# Optimal Bounds for Neuman Mean Using Arithmetic and Centroidal Means

Authors: Ying-Qing Song; Wei-Mao Qian; Yu-Ming Chu
Journal: Journal of Function Spaces
 (2016)

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## Abstract

We present the best possible parametersα1,α2,β1,β2∈R and α3,β3∈(1/2,1) such that the double inequalities α1A(a,b)+(1-α1)C(a,b)<NQA(a,b)<β1A(a,b)+(1-β1)C(a,b),Aα2(a,b)C1-α2(a,b)<NQA(a,b)<Aβ2(a,b)C1-β2(a,b), and C[α3a+(1-α3)b,α3b+(1-α3)a]<NQA(a,b)<C[β3a+(1-β3)b,β3b+(1-β3)a] hold for all a,b>0 with a≠b and give several sharp inequalities involving the hyperbolic and inverse hyperbolic functions. Here, N(a,b), A(a,b), Q(a,b), and C(a,b) are, respectively, the Neuman, arithmetic, quadratic, and centroidal means of a and b, and NQA(a,b)=N[Q(a,b),A(a,b)].


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## Body

## 1. Introduction

Leta,b>0 with a≠b. Then, the Schwab-Borchardt mean SB(a,b) [1–7] of a and b is given by (1)
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,where cosh-1(x)=log(x+x2-1) is the inverse hyperbolic cosine functions.It is well known that the Schwab-Borchardt meanSB(a,b) is strictly increasing in both a and b, nonsymmetric, and homogeneous of degree 1 with respect to a and b. Many symmetric bivariate means are special cases of the Schwab-Borchardt mean. For example, P(a,b)=(a-b)/[2arcsin((a-b)/(a+b))]=SB[G(a,b),A(a,b)] is the first Seiffert mean, T(a,b)=(a-b)/[2arctan((a-b)/(a+b))]=SB[A(a,b),Q(a,b)] is the second Seiffert mean, M(a,b)=(a-b)/[2sinh-1((a-b)/(a+b))]=SB[Q(a,b),A(a,b)] is the Neuman-Sándor mean, and L(a,b)=(a-b)/[2tanh-1((a-b)/(a+b))]=SB[A(a,b),G(a,b)] is the logarithmic mean, where sinh-1(x)=log(x+1+x2) is the inverse hyperbolic sine function, tanh-1(x)=log((1+x)/(1-x))/2 is the inverse hyperbolic tangent function, G(a,b)=ab, A(a,b)=(a+b)/2, and Q(a,b)=(a2+b2)/2 are, respectively, the geometric, arithmetic, and quadratic means of a and b, and the inequalities(2)Ha,b<Ga,b<La,b<Pa,b<Aa,b<Ma,b<Ta,b<Qa,bhold for all a,b>0 with a≠b (see [1]), where H(a,b)=2ab/(a+b) is the harmonic mean of a and b.Recently, the Schwab-Borchardt mean and its generated means have been the subject intensive research. In particular, many remarkable inequalities for these means can be found in the literature [1–5, 8–16].Liu and Meng [17] proved that the double inequality(3)αCa,b+1-αHa,b<Ta,b<βCa,b+1-βHa,bholds for all a,b>0 with a≠b if and only if α≤3/π and β≥1, where(4)Ca,b=2a2+ab+b23a+bis the centroidal mean of a and b.Very recently, Neuman [18] introduced the Neuman mean (5)Na,b=12a+b2SBa,b,presented the explicit formula for NQA(a,b) as (6)NQAa,b≜NQa,b,Aa,b=12Aa,b1+v2+sinh-1vv,and proved that the double inequality(7)Ma,b<NQAa,b<Ta,bholds for all a,b>0 with a≠b, where v=(a-b)/(a+b).From (2), (3), and (7), we clearly see that the double inequality(8)Aa,b<NQAa,b<Ca,bholds for all a,b>0 with a≠b.LetJ(x)=C[xa+(1-x)b,xb+(1-x)a]. Then, it is not difficult to verify that the function J(x) is continuous and strictly increasing on [1/2,1] for fixed a,b>0 with a≠b. Note that(9)J12=Aa,b<NQAa,b<Ca,b=J1for all a,b>0 with a≠b.Motivated by inequalities (8) and (9), it is natural to ask what are best possible parameters α1,α2,β1,β2∈R and α3,β3∈(1/2,1) such that the double inequalities (10)α1Aa,b+1-α1Ca,b<NQAa,b<β1Aa,b+1-β1Ca,b,Aα2a,bC1-α2a,b<NQAa,b<Aβ2a,bC1-β2a,b,Cα3a+1-α3b,α3b+1-α3a<NQAa,b<Cβ3a+1-β3b,β3b+1-β3ahold for all a,b>0 with a≠b? The main purpose of this paper is to answer these questions.


## 2. Lemmas

In order to prove our main results, we need several lemmas, which we present in this section.Lemma 1 (see [19–21]).
Fora,b∈R, let f,g:[a,b]→R be continuous on [a,b] and be differentiable on (a,b), and let g′(x)≠0 on (a,b). If f′(x)/g′(x) is increasing (decreasing) on (a,b), then so are (11)fx-fagx-ga,fx-fbgx-gb.
Iff′(x)/g′(x) is strictly monotone, then the monotonicity in the conclusion is also strict.Lemma 2 (see [22–24]).
Letf(x)=∑n=0∞anxn and g(x)=∑n=0∞bnxn be two real power series converging on R with bn>0 for all n∈N. Then, the following statements are true:(1)
If the nonconstant sequence{an/bn} is increasing (decreasing) for all n∈N, then the function x↦f(x)/g(x) is strictly increasing (decreasing) on (0,∞).(2)
If there existsm≥1 such that the nonconstant sequence {an/bn} is increasing (decreasing) for 0≤n≤m and decreasing (increasing) for n≥m, then there exists x0∈(0,∞) such that the function x↦f(x)/g(x) is strictly increasing (decreasing) on (0,x0) and strictly decreasing (increasing) on (x0,∞).Lemma 3.
The function(12)fx=6x+6xcosh2x+1/2sinh⁡4x-7sinh⁡2x-4x+4xcosh2x+sinh⁡4x-2sinh⁡2xis strictly increasing from (0,log(1+2)) onto (1/2,[3log(1+2)-2]/[log(1+2)+2]).Proof.
Making use of the power series formulassinh(x)=∑n=0∞x2n+1/(2n+1)! and cosh(x)=∑n=0∞x2n/(2n)!, we have(13)fx=6x+6x∑n=0∞22n/2n!x2n+1/2∑n=0∞42n+1/2n+1!x2n+1-7∑n=0∞22n+1/2n+1!x2n+1-4x+4x∑n=0∞22n/2n!x2n+∑n=0∞42n+1/2n+1!x2n+1-2∑n=0∞22n+1/2n+1!x2n+1=∑n=0∞22n+422n+1+3n+1/2n+3!x2n∑n=0∞22n+522n+1+n+1/2n+3!x2n.Let(14)cn=∑n=0∞22n+422n+1+3n+12n+3!,dn=∑n=0∞22n+522n+1+n+12n+3!.Then, we clearly see that(15)dn>0for all n≥0 and(16)cn+1dn+1-cndn=1+1-3n22n+122n+1+n+122n+3+n+2.Let(17)en=1+1-3n22n+1.Then, simple computations lead to(18)e0=3>0,en<1-22n+2<0for all n≥1.
It follows from (16)–(18) that the nonconstant sequence {cn/dn} is increasing for 0≤n≤1 and decreasing for n≥1. Then, from (13)–(15) and Lemma 2(2) we know that there exists x0∈(0,∞) such that f(x) is strictly increasing on (0,x0) and strictly decreasing on (x0,∞).
Differentiatingf(x) gives (19)f′x=6-8cosh⁡2x+12xsinh2x+2cosh⁡4x-4x+4xcosh2x+sinh⁡4x-2sinh⁡2x-6x+6xcosh2x+1/2sinh⁡4x-7sinh⁡2x-4+8xsinh2x+4cosh⁡4x-4x+4xcosh2x+sinh⁡4x-2sinh⁡2x2,f′log⁡1+2=6-24+242log⁡1+2+34-4log⁡1+2+12log⁡1+2+122-42-6log⁡1+2+18log⁡1+2+62-142-4+162log1+2+68-4log⁡1+2+12log⁡1+2+122-422=102-12log⁡1+2-32log21+22+log⁡1+22.Note that(20)102-12log⁡1+2-32log21+2=0.2698⋯>0,(21)f0+=c0d0=12,flog⁡1+2=3log⁡1+2-2log⁡1+2+2.
It follows from (19) and (20) together with the piecewise monotonicity of f(x) that log(1+2)<x0 and f(x) is strictly increasing on (0,log(1+2)).
Therefore, Lemma3 follows from (21) and the monotonicity of f(x) on the interval (0,log(1+2)).Lemma 4.
The function(22)gx=sinh⁡x2-cosh⁡x-xsinh3xis strictly increasing from (0,log(1+2)) onto (-1/3,2-2-log(1+2)).Proof.
Making use of the power series formulassinh(x)=∑n=0∞x2n+1/(2n+1)! and cosh(x)=∑n=0∞x2n/(2n+)! together with the identity 4sinh3(x)=sinh(3x)-3sinh(x), one has(23)gx=2sinh⁡x-1/2sinh⁡2x-x1/4sinh⁡3x-3sinh⁡x=2∑n=0∞1/2n+1!x2n+1-1/2∑n=0∞22n+1/2n+1!x2n+1-x1/4∑n=0∞32n+1/2n+1!x2n+1-3∑n=0∞1/2n+1!x2n+1=∑n=0∞2-22n+2/2n+3!x2n∑n=0∞32n+3-3/4×2n+3!x2n.Let(24)un=2-22n+22n+3!,vn=32n+3-34×2n+3!.Then,(25)un+1vn+1-unvn=810×22n-832n+1+22n+132n+4-132n+2-1>0,vn>0,for all n≥0. Note that(26)g0+=u0v0=-13,glog⁡1+2=2-2-log⁡1+2.
Therefore, Lemma4 follows easily from (23)–(26) and Lemma 2(1).


## 3. Main Results

Theorem 5.
The double inequality(27)α1Aa,b+1-α1Ca,b<NQAa,b<β1Aa,b+1-β1Ca,bholds for all a,b>0 with a≠b if and only if α1≥4-3[2+log(1+2)]/2=0.5566⋯ and β1≤1/2.Proof.
Since all the meansA(a,b), NQA(a,b), and C(a,b) are symmetric and homogeneous of degree 1, without loss of generality, we assume that a>b. Let v=(a-b)/(a+b)∈(0,1). Then, (4) leads to(28)Ca,b=1+13v2Aa,b.
From (6) and (28), we clearly see that inequality (27) is equivalent to(29)β1<2v2-31+v2+sinh-1v/v+62v2<α1.
Letx=sinh-1(v). Then, x∈(0,log(1+2)) and(30)2v2-31+v2+sinh-1v/v+62v2=sinh⁡3x-3sinh⁡2x+9sinh⁡x-6xsinh⁡3x-3sinh⁡x=1+3sinh⁡x2-cosh⁡x-x2sinh3x.
Therefore, Theorem5 follows easily from (29) and (30) together with Lemma 4.Theorem 6.
The double inequality(31)Aα2a,bC1-α2a,b<NQAa,b<Aβ2a,bC1-β2a,bholds for all a,b>0 with a≠b if and only if α2≥1-[log(2+log(1+2))-log2]/[2log2-log3]=0.5208⋯ and β2≤1/2.Proof.
Without loss of generality, we assume thata>b. Let v=(a-b)/(a+b)∈(0,1). Then, from (6) and (28) we clearly see that inequality (31) is equivalent to(32)β2<log⁡v2+3-log⁡1+v2+sinh-1v/v-log⁡3+log⁡2log⁡v2+3-log⁡3<α2.
Letx=sinh-1(v). Then, x∈(0,log(1+2)), and(33)log⁡v2+3-log⁡1+v2+sinh-1v/v-log⁡3+log⁡2log⁡v2+3-log⁡3=log⁡3+sinh2x-log⁡x+sinh⁡xcosh⁡x+logsinh⁡x-log3+log2log⁡3+sinh2x-log⁡3.Let(34)Gx=log⁡3+sinh2x-log⁡x+sinh⁡xcosh⁡x+log⁡sinh⁡x-log⁡3+log⁡2log⁡3+sinh2x-log⁡3,(35)g1x=log⁡3+sinh2x-log⁡x+sinh⁡xcosh⁡x+log⁡sinh⁡x-log⁡3+log⁡2,(36)g2x=log⁡3+sinh2x-log⁡3.Then, simple computations lead to(37)g10+=g20+=0,g1′xg2′x=6x+6xcosh2x+1/2sinh⁡4x-7sinh⁡2x-4x+4xcosh2x+sinh⁡4x-2sinh⁡2x.
It follows from (34)–(37) and Lemmas 1 and 3 that the function (38)Gx=g1xg2x=g1x-g10+g2x-g20+is strictly increasing on the interval (0,log(1+2)). Note that(39)G0+=g1′0+g2′0+=c0d0=12,Glog⁡1+2=1-log⁡2+log⁡1+2-log⁡22log⁡2-log⁡3,where cn and dn are defined by (14).
Therefore, Theorem6 follows easily from (32)–(34) and (39) together with the monotonicity of G(x) on the interval (0,log(1+2)).Theorem 7.
Letα3,β3∈(1/2,1). Then, the double inequality (40)Cα3a+1-α3b,α3b+1-α3a<NQAa,b<Cβ3a+1-β3b,β3b+1-β3aholds for all a,b>0 with a≠b if and only if α3≤1/2+6[2+log(1+2)]-12/4=0.8329⋯ and β3≥1/2+2/4=0.8535⋯.Proof.
Without loss of generality, we assume thata>b. Let v=(a-b)/(a+b)∈(0,1) and p∈(1/2,1). Then, (4) and (6) lead to(41)Cpa+1-pb,pb+1-pa-NQAa,b=13Aa,b3+2p-12v2-12Aa,b1+v2+sinh-1v/v=Aa,bv2-1+v2-sinh-1v2v3+132p-12v2.
Letx=sinh-1(v). Then, x∈(0,log(1+2)) and(42)v2-1+v2-sinh-1v2v3=sinh⁡x2-cosh⁡x-x2sinh3x=12gx,where g(x) is defined by Lemma 4.
Therefore, Theorem7 follows easily from (41) and (42) together with Lemma 4.Leta>b>0 and t=(loga-logb)/2>0. Then, we clearly see that (43)a-ba+b=tanh⁡t,1+a-ba+b2=cosh1/22tcosh⁡t,and (4) and (6) lead to(44)Ca,b=4cosh2t-13cosh2tAa,b,Cαa+1-αb,αb+1-αa=2α2-α+1cosh⁡2t+1+2α-2α23cosh2tAa,b,NQAa,b=12cosh1/22tcosh⁡t+sinh-1tanh⁡ttanh⁡tAa,b.From (44) and Theorems 5–7, we get Corollary 8 immediately.Corollary 8.
Letα1,α2,β1,β2∈R and α3,β3∈(1/2,1). Then, the double inequalities (45)8-2α13-21-α13cosh2t<cosh1/22tcosh⁡t+sinh-1tanh⁡ttanh⁡t<8-2β13-21-β13cosh2t,243-13cosh2t1-α2<cosh1/22tcosh⁡t+sinh-1tanh⁡ttanh⁡t<243-13cosh2t1-β2,4α32-α3+1cosh⁡2t+21+2α3-2α323cosh2t<cosh1/22tcosh⁡t+sinh-1tanh⁡ttanh⁡t<4β32-β3+1cosh⁡2t+21+2β3-2β323cosh2thold for all t>0 if and only if α1≥4-3[2+log(1+2)]/2, β1≤1/2, α2≥1-[log(2+log(1+2))-log2]/[2log2-log3], β2≤1/2, α3≤1/2+6[2+log(1+2)]-12/4, and β3≥1/2+2/4.Letx=tanh(t)∈(0,1). Then, we clearly see that (46)cosh⁡t=11-x2,cosh1/22t=1+x21-x2,Corollary 9.
Letα1,α2,β1,β2∈R and α3,β3∈(1/2,1). Then, the double inequalities (47)2+21-α13x2-1+x2<sinh-1xx<2+21-β13x2-1+x2,21+x231-α2-1+x2<sinh-1xx<21+x231-β2-1+x2,4α32-α3+131+x2+21+2α3-2α3231-x2-1+x2<sinh-1xx<4β32-β3+131+x2+21+2β3-2β3231-x2-1+x2hold for all x∈(0,1) if and only if α1≥4-3[2+log(1+2)]/2, β1≤1/2, α2≥1-[log(2+log(1+2))-log2]/[2log2-log3], β2≤1/2, α3≤1/2+6[2+log(1+2)]-12/4, and β3≥1/2+2/4.



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*Source: 5131907-2016-02-02.xml*