# Sharp Bounds by the Generalized Logarithmic Mean for the Geometric Weighted Mean of the Geometric and Harmonic Means

Authors: Wei-Mao Qian; Bo-Yong Long
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We present sharp upper and lower generalized logarithmic mean bounds for the geometric weighted mean of the geometric and harmonic means.


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## Body

## 1. Introduction

Forp∈ℝ the generalized logarithmic mean Lp(a,b) of two positive numbers a and b is defined by
(1.1)Lp(a,b)={a,a=b,[ap+1-bp+1(p+1)(a-b)]1/p,p≠0,p≠-1,a≠b,1e(bbaa)1/(b-a),p=0,a≠b,b-alog⁡b-log⁡a,p=-1,a≠b.It is well-known thatLp(a,b) is continuous and strictly increasing with respect to p∈ℝ for fixed a and b with a≠b. In the recent past, the generalized logarithmic mean has been the subject of intensive research. In particular, many remarkable inequalities for Lp can be found in the literature [1–23]. The generalized logarithmic mean has applications in convex function, economics, physics, and even in meteorology [24–27]. In [26] the authors study a variant of Jensen’s functional equation involving Lp, which appear in a heat conduction problem. Let A(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 it is well known that(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}.In [28–30], the authors present bounds for L and I in terms of G and A.Proposition 1.1.
For all positive real numbersa and b with a≠b, one has
(1.3)A1/3(a,b)G2/3(a,b)<L(a,b)<13A(a,b)+23G(a,b),13G(a,b)+23A(a,b)<I(a,b).The proof of the following Proposition1.2 can be found in [31].Proposition 1.2.
For all positive real numbersa and b with a≠b, we have
(1.4)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)).Forr∈ℝ the rth power mean Mr(a,b) of two positive numbers a and b is defined by(1.5)Mr(a,b)={(ar+br2)1/r,r≠0,ab,r=0.The main properties of these means are given in [32]. Several authors discussed the relationship of certain means to Mr. The following sharp bounds for L, I, (IL)1/2, and (I+L)/2 in terms of power means are proved in [31, 33–37].Proposition 1.3.
For all positive real numbersa and b with a≠b one has
(1.6)M0(a,b)<L(a,b)<M1/3(a,b),M2/3(a,b)<I(a,b)<Mlog⁡2(a,b),M0(a,b)<I1/2(a,b)L1/2(a,b)<M1/2(a,b),12[I(a,b)+L(a,b)]<M1/2(a,b).The following three results were established by Alzer and Qiu in [38].Proposition 1.4.
The inequalities(1.7)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
(1.8)α≤23,β≥2e=0.73575⋅⋅⋅.Proposition 1.5.
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).Proposition 1.6.
For all positive real numbersa and b with a≠b, one has
(1.11)Mc(a,b)<12(L(a,b)+I(a,b))
with the best possible parameter c=log2/(1+log2)=0.40938⋯In [39] the authors presented inequalities between the generalized logarithmic mean and the product Aα(a,b)Gβ(a,b)Hγ(a,b) for all a,b>0 with a≠b and α,β>0 with α+β<1.It is the aim of this paper to give a solution to the problem: forα∈(0,1), what are the greatest value p and the least value q, such that the inequality(1.12)Lp(a,b)≤Gα(a,b)H1-α(a,b)≤Lq(a,b)
holds for all a,b>0?


## 2. Main Result

Theorem 2.1.
Forα∈(0,1) and all a,b>0, one has the following:(1)
L3α-5(a,b)=Gα(a,b)H1-α(a,b)=L-(2/α)(a,b) for α=2/3,(2)
L3α-5(a,b)≥Gα(a,b)H1-α(a,b)≥L-(2/α)(a,b) for 0<α<2/3, and L3α-5(a,b)≤Gα(a,b)H1-α(a,b)≤L-(2/α)(a,b) for 2/3<α<1, with equality if and only if a=b, and the parameters 3α-5 and -2/α in each inequality cannot be improved.Proof.
(1) Ifα=2/3 and a=b, then (1.1) implies that L3α-5(a,b)=Gα(a,b)H1-α(a,b)=L-(2/α)(a,b)=a.
Ifα=2/3 and a≠b, then (1.1) leads to
(2.1)L3α-5(a,b)=L-2/α(a,b)=L-3(a,b)=[a-2-b-22(b-a)]-1/3=(ab)1/3(2aba+b)1/3=G2/3(a,b)H1/3(a,b)=Gα(a,b)H1-α(a,b).
(2) Ifa=b, then from (1.1) we clearly see that L3α-5(a,b)=Gα(a,b)H1-α(a,b)=L-(2/α)(a,b)=a for any α∈(0,1).
Ifa≠b, without loss of generality, we assume a>b. Let a/b=t>1 and
(2.2)f(t)=log⁡L3α-5(a,b)-log⁡[Gα(a,b)H1-α(a,b)].
Then (1.1) and simple computations yield
(2.3)f(t)=13α-5log⁡t3α-4-1(3α-4)(t-1)-α2log⁡t-(1-α)log⁡2t1+t,lim⁡t→1+f(t)=0,(2.4)f′(t)=-t4-3αt(t2-1)(t4-3α-1)g(t),
where g(t)=(2-α/2)t3α-2-((2-α)(2-3α)/5-3α)t3α-3+((1-α)(2-3α)/2(5-3α))t3α-4-((1-α)(2-3α)/2(5-3α))t2+((2-α)(2-3α)/(5-3α))t-(2-α)/2,
(2.5)g(1)=0,g′(t)=(2-α)(3α-2)2t3α-3-3(2-α)(2-3α)(α-1)5-3αt3α-4+(1-α)(2-3α)(3α-4)2(5-3α)t3α-5-(1-α)(2-3α)(5-3α)t+(2-α)(2-3α)(5-3α),g′(1)=0,g′′(t)=3(2-α)(3α-2)(α-1)2t3α-4-3(2-α)(2-3α)(α-1)(3α-4)5-3αt3α-5-(1-α)(2-3α)(3α-4)2t3α-6-(1-α)(2-3α)(5-3α),(2.6)g′′(1)=0,(2.7)g′′′(t)=32(1-α)(2-α)(4-3α)(3α-2)t3α-7(t-1)2.
If0<α<2/3, then (2.7) implies
(2.8)g′′′(t)<0
for t>1.
From (2.3)–(2.6) and (2.8) we know that f(t)>0 for t>1.
If2/3<α<1, then (2.7) leads to
(2.9)g′′′(t)>0
for t>1. Therefore f(t)<0 for t>1 follows from (2.3)–(2.6) and (2.9).
Let(2.10)h(t)=log⁡L-(2/α)(a,b)-log⁡[Gα(a,b)H1-α(a,b)]
for t=a/b>1; then (1.1) and elementary calculations lead to
(2.11)h(t)=-α2log⁡⁡t(α-2)/α-1((α-2)/α)(t-1)-α2log⁡t-(1-α)log⁡2t1+t,lim⁡t→1+h(t)=0,(2.12)h′(t)=-t(2-α)/αt(t2-1)(t(2-α)/α-1)v(t),
where v(t)=((2-α)/2)t(3α-2)/α+((3α-2)/2)t(2α-2)/α-((3α-2)/2)t-(2-α)/2,
(2.13)v(1)=0,v′(t)=(2-α)(3α-2)2αt(2α-2)/α+(3α-2)(α-1)αt(α-2)/α-3α-22,(2.14)v′(1)=0,(2.15)v′′(t)=(2-α)(1-α)(2-3α)α2t-2/α(t-1).
Ifα∈(0,2/3), then (2.15) implies
(2.16)v′′(t)>0
for t>1.
From (2.11)–(2.14) and (2.16) we know that h(t)<0 for t>1.
Ifα∈(2/3,1), then (2.15) leads to
(2.17)v′′(t)<0
for t>1. Therefore, h(t)>0 for t>1 follows from (2.11)–(2.14) and (2.17).
Next, we prove that the parameters–(2/α) and 3α-5 in either case cannot be improved. The proof is divided into two cases.
Case 1 (α∈(0,2/3)).
For anyϵ>0 and x∈(0,1), from (1.1) one has
(2.18)[Gα(1,1+x)H1-α(1,1+x)]5-3α+ϵ-[L3α-5-ϵ(1,1+x)]5-3α+ϵ=f1(x)(1+x/2)(1-α)(5-3α+ϵ)[(1+x)4-3α+ϵ-1],
where f1(x)=(1+x)(1-α/2)(5-3α+ϵ)[(1+x)4-3α+ϵ-1]-(4-3α+ϵ)x(1+x)4-3α+ϵ(1+x/2)(1-α)(5-3α+ϵ).
Letx→0; making use of the Taylor expansion, we get
(2.19)f1(x)=ϵ(4-3α+ϵ)(5-3α+ϵ)24x3+o(x3).
Equations (2.18) and (2.19) imply that for any α∈(0,2/3) and ϵ>0 there exists δ=δ(ϵ,α)∈(0,1), such that L3α-5-ϵ(1,1+x)<Gα(1,1+x)H1-α(1,1+x) for x∈(0,δ).
On the other hand, for anyϵ∈(0,(2/α)-1) we have
(2.20)L-(2/α)+ϵ(1,t)-Gα(1,t)H1-α(1,t)=tα/(2-ϵα){[1-t-2/α+ϵ+1(2/α-ϵ-1)(1-1/t)]-α/(2-ϵα)-t-ϵα2/2(2-ϵα)(2t1+t)1-α},lim⁡t→+∞{[1-t-2/α+ϵ+1(2/α-ϵ-1)(1-1/t)]-α/(2-ϵα)-t-ϵα2/2(2-ϵα)(2t1+t)1-α}=(2α-ϵ-1)α/(2-ϵα)>0.
From (2.20) we know that for any α∈(0,2/3) and ϵ∈(0,2/α-1) there exists T=T(ϵ,α)>1, such that L-2/α+ϵ(1,t)>Gα(1,t)H1-α(1,t) for t∈(T,∞).Case 2 (α∈(2/3,1)).
For anyϵ∈(0,4-3α) and x∈(0,1), from (1.1) one has
(2.21)[L3α-5+ϵ(1,1+x)]5-3α-ϵ-[Gα(1,1+x)H1-α(1,1+x)]5-3α-ϵ=f2(x)(1+x/2)(1-α)(5-3α-ϵ)[(1+x)4-3α-ϵ-1],
where f2(x)=(4-3α-ϵ)x(1+x)4-3α-ϵ(1+x/2)(1-α)(5-3α-ϵ)-(1+x)(1-α/2)(5-3α-ϵ)[(1+x)4-3α-ϵ-1].
Letx→0; making use of the Taylor expansion, we have
(2.22)f2(x)=ϵ24(4-3α-ϵ)(5-3α-ϵ)x3+o(x3).
Equations (2.21) and (2.22) imply that for any α∈(2/3,1) and ϵ∈(0,4-3α) there exists δ=δ(ϵ,α)∈(0,1), such that L3α-5+ϵ(1,1+x)>Gα(1,1+x)H1-α(1,1+x) for x∈(0,δ).
On the other hand, for anyϵ>0, we have
(2.23)Gα(1,t)H1-α(1,t)-L-(2/α)-ϵ(1,t)=tα/2{(2t1+t)1-α-t-ϵα2/2(2+ϵα)[1-t-(2/α+ϵ-1)(2/α+ϵ-1)(1-1/t)]-α/(2+ϵα)},lim⁡t→+∞{(2t1+t)1-α-t-ϵα2/2(2+ϵα)[1-t-(2/α+ϵ-1)(2/α+ϵ-1)(1-1/t)]-α/(2+ϵα)}=21-α>0.
From (2.23) we know that for any α∈(2/3,1) and ϵ>0 there exists T=T(ϵ,α)>1, such that L-(2/α)-ϵ(1,t)<Gα(1,t)H1-α(1,t) for t∈(T,∞).



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