# Best Possible Inequalities between Generalized Logarithmic Mean and Classical Means

Authors: Yu-Ming Chu; Bo-Yong Long
Journal: Abstract and Applied Analysis
 (2010)

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## Abstract

We answer the question: forα,β,γ∈(0,1) with α+β+γ=1, what are the greatest value p and the least value q, such that the double inequality Lp(a,b)<Aα(a,b)Gβ(a,b)Hγ(a,b)<Lq(a,b) holds for all a,b>0 with a≠b? Here Lp(a,b), A(a,b), G(a,b), and H(a,b) denote the generalized logarithmic, arithmetic, geometric, and harmonic means of two positive numbers a and b, respectively.


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## Body

## 1. Introduction

Forp∈ℝ the generalized logarithmic mean Lp(a,b) of two positive numbers a and b with a≠b is defined by(1.1)Lp(a,b)={[ap+1-bp+1(p+1)(a-b)]1/p,p≠0,p≠-1,1e(bbaa)1/(b-a),p=0,b-alogb-loga,p=-1.It is well known thatLp(a,b) is continuous and strictly increasing with respect to p∈ℝ for fixed a,b>0 with a≠b. Recently, the generalized logarithmic mean has been the subject of intensive research. Many remarkable inequalities and monotonicity results for the generalized logarithmic mean can be found in the literature [1–9]. It might be surprising that the generalized logarithmic mean has applications in physics, economics, and even in meteorology [10–13].LetA(a,b)=(a+b)/2, I(a,b)=(1/e)(bb/aa)1/(b-a), L(a,b)=(b-a)/(logb-loga), G(a,b)=ab, and H(a,b)=2ab/(a+b) be the arithmetic, identric, logarithmic, geometric, and harmonic means of two positive numbers a and b with a≠b, respectively. Then(1.2)min{a,b}<H(a,b)<G(a,b)=L-2(a,b)<L(a,b)=L-1(a,b)<I(a,b)=L0(a,b)<A(a,b)=L1(a,b)<max{a,b}.Forp∈ℝ, the pth power mean Mp(a,b) of two positive numbers a and b with a≠b is defined by(1.3)Mp(a,b)=(ap+bp2)1/p(p≠0),M0(a,b)=(ab)1/2.In [14], Alzer and Janous established the following sharp double inequality (see also [15,  page 350]):(1.4)Mlog2/log3(a,b)<23A(a,b)+13G(a,b)<M2/3(a,b)
for all a,b>0 with a≠b.Forα∈(0,1), Janous [16] found the greatest value p and the least value q such that(1.5)Mp(a,b)<αA(a,b)+(1-α)G(a,b)<Mq(a,b)
for all a,b>0 with a≠b.In [17–19], the authors presented bounds for L(a,b) and I(a,b) in terms of G(a,b) and A(a,b).Theorem 1 A.
For all positive real numbersa and b with a≠b, we have
(1.6)L(a,b)<13A(a,b)+23G(a,b),13G(a,b)+23A(a,b)<I(a,b).The proof of the following Theorem B can be found in [20].Theorem 1 B.
For all positive real numbersa and b with a≠b, we have
(1.7)G(a,b)A(a,b)<L(a,b)I(a,b)<12(L(a,b)+I(a,b))<12(G(a,b)+A(a,b)).The following Theorems C–E were established by Alzer and Qiu in [21].Theorem 1 C.
The inequalities(1.8)αA(a,b)+(1-α)G(a,b)<I(a,b)<βA(a,b)+(1-β)G(a,b)
hold for all positive real numbers a and b with a≠b if and only if α≤2/3 and β≥2/e=0.73575⋯.Theorem 1 D.
Leta and b be real numbers with a≠b. If 0<a,b≤e, then
(1.9)[G(a,b)]A(a,b)<[L(a,b)]I(a,b)<[A(a,b)]G(a,b).
And if a,b≥e, then
(1.10)[A(a,b)]G(a,b)<[I(a,b)]L(a,b)<[G(a,b)]A(a,b).Theorem 1 E.
For all positive real numbersa and b with a≠b, we have
(1.11)Mp(a,b)<12(L(a,b)+I(a,b))
with the best possible parameter p=log2/(1+log2)=0.40938⋯.However, the following problem is still open: forα,β,γ∈(0,1) with α+β+γ=1, what are the greatest value p and the least value q, such that the double inequality(1.12)Lp(a,b)<Aα(a,b)Gβ(a,b)Hγ(a,b)<Lq(a,b)
holds for all a,b>0 with a≠b? The purpose of this paper is to give the solution to this open problem.


## 2. Lemmas

In order to establish our main result, we need four lemmas, which we present in this section.Lemma 2.1.
Ift>1, then
(2.1)t-1logt-[t(t+1)2]1/3>0.Proof.
Ift>1, then it follows from [22, 3.6.16] that(2.2)t-1logt>t+t1/31+t1/3.
It is not difficult to verify that
(2.3)t+t1/31+t1/3>[t(t+1)2]1/3
for t>1.
Therefore, Lemma2.1 follows from inequalities (2.2) and (2.3).Lemma 2.2.
Ift>1, then
(2.4)tt-1logt-16logt-23log1+t2-1>0.Proof.
Letf(t)=(t/(t-1))logt-(1/6)logt-(2/3)log((1+t)/2)-1. Then simple computations lead to
(2.5)limt→1f(t)=0,(2.6)f'(t)=g(t)6t(t-1)2(t+1),
where g(t)=-6t(t+1)logt+t3+9t2-9t-1,(2.7)g(1)=0,(2.8)g'(t)=-6(2t+1)logt+3t2+12t-15,(2.9)g'(1)=0,(2.10)g′′(t)=h(t)t,
where h(t)=-12logt+6t2-6,(2.11)g′′(1)=h(1)=0,(2.12)h'(t)=-12logt+12t-12,(2.13)h'(1)=0,(2.14)h′′(t)=12t(t-1)>0
for t>1.
Therefore, Lemma2.2 follows from (2.5)–(2.7), (2.9)–(2.11), (2.13), and (2.14).Lemma 2.3.
Lett>1 and g(t)=(1-(1/2)λ)(3λ-5)t3λ-2+[(λ-1)(3λ-5)-1]t3λ-3-[(1/2)λ(3λ-5)+1]t3λ-4+[(1/2)λ(3λ-5)+1]t2-[(λ-1)(3λ-5)-1]t-(1-(1/2)λ)(3λ-5). Then(1)
g(t)>0 for λ∈(0,2/3)∪(1,4/3)∪(5/3,2) and(2)
g(t)<0 for λ∈(2/3,1)∪(4/3,5/3).Proof.
Simple computations yield(2.15)limt→1g(t)=0,(2.16)g'(t)=(1-12λ)(3λ-5)(3λ-2)t3λ-3+3(λ-1)[(λ-1)(3λ-5)-1]t3λ-4-(3λ-4)[12λ(3λ-5)+1]t3λ-5+[λ(3λ-5)+2]t-(λ-1)(3λ-5)+1,(2.17)g'(1)=0,(2.18)g′′(t)=3(1-12λ)(λ-1)(3λ-5)(3λ-2)t3λ-4+3(λ-1)(3λ-4)[(λ-1)(3λ-5)-1]t3λ-5-(3λ-4)(3λ-5)[12λ(3λ-5)+1]t3λ-6+λ(3λ-5)+2,(2.19)g′′(1)=0,(2.20)g′′′(t)=t3λ-7h(t),
where h(t)=3(1-(1/2)λ)(λ-1)(3λ-5)(3λ-4)(3λ-2)t2+3(λ-1)(3λ-4)(3λ-5)[(λ-1)(3λ-5)-1]t-3(λ-2)(3λ-4)(3λ-5)[(1/2)λ(3λ-5)+1],
(2.21)h(1)=0,(2.22)h'(t)=81(2-λ)(λ-23)(λ-1)(λ-43)(λ-53)(t-1).
(1) If λ∈(0,2/3)∪(1,4/3)∪(5/3,2), then (2.22) implies
(2.23)h'(t)>0
for t>1.
Therefore, Lemma2.3(1) follows from (2.15), (2.17), (2.19), (2.20), (2.21), and (2.23).
(2) If λ∈(2/3,1)∪(4/3,5/3), then (2.22) leads to
(2.24)h'(t)<0
for t>1.
Therefore, Lemma2.3(2) follows from (2.15), (2.17), (2.19), (2.20), (2.21), and (2.24).Lemma 2.4.
Lett>1 and G(t)=(1-2/λ)t3-2/λ-(3-2/λ)t2-2/λ+(3-2/λ)t+(2/λ)-1. Then(1)
G(t)>0 for λ∈(2/3,1) and(2)
G(t)<0 for λ∈(0,2/3)∪(1,2).Proof.
From the expression ofG(t), we clearly see that
(2.25)limt→1G(t)=0,(2.26)G'(t)=(1-2λ)(3-2λ)t2-2/λ-2(3-2λ)(1-1λ)t1-2/λ+3-2λ,(2.27)G'(1)=0,(2.28)G′′(t)=6λ3(λ-23)(λ-1)(λ-2)t-2/λ(t-1).
From (2.28), we have
(2.29)G′′(t)>0
for λ∈(2/3,1) and t>1, and
(2.30)G′′(t)<0
for λ∈(0,2/3)∪(1,2) and t>1.
Therefore, Lemma2.4(1) follows from (2.25) and (2.27) together with (2.29), and Lemma 2.4(2) follows from (2.25) and (2.27) together with (2.30).


## 3. Main Result

Theorem 3.1.
Leta,b>0 with a≠b and α,β>0 with α+β<1. Then(1)
L6α+3β-5(a,b)=Aα(a,b)Gβ(a,b)H1-α-β(a,b)=L-2/(2α+β)(a,b) for 2α+β∈{2/3,1};(2)
L6α+3β-5(a,b)>Aα(a,b)Gβ(a,b)H1-α-β(a,b)>L-2/(2α+β)(a,b) for 2α+β∈(0,2/3)∪(1,2) and L6α+3β-5(a,b)<Aα(a,b)Gβ(a,b)H1-α-β(a,b)<L-2/(2α+β)(a,b) for 2α+β∈(2/3,1), and the parameters 6α+3β-5 and -2/(2α+β) cannot be improved in either case.Proof.
(1) We divide the proof into two cases. Case 1.
If2α+β=2/3, then simple computations lead to
(3.1)L6α+3β-5(a,b)=L-2/(2α+β)(a,b)=L-3(a,b)=(2a2b2a+b)1/3=21-(2α+β)(ab)1-(2α+β)/2(a+b)2α+β-1=Aα(a,b)Gβ(a,b)H1-α-β(a,b).Case 2.
If2α+β=1, then we clearly see that
(3.2)L6α+3β-5(a,b)=L-2/(2α+β)(a,b)=L-2(a,b)=(ab)1/2=Aα(a,b)Gβ(a,b)H1-α-β(a,b).
(2)  Without loss of generality, we assume that a>b and we put t=a/b>1 and λ=2α+β.
Firstly, we compareAα(a,b)Gβ(a,b)H1-α-β(a,b) with L3λ-5(a,b). We divide the proof into five cases.Case 1.
Ifλ=4/3, then (1.1) leads to
(3.3)L3λ-5(a,b)-Aα(a,b)Gβ(a,b)H1-α-β(a,b)=a-bloga-logb-(a+b2)α(ab)β/2(2aba+b)1-α-β=b[t-1logt-(t(t+1)2)1/3].
Therefore,L3λ-5(a,b)>Aα(a,b)Gβ(a,b)H1-α-β(a,b) follows from (3.3) and Lemma 2.1.Case 2.
Ifλ=5/3, then from (1.1) we have
(3.4)log[L3λ-5(a,b)]-log[Aα(a,b)Gβ(a,b)H1-α-β(a,b)]=tt-1logt-16logt-23log1+t2-1.
From (3.4) and Lemma 2.2, we clearly see that L3λ-5(a,b)>Aα(a,b)Gβ(a,b)H1-α-β(a,b).Case 3.
Ifλ∈(0,2/3)∪(1,4/3)∪(5/3,2), then (1.1) yields
(3.5)log[L3λ-5(a,b)]-log[Aα(a,b)Gβ(a,b)H1-α-β(a,b)]=13λ-5logt3λ-4-1(3λ-4)(t-1)-(λ-1)log1+t2-(1-λ2)logt.
Letf(t)=(1/(3λ-5))log((t3λ-4-1)/(3λ-4)(t-1))-(λ-1)log((1+t)/2)-(1-λ/2)logt. Then simple computations lead to
(3.6)limt→1f(t)=0,(3.7)f'(t)=g(t)(3λ-5)(t3λ-4-1)(t2-1)t,
where g(t)=(1-(1/2)λ)(3λ-5)t3λ-2+[(λ-1)(3λ-5)-1]t3λ-3-[(1/2)λ(3λ-5)+1]t3λ-4+[(1/2)λ(3λ-5)+1]t2-[(λ-1)(3λ-5)-1]t-(1-(1/2)λ)(3λ-5).
We clearly see that
(3.8)(3λ-5)(t3λ-4-1)>0.
Therefore,L3λ-5(a,b)>Aα(a,b)Gβ(a,b)H1-α-β(a,b) follows from (3.5)–(3.8) and Lemma 2.3(1).Case 4.
Ifλ∈(2/3,1), then (3.5)–(3.8) again hold, and from (3.5)–(3.8) and Lemma 2.3(2), we know that L3λ-5(a,b)<Aα(a,b)Gβ(a,b)H1-α-β(a,b).Case 5.
Ifλ∈(4/3,5/3), then (3.5)–(3.7) hold and
(3.9)(3λ-5)(t3λ-4-1)<0.
Therefore,L3λ-5(a,b)>Aα(a,b)Gβ(a,b)H1-α-β(a,b) follows from (3.5)–(3.7) and (3.9) together with Lemma 2.3(2).
Secondly, we compareAα(a,b)Gβ(a,b)H1-α-β(a,b) with L-2/λ(a,b).
From (1.1), we have
(3.10)log[L-2/λ(a,b)]-log[Aα(a,b)Gβ(a,b)H1-α-β(a,b)]=-λ2logt1-λ/2-1(1-λ/2)(t-1)-(λ-1)log1+t2-(1-λ2)logt.
LetF(t)=-λ/2log((t1-(λ/2)-1)/(1-λ/2)(t-1))-(λ-1)log((1+t)/2)-(1-λ/2)logt. Then simple computations yield
(3.11)limt→1F(t)=0,(3.12)F'(t)=λG(t)2t(1-t1-2/λ)(t2-1),
where G(t)=(1-2/λ)t3-2/λ-(3-2/λ)t2-2/λ+(3-2/λ)t+(2/λ)-1.
We clearly see that
(3.13)1-t1-2/λ>0
for λ∈(0,2).
Therefore,Aα(a,b)Gβ(a,b)H1-α-β(a,b)<L-2/λ(a,b) for λ∈(2/3,1) follows from (3.10)–(3.13) and Lemma 2.4(1), and Aα(a,b)Gβ(a,b)H1-α-β(a,b)>L-2/λ(a,b) for λ∈(0,2/3)∪(1,2) follows from (3.10)–(3.13) and Lemma 2.4(2).
At last, we prove that the parameters-2/λ and 3λ-5 cannot be improved in either case.
The following two cases will complete the proof for the optimality of parameter-2/λ.
Case I.
Ifλ∈(0,2/3)∪(1,2), then for any ϵ∈(0,2/λ-1), one has
(3.14)limt→+∞L-2/λ+ϵ(1,t)Aα(1,t)Gβ(1,t)H1-α-β(1,t)=limt→+∞{[2/λ-1-ϵ1-t1+ϵ-2/λ(1-1t)]λ/(2-ϵλ)(1+t2t)1-λtϵλ2/2(2-ϵλ)}=+∞.
Equation (3.14) implies that for any ϵ∈(0,2/λ-1), there exists a sufficiently large T1=T1(ϵ,α,β)>1, such that L-2/λ+ϵ(1,t)>Aα(1,t)Gβ(1,t)H1-α-β(1,t) for t∈(T1,+∞).Case II.
Ifλ∈(2/3,1), then for any ϵ>0, one has
(3.15)limt→+∞Aα(1,t)Gβ(1,t)H1-α-β(1,t)L-2/λ-ϵ(1,t)=limt→+∞{(2t1+t)1-λ[1-t1-2/λ-ϵ(2/λ+ϵ-1)(1-1/t)]λ/(2+ϵλ)tϵλ2/2(2+ϵλ)}=+∞.
Equation (3.15) implies that for any ϵ>0, there exists a sufficiently large T2=T2(ϵ,α,β)>1, such that Aα(1,t)Gβ(1,t)H1-α-β(1,t)>L-2/λ-ϵ(1,t) for t∈(T2,+∞).
The following seven cases will complete the proof for the optimality of parameter3λ-5.
Case A.
Ifλ=4/3, then for any ϵ>0 and x>0, one has
(3.16)[L3λ-5-ϵ(1,1+x)]1+ϵ-[Aα(1,1+x)Gβ(1,1+x)H1-α-β(1,1+x)]1+ϵ=g1(x)(1+x)ϵ-1,
where g1(x)=ϵx(1+x)ϵ-(1+x/2)(1+ϵ)/3(1+x)(1+ϵ)/3[(1+x)ϵ-1].
Upon lettingx→0, the Taylor expansion leads to
(3.17)g1(x)=-124ϵ2(ϵ+1)x3+o(x3).
Equations (3.16) and (3.17) imply that for ϵ>0, there exists a sufficiently small δ1=δ1(ϵ)>0, such that L3λ-5-ϵ(1,1+x)<Aα(1,1+x)Gβ(1,1+x)H1-α-β(1,1+x) for x∈(0,δ1).Case B.
Ifλ=5/3, then for any ϵ∈(0,1) and x>0(x→0), one has
(3.18)[L3λ-5-ϵ(1,1+x)]ϵ-[Aα(1,1+x)Gβ(1,1+x)H1-α-β(1,1+x)]ϵ=(1-ϵ)x(1+x)1-ϵ-1-(1+x)ϵ/6(1+x2)(2/3)ϵ=1(1+x)1-ϵ-1[-124(1-ϵ)ϵ2x3+o(x3)].
Equation (3.18) implies that for any ϵ∈(0,1), there exists a sufficiently small δ2=δ2(ϵ)>0, such that L3λ-5-ϵ(1,1+x)<Aα(1,1+x)Gβ(1,1+x)H1-α-β(1,1+x) for x∈(0,δ2).Case C.
Ifλ∈(0,2/3), then for any ϵ>0 and x>0(x→0), one has
(3.19)[L3λ-5-ϵ(1,1+x)]5+ϵ-3λ-[Aα(1,1+x)Gβ(1,1+x)H1-α-β(1,1+x)]5+ϵ-3λ=(4+ϵ-3λ)x(1+x)4+ϵ-3λ(1+x)4+ϵ-3λ-1-[(1+x)1-λ/2(1+(x2))λ-1]5+ϵ-3λ=1(1+x)4+ϵ-3λ-1[-ϵ24(5+ϵ-3λ)(4+ϵ-3λ)x3+o(x3)].
Equation (3.19) implies that for any ϵ>0, there exists a sufficiently small δ3=δ3(ϵ,α,β)>0, such that L3λ-5-ϵ(1,1+x)<Aα(1,1+x)Gβ(1,1+x)H1-α-β(1,1+x) for x∈(0,δ3).Case D.
Ifλ∈(2/3,1), then for any ϵ∈(0,4-3λ) and x>0(x→0), one has
(3.20)[L3λ-5+ϵ(1,1+x)]5-3λ-ϵ-[Aα(1,1+x)Gβ(1,1+x)H1-α-β(1,1+x)]5-3λ-ϵ=(4-3λ-ϵ)x(1+x)4-3λ-ϵ(1+x)4-3λ-ϵ-1-[(1+x)1-λ/2(1+x/2)1-λ]5-3λ-ϵ=(ϵ/24)(5-3λ-ϵ)(4-3λ-ϵ)x3+o(x3)[(1+x)4-3λ-ϵ-1](1+x/2)(1-λ)(5-3λ-ϵ).
Equation (3.20) implies that for any ϵ∈(0,4-3λ), there exists a sufficiently small δ4=δ4(ϵ,α,β)>0, such that L3λ-5+ϵ(1,1+x)>Aα(1,1+x)Gβ(1,1+x)H1-α-β(1,1+x) for x∈(0,δ4).Case 3 E.
Ifλ∈(1,4/3), then for any ϵ>0 and x>0(x→0), one has
(3.21)[L3λ-5-ϵ(1,1+x)]5+ϵ-3λ-[Aα(1,1+x)Gβ(1,1+x)H1-α-β(1,1+x)]5+ϵ-3λ=(4+ϵ-3λ)x(1+x)4+ϵ-3λ(1+x)4+ϵ-3λ-1-[(1+x)1-λ/2(1+(x2))λ-1]5+ϵ-3λ=1(1+x)4+ϵ-3λ-1[-(ϵ/24)(5+ϵ-3λ)(4+ϵ-3λ)x3+o(x3)].
Equation (3.21) implies that for any ϵ>0, there exists a sufficiently small δ5=δ5(ϵ,α,β)>0, such that L3λ-5+ϵ(1,1+x)<Aα(1,1+x)Gβ(1,1+x)H1-α-β(1,1+x) for x∈(0,δ5).Case F.
Ifλ∈(4/3,5/3), then for any ϵ∈(0,3λ-4) and x>0(x→0), one has
(3.22)[L3λ-5-ϵ(1,1+x)]5+ϵ-3λ-[Aα(1,1+x)Gβ(1,1+x)H1-α-β(1,1+x)]5+ϵ-3λ=(3λ-4-ϵ)x(1+x)3λ-4-ϵ-1-[(1+x)1-λ/2(1+(x2))λ-1]5+ϵ-3λ=1(1+x)3λ-4-ϵ-1[-ϵ24(5+ϵ-3λ)(3λ-4-ϵ)x3+o(x3)].
Equation (3.22) implies that for any ϵ∈(0,3λ-4), there exists a sufficiently small δ6=δ6(ϵ,α,β)>0, such that L3λ-5+ϵ(1,1+x)<Aα(1,1+x)Gβ(1,1+x)H1-α-β(1,1+x) for x∈(0,δ6).Case G.
Ifλ∈(5/3,2), then for any ϵ∈(0,3λ-5) and x>0(x→0), one has
(3.23)[L3λ-5-ϵ(1,1+x)]3λ-5-ϵ-[Aα(1,1+x)Gβ(1,1+x)H1-α-β(1,1+x)]3λ-5-ϵ=(1+x)3λ-4-ϵ-1(3λ-4-ϵ)x-[(1+x)1-λ/2(1+(x2))λ-1]3λ-5-ϵ=1(3λ-4-ϵ)x[-ϵ24(3λ-4-ϵ)(3λ-5-ϵ)x3+o(x3)].
Equation (3.23) implies that for any ϵ∈(0,3λ-5), there exists a sufficiently small δ7=δ7(ϵ,α,β)>0, such that L3λ-5+ϵ(1,1+x)<Aα(1,1+x)Gβ(1,1+x)H1-α-β(1,1+x) for x∈(0,δ7).



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