Journal of Inequalities and Applications

J Inequal Appl. 2017; 2017(1): 201.
Published online 2017 August 29.
PMCID: PMC5575000

# Optimal convex combination bounds of geometric and Neuman means for Toader-type mean

## Abstract

In this paper, we prove that the double inequalities

$αNQA(a,b)+(1−α)G(a,b)

hold for all ab > 0 with a ≠ b if and only if α ≤ 3/8, $β≥4/[π(log(1+2)+2)]=0.5546⋯$, λ ≤ 3/10 and μ ≥ 8/[π(π + 2)] = 0.4952 ⋯ , where TD(ab), G(ab), A(ab) and NQA(ab), NAQ(ab) are the Toader, geometric, arithmetic and two Neuman means of a and b, respectively.

Keywords: Toader mean, geometric mean, Neuman mean

## Introduction

For xyz ≥ 0 with xyxzyz ≠ 0 and r ∈ (0, 1), the symmetric integrals RF(xyz) and RG(xyz) [1] of the first and second kinds, and the complete elliptic integrals 𝒦(r) and ℰ(r) of the first and second kinds are defined by

$RF(x,y,z)=12∫0∞[(t+x)(t+y)(t+z)]−1/2dt,RG(x,y,z)=14∫0∞[(t+x)(t+y)(t+z)]−1/2(xt+x+yt+y+zt+z)tdt,K(r)=∫0π/2[1−r2sin2(t)]−1/2dt,E(r)=∫0π/2[1−r2sin2(t)]1/2dt,$

respectively.

The well-known identities

𝒦(r) = RF(0, 1 − r2, 1),  ℰ(r) = 2RG(0, 1 − r2, 1)

were established by Carlson in [1].

Let ab > 0 with a ≠ b. Then the Toader mean TD(ab) [2] and the Schwab-Borchardt mean SB(ab) [35] are respectively defined by

$TD(a,b)=2π∫0π/2a2cos2(t)+b2sin2(t)dt={2aE(1−(b/a)2)/π,a>b,2bE(1−(a/b)2)/π,a
1.1

and

$SB(a,b)={b2−a2cos−1(a/b),ab,$

where cos−1(x) and $cosh−1(x)=log(x+x2−1)$ are the inverse cosine and inverse hyperbolic cosine functions, respectively.

Very recently, Neuman [6] introduced the Neuman mean N(ab) of the second kind as follows:

$N(a,b)=12[a+b2SB(a,b)].$

It is well known that the Toader mean TD(ab), the Schwab-Borchardt mean SB(ab) and the Neuman mean of the second kind N(ab) satisfy the identities (see [6, 7])

$TD(a,b)=4πRG(a2,b2,0)=1π∫0∞[(t+a2)(t+b2)]−1/2(a2t+a2+b2t+b2)tdt,SB(a,b)=1/RF(a2,b2,b2)=2/∫0∞[(t+a2)(t+b2)(t+b2)]−1/2dt,N(a,b)=RG(a2,b2,b2)=14∫0∞[(t+a2)(t+b2)(t+b2)]−1/2(a2t+a2+b2t+b2+b2t+b2)tdt.$

Let p ∈ ℝ and ab > 0. Then the pth power mean Mp(ab) is defined by

$Mp(a,b)=[(ap+bp)/2]1/p(p≠0),M0(a,b)=ab.$
1.2

We clearly see that Mp(ab) is symmetric and homogeneous of degree one with respect to a and b, strictly increasing with respect to p ∈ ℝ for fixed ab > 0 with a ≠ b, and the inequalities

G(ab) = M0(ab) < A(ab) = M1(ab) < Q(ab) = M2(ab)

hold for ab > 0 with a ≠ b, where $G(a,b)=ab$, A(ab) = (ab)/2 and $Q(a,b)=(a2+b2)/2$ are the geometric, arithmetic and quadratic means of a and b, respectively.

In [6], Neuman presented the explicit formula for NQA(ab) ≡ N[Q(ab), A(ab)] and NAQ(ab) ≡ N[A(ab), Q(ab)] as follows:

$NQA(a,b)=12A(a,b)[1+v2+sinh−1(v)v],$
1.3

$NAQ(a,b)=12A(a,b)[1+(1+v2)tan−1(v)v]$
1.4

and proved that the inequalities

A(ab) < NQA(ab) < NAQ(ab) < Q(ab)
1.5

hold for ab > 0 with a ≠ b, where v = (a − b)/(ab).

Recently, the Toader mean has been the subject of intensive research. In particular, many remarkable inequalities for Toader mean and other related means can be found in the literature [841].

In [42], Vuorinen conjectured that

TD(ab) > M3/2(ab)

for all ab > 0 with a ≠ b. This conjecture was proved by Qiu and Shen [43], and Barnard et al. [44], respectively, and Alzer and Qiu [45] presented the best possible upper power mean bound for the Toader mean as follows:

TD(ab) < Mlog2/log(π/2)(ab)

for all ab > 0 with a ≠ b.

Li, Qian and Chu [46] proved that the inequality

αNAQ(ab) + (1 − α)A(ab) < TD(ab) < βNAQ(ab) + (1 − β)A(ab)

holds for all ab > 0 with a ≠ b if and only if α ≤ 3/4 and β ≥ 4(4 − π)/[π(π − 2)] = 0.9573 ⋯ .

Note that

G(ab) < TD[A(ab), G(ab)] < A(ab)
1.6

for all ab > 0 with a ≠ b.

From inequalities (1.5) and (1.6) we clearly see that

G(ab) < TD[A(ab), G(ab)] < NQA(ab) < NAQ(ab)

for all ab > 0 with a ≠ b.

The main purpose of this paper is to find the greatest values α, λ and the least values β, μ such that the double inequalities

$αNQA(a,b)+(1−α)G(a,b)

hold for all ab > 0 with a ≠ b. As applications, we get two new bounds for the complete elliptic integral of the second kind in terms of elementary functions.

## Lemmas

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

For r ∈ (0, 1), we clearly see that

𝒦(0+) = ℰ(0+) = π/2,  𝒦(1) =  + ∞,  ℰ(1) = 1,

and 𝒦(r) and ℰ(r) satisfy the formulas (see[21], Appendix E, pp.474-475)

$dK(r)dr=E(r)−(1−r2)K(r)r(1−r2),dE(r)dr=E(r)−K(r)r,d[E(r)−K(r)]dr=−rE(r)1−r2.$

### Lemma 2.1

see [21], Theorem 1.25

For −∞ < a < b <  + ∞, let fg:[ab] → ℝ be continuous on [ab] and differentiable on (ab), and g(x) ≠ 0 on (ab). If f(x)/g(x) is increasing (decreasing) on (ab), then so are

$f(x)−f(a)g(x)−g(a)andf(x)−f(b)g(x)−g(b).$

If f(x)/g(x) is strictly monotone, then the monotonicity in the conclusion is also strict.

### Lemma 2.2

see [21], Theorem 3.21(1), Exercise 3.43(11) and Exercise 3.43(29)

1. The function r ↦ [ℰ(r)−(1 − r2)𝒦(r)]/r2 is strictly increasing from (0, 1) onto (π/4, 1);
2. The function r ↦ [𝒦(r)−ℰ(r)]/r2 is strictly increasing from (0, 1) onto (π/4,  + ∞);
3. The function r ↦ [(2 − r2)𝒦(r)−2ℰ(r)]/r4 is strictly increasing from (0, 1) onto (π/16,  + ∞).

### Lemma 2.3

The function $r↦φ1(r)={2π1−r2[2E(r)−K(r)]+2r2−1}/r2$ is strictly increasing from (0, 1) onto (3/4, 1).

### Proof

φ1(0+) = ¾,  φ1(1) = 1,
2.1

$φ1′(r)=2πr3γ1(r),$
2.2

where

$γ1(r)=K(r)−3E(r)1−r2+π,$
2.3

γ1(0+) = 0,
2.4

$γ1′(r)=r3(1−r2)3/2(2−r2)K(r)−2E(r)r4.$
2.5

From (2.5) and Lemma 2.2(3) we get

$γ1′(r)>πr316(1−r2)3/2>0.$
2.6

Therefore, Lemma 2.3 follows easily from (2.1), (2.2), (2.4) and (2.6).

### Lemma 2.4

The function $r↦φ2(r)=(2r2+1−r4−1)/r2$ is strictly decreasing from (0, 1) onto (1, 2).

### Proof

It is easy to verify that

φ2(0+) = 2,  φ2(1) = 1,
2.7

$φ2′(r)=2(1−r4−1)r31−r4<0$
2.8

for r ∈ (0, 1).

Therefore, Lemma 2.4 follows easily from (2.7) and (2.8).

### Lemma 2.5

The function $r↦φ3(r)=[2r2K(r)−5E(r)]/1−r2$ is strictly increasing from (0, 1) onto (−5π/2,  + ∞).

### Proof

It is not difficult to verify that

$φ3(0+)=−52π,φ3(1−)=+∞,$
2.9

$φ3′(r)=r(1−r2)3/2[(5−3r2)K(r)−E(r)r2−E(r)].$
2.10

From (2.10) and Lemma 2.2(2) together with the monotonicity of ℰ(r) on (0, 1) we clearly see that

$φ3′(r)>r(1−r2)3/2[(5−3r2)×π4−π2]=3π4r1−r2>0$
2.11

for r ∈ (0, 1).

Therefore, Lemma 2.5 follows from (2.9) and (2.11).

### Lemma 2.6

The function $r↦φ4(r)={2π1−r2[2E(r)−(1+r2)K(r)]+3r2−1}/r2$ is strictly increasing from (0, 1) onto (3/4, 2).

### Proof

Let $ϕ1(r)=2π1−r2[2E(r)−(1+r2)K(r)]+3r2−1$, ϕ2(r) = r2. Then simple computations give

ϕ1(0+) = ϕ2(0) = 0,  φ4(r) = ϕ1(r)/ϕ2(r),
2.12

φ4(1) = 2,
2.13

$ϕ1′(r)ϕ2′(r)=3+1π1−r2[E(r)−(1−r2)K(r)r2]+1πφ3(r).$
2.14

It follows from Lemma 2.2(1), Lemma 2.5 and the function $r↦1−r2$ strictly decreasing that $ϕ1′(r)/ϕ2′(r)$ is strictly increasing on (0, 1) and

$φ4(0+)=limr→0+ϕ1′(r)ϕ2′(r)=34.$
2.15

Therefore, Lemma 2.6 follows from Lemma 2.1, (2.12), (2.13) and (2.15) together with the monotonicity of $ϕ1′(r)/ϕ2′(r)$.

### Lemma 2.7

The function $φ5(r)=[3r2+1−r2−1]/r2$ is strictly decreasing from (0, 1) onto (2, 5/2).

### Proof

We clearly see that

$φ5(0+)=52,φ5(1−)=2,$
2.16

$φ5′(r)=−(1−1−r2)2r31−r2<0$
2.17

for r ∈ (0, 1).

Therefore, Lemma 2.7 follows easily from (2.16) and (2.17).

## Main results

### Theorem 3.1

The double inequality

αNQA(ab) + (1 − α)G(ab) < TD[A(ab), G(ab)] < βNQA(ab) + (1 − β)G(ab)
3.1

holds for all ab > 0 with a ≠ b if and only if α ≤ 3/8 and $β≥4/[π(log(1+2)+2)]=0.5546⋯$.

### Proof

Since G(ab), TD(ab) and NQA(ab) are symmetric and homogenous of degree 1, without loss of generality, we assume that a > b > 0 and let r = (a − b)/(ab) ∈ (0, 1). Then (1.1)-(1.3) lead to

$TD[A(a,b),G(a,b)]=2πA(a,b)E(r),$
3.2

$G(a,b)=A(a,b)1−r2,NQA(a,b)=12A(a,b)[1+r2+sinh−1(r)r].$
3.3

It follows from (3.2)-(3.3) that

$T[A(a,b),G(a,b)]−G(a,b)NQA(a,b)−G(a,b)=2πε(r)−1−r212[1+r2+sinh−1(r)r]−1−r2=4πrε(r)−2r1−r2sinh−1(r)+(r1+r2−2r1−r2).$
3.4

Let $f1(r)=4πrε(r)−2r1−r2$, $f2(r)=sinh−1(r)+(r1+r2−2r1−r2)$ and

$f(r)=4πrε(r)−2r1−r2sinh−1(r)+(r1+r2−2r1−r2).$
3.5

f1(0+) = f2(0) = 0,
3.6

$f1′(r)f2′(r)=2π1−r2[2ε(r)−κ(r)]+2r2−12r2+1−r4−1=φ1(r)φ2(r),$
3.7

where φ1(r) and φ2(r) are defined as in Lemmas 2.3 and 2.4.

It follows from Lemmas 2.3-2.4 and (3.7) that $f1′(r)/f2′(r)$ is strictly increasing on (0, 1). Then (3.5), (3.6) and Lemma 2.1 lead to the conclusion that f(r) is strictly increasing.

Moreover,

$limr→0+4πrε(r)−2r1−r2sinh−1(r)+(r1+r2−2r1−r2)=38,$
3.8

$limr→1−4πrε(r)−2r1−r2sinh−1(r)+(r1+r2−2r1−r2)=4π[π(log(1+2)+2)].$
3.9

Therefore, Theorem 3.1 follows easily from (3.4), (3.8) and (3.9) together with the monotonicity of f(r).

### Theorem 3.2

The double inequality

$λNAQ(a,b)+(1−λ)G(a,b)
3.10

holds for all ab > 0 with a ≠ b if and only if λ ≤ 3/10 and μ ≥ 8/[π(π + 2)] = 0.4952 ⋯ .

### Proof

Without loss of generality, we assume that a > b > 0 and let r = (a − b)/(ab) ∈ (0, 1). Then from (1.4) we get

$NAQ(a,b)=12A(a,b)[1+(1+r2)tan−1(r)r].$
3.11

It follows from (3.2), (3.11) and $G(a,b)=A(a,b)1−r2$ that

$TD[A(a,b),G(a,b)]−G(a,b)NAQ(a,b)−G(a,b)2πE(r)−1−r212[1+(1+r2)tan−1(r)r]−1−r2=[4πrE(r)−2r1−r2]/(1+r2)tan−1(r)+(r−2r1−r2)/(1+r2).$
3.12

Let $g1(r)=[4πrE(r)−2r1−r2]/(1+r2)$, $g2(r)=tan−1(r)+(r−2r1−r2)/(1+r2)$ and

$g(r)=[4πrE(r)−2r1−r2]/(1+r2)tan−1(r)+(r−2r1−r2)/(1+r2).$
3.13

g1(0+) = g2(0) = 0,
3.14

$g1′(r)g2′(r)=2π1−r2[2ε(r)−(1+r2)κ(r)]+3r2−13r2+1−r2−1=φ4(r)φ5(r),$
3.15

where φ4(r) and φ5(r) are defined as in Lemmas 2.6 and 2.7.

It follows from Lemmas 2.6-2.7 and (3.15) that $g1′(r)/g2′(r)$ is strictly increasing on (0, 1). Then (3.13), (3.14) and Lemma 2.1 lead to the conclusion that g(r) is strictly increasing.

Moreover,

$limr→0+[4πrε(r)−2r1−r2]/(1+r2)tan−1(r)+(r−2r1−r2)/(1+r2)=310,$
3.16

$limr→1−[4πrε(r)−2r1−r2]/(1+r2)tan−1(r)+(r−2r1−r2)/(1+r2)=8π(π+2).$
3.17

Therefore, Theorem 3.2 follows from (3.12), (3.16) and (3.17) together with the monotonicity of g(r).

From Theorems 3.1-3.2 we get the following Corollary 3.3 immediately.

### Corollary 3.3

Let α = 3/8, $β=4/[π(log(1+2)+2)]=0.5546⋯$, λ = 3/10 and μ = 8/[π(π + 2)] = 0.4952 ⋯ . Then the double inequalities

$14πα[1+r2+sinh−1(r)r]+12π(1−α)1−r2

hold for all r ∈ (0, 1).

## Results and discussion

In this paper, we provide the sharp bounds for the Toader-type mean in terms of the convex combination of geometric and Neuman means. As applications, we find new bounds for the complete elliptic integral of the second kind.

## Conclusion

In the article, we present the optimal convex combination bounds of the geometric and Neuman means for the Toader-type mean, and give several new upper and lower bounds for the complete elliptic integral of the second kind. The given results are the improvements of some previously known results.

## Acknowledgements

This research was supported by the Natural Science Foundation of Zhejiang Province under Grant LY13A010004 and the Natural Science Foundation of Zhejiang Broadcast and TV University under Grant XKT-15G17.

## Footnotes

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 final manuscript.

Publisher’s Note

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## Contributor Information

Yue-Ying Yang, moc.361@zh8001yyy.

Wei-Mao Qian, umoc.621@779166mwq.

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