# Optimal Bounds for Neuman-Sándor Mean in Terms of the Convex Combinations of Harmonic, Geometric, Quadratic, and Contraharmonic Means

Authors: Tie-Hong Zhao; Yu-Ming Chu; Bao-Yu Liu
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We present the best possible lower and upper bounds for the Neuman-Sándor mean in terms of the convex combinations of either the harmonic and quadratic means or the geometric and quadratic means or the harmonic and contraharmonic means.


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## Body

## 1. Introduction

Fora,b>0 with a≠b, the Neuman-Sándor mean M(a,b) [1] is defined by
(1.1)M(a,b)=a-b2sinh-1[(a-b)/(a+b)],
where sinh-1(x)=log(x+1+x2) is the inverse hyperbolic sine function.Recently, the theory of bivariate means have been the subject of intensive research [2–17]. In particular, many remarkable inequalities for the Neuman-Sándor mean M(a,b) can be found in the literature [1, 18–20].LetH(a,b)=2ab/(a+b), G(a,b)=ab, L(a,b)=(b-a)/(logb-loga), P(a,b)=(a-b)/(4arctana/b-π), A(a,b)=(a+b)/2, T(a,b)=(a-b)/[2arctan((a-b)/(a+b))], Q(a,b)=(a2+b2)/2 and C(a,b)=(a2+b2)/(a+b) be the harmonic, geometric, logarithmic, first Seiffert, arithmetic, second Seiffert, quadratic, and contraharmonic means of a and b, respectively. Then it is well known that the inequalities
(1.2)H(a,b)<G(a,b)<L(a,b)<P(a,b)<A(a,b)<M(a,b)<T(a,b)<Q(a,b)<C(a,b)
hold for all a,b>0 with a≠b.In [1, 18], Neuman and Sándor proved that the double inequalities
(1.3)A(a,b)<M(a,b)<T(a,b),P(a,b)M(a,b)<A2(a,b),A(a,b)T(a,b)<M2(a,b)<A2(a,b)+T2(a,b)2
hold for all a,b>0 with a≠b.Let0<a, b<1/2 with a≠b, a′=1-a and b′=1-b. Then the following Ky Fan inequalities
(1.4)G(a,b)G(a′,b′)<L(a,b)L(a′,b′)<P(a,b)P(a′,b′)<A(a,b)A(a′,b′)<M(a,b)M(a′,b′)<T(a,b)T(a′,b′)
were presented in [1].The double inequalityLp0(a,b)<M(a,b)<L2(a,b) for all a,b>0 with a≠b was established by Li et al. in [19], where Lp(a,b)=[(bp+1-ap+1)/((p+1)(b-a))]1/p (p≠-1,0), L0(a,b)=1/e(bb/aa)1/(b-a) and L-1(a,b)=(b-a)/(logb-loga) is the pth generalized logarithmic mean of a and b, and p0=1.843… is the unique solution of the equation (p+1)1/p=2log(1+2).Neuman [20] proved that the double inequalities
(1.5)αQ(a,b)+(1-α)A(a,b)<M(a,b)<βQ(a,b)+(1-β)A(a,b),λQ(a,b)+(1-λ)A(a,b)<M(a,b)<μQ(a,b)+(1-μ)A(a,b)
hold for all a,b>0 with a≠b if and only if α≤1-log(2+1)/[(2-1)log(2+1)]=0.3249…, β≥1/3, λ≤1-log(2+1)/log(2+1)=0.1345… and μ≥1/6.The main purpose of this paper is to find the least valuesα1, α2, α3, and the greatest values β1, β2, β3, such that the double inequalities
(1.6)α1H(a,b)+(1-α1)Q(a,b)<M(a,b)<β1H(a,b)+(1-β1)Q(a,b),α2G(a,b)+(1-α2)Q(a,b)<M(a,b)<β2G(a,b)+(1-β2)Q(a,b),α3H(a,b)+(1-α3)C(a,b)<M(a,b)<β3H(a,b)+(1-β3)C(a,b)
hold true for all a,b>0 with a≠b.Our main results are presented in Theorems1.1–1.3.Theorem 1.1.
The double inequality(1.7)α1H(a,b)+(1-α1)Q(a,b)<M(a,b)<β1H(a,b)+(1-β1)Q(a,b)
holds for all a,b>0 with a≠b if and only if α1≥2/9=0.2222… and β1≤1-1/[2log(1+2)]=0.1977….Theorem 1.2.
The double inequality(1.8)α2G(a,b)+(1-α2)Q(a,b)<M(a,b)<β2G(a,b)+(1-β2)Q(a,b)
holds for all a,b>0 with a≠b if and only if α2≥1/3=0.3333… and β2≤1-1/[2log(1+2)]=0.1977….Theorem 1.3.
The double inequality(1.9)α3H(a,b)+(1-α3)C(a,b)<M(a,b)<β3H(a,b)+(1-β3)C(a,b)
holds for all a,b>0 with a≠b if and only if α3≥1-1/[2log(1+2)]=0.4327… and β3≤5/12=0.4166….


## 2. Lemmas

In order to prove our main results we need two Lemmas, which we present in this section.Lemma 2.1 (see [21, Lemma 1.1]).
Suppose that the power seriesf(x)=∑n=0∞anxn and g(x)=∑n=0∞bnxn have the radius of convergence r>0 and bn>0 for all n∈{0,1,2,…}. Let h(x)=f(x)/g(x), then the following statements are true.(1)
If the sequence{an/bn}n=0∞ is (strictly) increasing (decreasing), then h(x) is also (strictly) increasing (decreasing) on (0,r).(2)
If the sequence{an/bn} is (strictly) increasing (decreasing) for 0<n≤n0 and (strictly) decreasing (increasing) for n>n0, then there exists x0∈(0,r) such that h(x) is (strictly) increasing (decreasing) on (0,x0) and (strictly) decreasing (increasing) on (x0,r).Lemma 2.2.
Letp∈(0,1), λ0=1-1/[2log(1+2)]=0.1977… and
(2.1)fp(x)=sinh-1(x)-x1+x2-p(1+x2-1-x2).
Then f1/3(x)<0 and fλ0(x)>0 for all x∈(0,1).Proof.
From (2.1), one has
(2.2)fp(0)=0,(2.3)fp(1)=log(1+2)-12(1-p),(2.4)fp′(x)=gp(x)1-x4(1+x2+p(1-x2-1+x2))2,
where
(2.5)gp(x)=1-x2(1+x2+p(1-x2-1+x2))2-1-x2-p(1+x2-1-x2).
We divide the proof into two cases.
Case
1. (p=1/3). Then (2.5) leads to
(2.6)g1/3(0)=0,g1/3(1)=-23<0,(2.7)g1/3′(x)=x31-x4h1/3(x),
where
(2.8)h1/3(x)=149(1+x2+1-x2)-(1+x2+1-x2)-1-x23.
We clearly see that the function 1+x2+1-x2 is strictly decreasing in (0,1). Then from (2.8), we get
(2.9)h1/3(x)<h1/3(1)=-229<0
for x∈(0,1).
Therefore,f1/3(x)<0 for all x∈(0,1) follows easily from (2.2), (2.4), (2.6), (2.7), and (2.9).
Case
2. (p=λ0). Then (2.3) and (2.5) yield
(2.10)fλ0(1)=gλ0(0)=0,gλ0(1)=-2λ0<0,(2.11)gλ0′(x)=x1-x4hλ0(x),
where
(2.12)hλ0(x)=[(2-3λ0-2λ02)-(3-6λ0)x2]1+x2-[(3λ0-2λ02)+(6λ0-6λ02)x2]1-x2.
We divide the discussion of this case into two subcases, and all computations are carried out using MATHEMATICA software.
Subcase A.
x
∈
(
0.9,1
). Then from (2.12) and the fact that
(2.13)(2-3λ0-2λ02)-(3-6λ0)x2<(2-3λ0-2λ02)-(3-6λ0)×(0.9)2=-0.1404⋯<0,
we know that
(2.14)hλ0(x)<0
forx∈(0.9,1).
Subcase B.
x
∈
(
0,0.9
]. Then from (2.12), one has
(2.15)hλ0(0)=0.8137⋯>0,hλ0(0.9)=-0.7494⋯<0,hλ0′(x)=x1-x4μ(x),
where
(2.16)μ(x)=[(18λ0-18λ02)x2-(9λ0-10λ02)]1+x2-[(9-18λ0)x2+(4-9λ0+2λ02)]1-x2.
We conclude that
(2.17)μ(x)<0
for all x∈(0,0.9]. Indeed, if x∈(0,1/2), then (2.17) follows from (2.16) and the inequality
(2.18)(18λ0-18λ02)x2-(9λ0-10λ02)<5.5λ02-4.5λ0=-0.6747⋯<0.
If x∈[1/2,0.9], then (2.17) follows from (2.16) and the inequalities
(2.19)(18λ0-18λ02)x2-(9λ0-10λ02)≤(18λ0-18λ02)×(0.9)2-(9λ0-10λ02)=5.58λ0-4.58λ02=0.9242…,(9-18λ0)x2+(4-9λ0+2λ02)≥14(9-18λ0)+(4-9λ0+2λ02)=6.25-13.5λ0+2λ02=3.6589…,[(18λ0-18λ02)x2-(9λ0-10λ02)]1+x2-[(9-18λ0)x2+(4-9λ0+2λ02)]1-x2≤(5.58λ0-4.58λ02)1+(0.9)2-(6.25-13.5λ0+2λ02)1-(0.9)2=-0.3514⋯<0.
From (2.15) together with (2.17) we clearly see that there exists x0∈(0,0.9) such that hλ0(x)>0 for x∈[0,x0) and hλ0(x)<0 for (x0,0.9].
Subcases A and B lead to the conclusion thathλ0(x)>0 for x∈[0,x0) and hλ0(x)<0 for x∈(x0,1). Thus from (2.11), we know that gλ0(x) is strictly increasing in (0,x0] and strictly decreasing in [x0,1).
It follows from (2.4) and (2.10) together with the piecewise monotonicity of gλ0(x) that there exists x1∈(0,1) such that fλ0(x) is strictly increasing in [0,x1) and strictly decreasing in [x1,1).
Therefore,fλ0(x)>0 for x∈(0,1) follows from (2.2) and (2.10) together with the piecewise monotonicity of fλ0(x).


## 3. Proof of Theorems1.1–1.3

Proof of Theorem1.1.
SinceH(a,b), M(a,b) and Q(a,b) are symmetric and homogeneous of degree 1. Hence, without loss of generality, we assume that a>b. Let x=(a-b)/(a+b) and t=sinh-1(x). Then x∈(0,1), t∈(0,log(1+2)), M(a,b)/A(a,b)=x/sinh-1(x)=sinh(t)/t, H(a,b)/A(a,b)=1-x2=1-sinh2(t)=[3-cosh(2t)]/2, Q(a,b)/A(a,b)=1+x2=cosh(t) and
(3.1)Q(a,b)-M(a,b)Q(a,b)-H(a,b)=1+x2sinh-1(x)-x[1+x2-(1-x2)]sinh-1(x)=tcosh(t)-sinh(t)t[(1/2)cosh(2t)+cosh(t)-(3/2)]:=φ(t).
Making use of power seriessinh(t)=∑n=0∞‍t2n+1/(2n+1)! and cosh(t)=∑n=0∞‍t2n/(2n)!, we can express (3.1) as follows:
(3.2)φ(t)=∑n=1∞[2n/((2n+1)(2n)!)]t2n+1∑n=1∞[(22n-1+1)/(2n)!]t2n+1.
Let an=2n/((2n+1)(2n)!) and bn=(22n-1+1)/(2n)! Then an/bn=2n/[(2n+1)(22n-1+1)]. Moreover, by a simple calculation, we see that
(3.3)an+1bn+1-anbn=2+(2-18n-12n2)22n-1(2n+1)(2n+3)(22n-1+1)(22n+1+1)<0
for n≥1.
Equations (3.1) and (3.2) together with inequality (3.3) and Lemma 2.1 lead to the conclusion that φ(t) is strictly decreasing in (0,log(1+2)). This in turn implies that
(3.4)limt→0+φ(t)=29,limt→log(1+2)φ(t)=1-12log(1+2).
Therefore, Theorem1.1 follows from (3.1) and (3.4) together with the monotonicity of φ(t).Proof of Theorem1.2.
SinceG(a,b), M(a,b) and Q(a,b) are symmetric and homogeneous of degree 1. Hence, without loss of generality, we assume that a>b. Let x=(a-b)/(a+b), p∈(0,1) and λ0=1-1/[2log(1+2)]. Then making use of G(a,b)/A(a,b)=1-x2 gives
(3.5)Q(a,b)-M(a,b)Q(a,b)-G(a,b)=1+x2sinh-1(x)-x(1+x2-1-x2)sinh-1(x).
Moreover, we obtain
(3.6)limx→0+1+x2sinh-1(x)-x(1+x2-1-x2)sinh-1(x)=13,(3.7)limx→1-1+x2sinh-1(x)-x(1+x2-1-x2)sinh-1(x)=1-12log(1+2)=λ0.
We take the difference between the additive convex combination ofG(a,b), Q(a,b), and M(a,b) as follows:
(3.8)pG(a,b)+(1-p)Q(a,b)-M(a,b)=A(a,b)[p1-x2+(1-p)1+x2-xsinh-1(x)]=A(a,b)[p1-x2+(1-p)1+x2]sinh-1(x)fp(x),
where fp(x) is defined as in Lemma 2.2.
Therefore,(1/3)G(a,b)+(2/3)Q(a,b)<M(a,b)<λ0G(a,b)+(1-λ0)Q(a,b) for all a,b>0 with a≠b follows from (3.8) and Lemma 2.2. This conjunction with the following statement gives the asserted result.  (i)
Ifp<1/3, then (3.5) and (3.6) imply that there exists 0<δ1<1 such that M(a,b)<pG(a,b)+(1-p)Q(a,b) for all a,b>0 with (a-b)/(a+b)∈(0,δ1).(ii)
Ifp>λ0, then (3.5) and (3.7) imply that there exists 0<δ2<1 such that M(a,b)>pG(a,b)+(1-p)Q(a,b) for all a,b>0 with (a-b)/(a+b)∈(1-δ2,1).Proof of Theorem1.3.
We will follow, to some extent, lines in the proof of Theorem1.1. First we rearrange terms of (1.9) to obtain
(3.9)β3<C(a,b)-M(a,b)C(a,b)-H(a,b)<α3.
Use of C(a,b)/A(a,b)=1+x2 followed by a substitution x=sinh(t) gives
(3.10)β3<ϕ(t)<α3,
where
(3.11)ϕ(t)=t[cosh(2t)+1]-2sinh(t)2t[cosh(2t)-1],|t|<log(1+2).
Since the functionϕ(t) is an even function, it suffices to investigate its behavior on the interval (0,log(1+2)).
Using power series ofsinh(t) and cosh(t), then (3.11) can be rewritten as
(3.12)ϕ(t)=∑n=1∞[22n/(2n)!-2/(2n+1)!]t2n+1∑n=1∞[22n+1/(2n)!]t2n+1.
Letcn=22n/(2n)!-2/(2n+1)! and dn=22n+1/(2n)!. Then
(3.13)cndn=12-1(2n+1)22n.
It follows from (3.13) that the sequence {cn/dn} is strictly increasing for n≥1.
Equations (3.12) and (3.13) together with Lemma 2.1 and the monotonicity of {cn/dn} lead to the conclusion that ϕ(t) is strictly increasing in (0,log(1+2)). Moreover,
(3.14)limt→0+ϕ(t)=c1d1=512,limt→log(1+2)ϕ(t)=1-12log(1+2).
Making use of (3.14) and (3.10) together with the monotonicity of ϕ(t) gives the asserted result.



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