)

et/(p) >

+ t

/(p)

(t ),

et et/(p)

[parenleftbigg] + ept

which contradicts (.).(iii) Let p = , /, /, /, +. Then parts (i) and (ii) together with the monotonicity of the function p [cosh(pt)]/(p) lead to (.).

Theorem . Let [, /]. Then the inequality

I(t) >

cosh(t cos ) + cosh(t sin ) (.)

or

TQ(a, b) > Acoscos(a, b)Gcos(a, b) + Asinsin(a, b)Gsin(a, b)

holds for all t > or all a, b > with a = b if and only if [/, /]. In particular, the

inequalities

I(t) >

[bracketleftbigg]cosh[parenleftbigg]

[radicalbig]

[parenrightbigg]

[parenleftbigg][radicalbig] +

[parenrightbigg][bracketrightbigg]

t

+ cosh

t

>

[bracketleftbigg]cosh[parenleftbigg]

[parenrightbigg]

[parenleftbigg]

t

[parenrightbigg][bracketrightbigg]

[parenleftbigg]

[parenrightbigg]

t

+ cosh

> cosh

t

(.)

Yang and Chu Journal of Inequalities and Applications (2016) 2016:40 Page 16 of 21

or

/

/(a, b)G

/(a, b) + A

+/

+/(a, b)G

+/(a, b)

TQ(a, b) >

A

/

/(a, b)G

>

A

/ + A//(a, b)G/(a, b)

> A

/

/(a, b)G

/(a, b)

hold for all t > or all a, b > with a = b.

Proof If inequality (.) holds for all t > , then we have

lim

t

+

I(t) cosh(tcos)+cosh(tsin)

t =

cos() ,

which implies that [/, /].

Next, we prove the suciency of inequality (.). Simple computations lead to

[cosh(t cos ) + cosh(t sin )] =

t sin()

[bracketrightbigg]

sinh(t sin )t sin

sinh(t cos )

t cos

, (.)

[parenrightbigg]

[parenrightbigg]

sinh x x

=

x

[parenleftbigg]cosh x

sinh x x

> (.)

for x > .

Equation (.) and inequality (.) imply that the function [cosh(t cos ) +

cosh(t sin )] is decreasing on [, /] and increasing on [/, /] for any xed t > . Hence, it suces to prove that inequality (.) holds for all t > and = = /.

Let

n = (

)n + (+

)n (n)! ,

n =

n(n!) ,

cosh(t cos ) + cosh(t sin )

I(t) . (.)

Then simple computations lead to

R(t) =

R(t) =

n= ntn [summationtext]

n= ntn , (.)

=

=

[summationtext]

=

= , (.)

n+ n+

n

n

=

[(n + )( )n + (n )( + )n]

(n + )[( )n + ( + )n] < (.)

for n .

It follows from Lemma . and (.)-(.) that R(t) is strictly decreasing on (, ).

Therefore,

I(t) >

cosh(t cos ) + cosh(t sin )

(.)

follows from (.) and the monotonicity of R(t) together with R(+) = / = .

Yang and Chu Journal of Inequalities and Applications (2016) 2016:40 Page 17 of 21

Let = /, /, /. Then inequality (.) follows easily from (.) and the monotonicity of the function [cosh(t cos ) + cosh(t sin )].

Theorem . The inequality

I(t) >

sinh t

t +

( )(t sinh t cosh t + )t (.)

holds for all t > .

Proof It is easy to verify that

d dx

[parenleftbigg]

x

[parenrightbigg] = + x

( x)/ > ,

cosh(tx)

cosh(tx) >

for all t > and x (, ), which implies that the two functions / x and cosh(tx) are convex with respect to x on the interval (, ). Then from Lemma . and (.) we have

I(t)

x = t

sinh t t

= [integraldisplay]

cosh(tx)

x dx

[integraldisplay]

dx x

[integraldisplay]

cosh(tx) dx

> [integraldisplay]

x x dx

[integraldisplay]

x

[parenrightbigg] cosh(tx) dx

[parenleftbigg]

)(t sinh t cosh t + )

t . (.)

Therefore, inequality (.) follows from (.).

Remark . The inequality I(t) > sinh(t)/t in (.) is equivalent to the rst inequality TQ(a, b) > L(a, b) in (.). Therefore, Theorem . is an improvement of the rst inequality in (.).

Let p

= (

R and M(a, b) be a bivariate mean of two positive a and b. Then the pth power-type mean Mp(a, b) is dened by

Mp(a, b) = M/p[parenleftbig]ap, bp[parenrightbig] (p = ), M(a, b) = ab.

We clearly see that

Mp(a, b) = M/p[parenleftbig]a, b[parenrightbig]

for all , p

R and a, b > if M is a bivariate mean.

Theorem . The inequality

TQ(a, b) < Ip(a, b)

holds for all a, b > with a = b if and only if p /.

Yang and Chu Journal of Inequalities and Applications (2016) 2016:40 Page 18 of 21

Proof The second inequality (.) can be rewritten as

TQ(a, b) < A/(a, b). (.)

In [, ], the authors proved that the inequality

I(a, b) > A/(a, b) (.)

holds for all distinct positive real numbers a and b with the best possible constant /.Inequalities (.) and (.) lead to

TQ(a, b) < A/(a, b) = A//[parenleftbig]a/, b/[parenrightbig] < I/[parenleftbig]a/, b/[parenrightbig] = I/(a, b) (.)

for all a, b > with a = b.

If p /, then TQ(a, b) < I/(a, b) Ip(a, b) follows from (.) and the function p Ip(a, b) is strictly increasing [].

If TQ(a, b) < Ip(a, b) for all a, b > with a = b. Then

I(t) et/tanh(pt)/p < (.)

for all t > .Inequality (.) leads to

lim

t

+

[parenleftbigg]

p

[parenrightbigg]

I(t) et/tanh(pt)/pt =

,

which implies that p /.

Remark . For all a, b > with a = b, the Toader mean T(a, b) satises the double inequality [, ]

A/(a, b) < T(a, b) < Alog/(loglog)(a, b) (.)

with the best possible constants / and log /(log log ), and the one-sided inequality []

T(a, b) < I/(a, b). (.)

It follows from (.) and (.) that

A//(a, b) = A/[parenleftbig]a/, b/[parenrightbig] < T[parenleftbig]a/, b/[parenrightbig]

= T//(a, b) < I/[parenleftbig]a/, b/[parenrightbig] = I//(a, b),

which can be rewritten as

A/(a, b) < T/(a, b) < I/(a, b). (.)

Yang and Chu Journal of Inequalities and Applications (2016) 2016:40 Page 19 of 21

Inequalities (.) and (.) lead to the inequalities

TQ(a, b) < A/(a, b) < T/(a, b) < I/(a, b) (.)

for all a, b > with a = b.

Remark . For all a, b > with a = b, Theorem . shows that

L/(a, b)A/(a, b) < TQ(a, b) < L(a, b) + A(a, b) . (.)

It follows from L(a, b) < A(a, b)/ + G(a, b)/, given by Carlson in [], and A(a, b) > L(a, b) that

L(a, b) < L/(a, b)A/(a, b), A(a, b) + G(a, b) >

L(a, b) + A(a, b)

.

Therefore, inequality (.) is an improvement of the rst and second inequalities of (.).

Remark . In [, , ], the authors proved that the inequalities

L(a, b) < AGM(a, b) < L/(a, b)A/(a, b) < L/(a, b) (.)

hold for all a, b > with a = b.

Inequalities (.)-(.) lead to the chain of inequalities

L(a, b) < AGM(a, b) < L/(a, b)A/(a, b)

< TQ(a, b) < A/(a, b) < T/(a, b) < I/(a, b) (.)

for all a, b > with a = b.

Motivated by the rst inequality in (.) and the third inequality in (.), we propose Conjecture ..

Conjecture . The inequality

TQ(a, b) > L/(a, b)

holds for all a, b > with a = b.

For all a, b > with a = b, inspired by the double inequality

[radicalbig]A(a, b)G(a, b) < TQ(a, b) < A(a, b) + G(a, b)

given in Corollary . and the inequalities

[radicalbig]A(a, b)G(a, b) < [radicalbig]I(a, b)L(a, b) < I(a, b) + L(a, b)

<

proved by Alzer in [] we propose Conjecture ..

A(a, b) + G(a, b)

Yang and Chu Journal of Inequalities and Applications (2016) 2016:40 Page 20 of 21

Conjecture . The inequality

TQ(a, b) < [radicalbig]I(a, b)L(a, b)

holds for all a, b > with a = b.

Remark . Let Wn be the Wallis ratio dened by (.), and cn, dn, and sn be dened by (.). Then it follows from Lemma . and the proof of Theorem . that the sequence

{sn}n= is strictly decreasing and limn sn = /, and the sequence {cn/dn}n= is strictly increasing for n = , and strictly decreasing for n . Hence, we have

< sn = (n + )Wn s =

(.)

and

= min

c d ,

lim

n

cn

dn

< cn

dn =

nsn

n

c

d =

(.)

N.

Inequalities (.) and (.) lead to the Wallis ratio inequalities

[radicalBig](n + )

for all n

< Wn

[radicalBig]n +

and

n( ) +

(n + ) < Wn

+ n

(n + )

for all n

N.

Competing interests

The authors declare that they have no competing interests.

Authors contributions

All authors contributed equally to the writing of this paper. All authors read and approved the nal manuscript.

Acknowledgements

The authors would like to thank the anonymous referee for his/her valuable comments and suggestions. The research was supported by the Natural Science Foundation of China under Grants 11371125, 11401191, and 61374086.

Received: 14 October 2015 Accepted: 21 January 2016

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