# The Optimal Upper and Lower Power Mean Bounds for a Convex Combination of the Arithmetic and Logarithmic Means

Authors: Wei-Feng Xia; Yu-Ming Chu; Gen-Di Wang
Journal: Abstract and Applied Analysis
 (2010)

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## Abstract

Forp∈ℝ, the power mean Mp(a,b) of order p, logarithmic mean L(a,b), and arithmetic mean A(a,b) of two positive real values a and b are defined by Mp(a,b)=((ap+bp)/2)1/p, for p≠0 and Mp(a,b)=ab, for p=0, L(a,b)=(b-a)/(logb-loga), for a≠b and L(a,b)=a, for a=b and A(a,b)=(a+b)/2, respectively. In this paper, we answer the question: for α∈(0,1), what are the greatest value p and the least value q, such that the double inequality Mp(a,b)≤αA(a,b)+(1-α)L(a,b)≤Mq(a,b) holds for all a,b>0?


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## Body

## 1. Introduction

Forp∈ℝ, the power mean Mp(a,b) of order p and logarithmic mean L(a,b) of two positive real values a and b are defined by(1.1)Mp(a,b)={(ap+bp2)1/p,p≠0,ab,p=0,(1.2)L(a,b)={b-alogb-loga,a≠b,a,a=b,
respectively. In the recent past, both mean values have been the subject of intensive research. In particular, many remarkable inequalities for power mean or logarithmic mean can be found in the literature [1–15]. It might be surprising that the logarithmic mean has applications in physics, economics, and even in meteorology [16–18]. In [16] the authors study a variant of Jensen’s functional equation involving L, which appears in a heat conduction problem. A representation of L as an infinite product and an iterative algorithm for computing the logarithmic mean as the common limit of two sequences of special geometric and arithmetic means are given in [11]. In [19, 20] it is shown that L can be expressed in terms of Gauss's hypergeometric function F21. And, in [20] the authors prove that the reciprocal of the logarithmic mean is strictly totally positive, that is, every n×n determinant with elements 1/L(ai,bi), where 0<a1<a2<⋯<an and 0<b1<b2<⋯<bn, is positive for all n≥1.LetA(a,b)=(1/2)(a+b), G(a,b)=ab, and H(a,b)=2ab/(a+b) be the arithmetic, geometric, and harmonic means of two positive numbers a and b, respectively. Then it is well known that(1.3)min{a,b}≤H(a,b)=M-1(a,b)≤G(a,b)=M0(a,b)≤L(a,b)≤A(a,b)=M1(a,b)≤max{a,b},
and all inequalities are strict for a≠b.In [21], Alzer and Janous established the following best possible inequality:(1.4)Mlog2/log3(a,b)≤23A(a,b)+13G(a,b)≤M2/3(a,b)
for all a,b>0.In [11, 13, 22] the authors present bounds for L in terms of G and A(1.5)G2/3(a,b)A1/3(a,b)<L(a,b)<23G(a,b)+13A(a,b)
for all a,b>0 with a≠b.The following sharp bounds forL in terms of power means are proved by Lin [12](1.6)M0(a,b)<L(a,b)<M1/3(a,b).The main purpose of this paper is to answer the question: forα∈(0,1), what are the greatest value p and the least value q, such that the double inequality Mp(a,b)≤αA(a,b)+(1-α)L(a,b)≤Mq(a,b) holds for all a,b>0?


## 2. Lemmas

In order to establish our results we need several lemmas, which we present in this section.Lemma 2.1.
Ifα∈(0,1), then (1+2α)(log2-logα)>3log2.Proof.
Forα∈(0,1), let f(α)=(1+2α)(log2-logα), then simple computations lead to
(2.1)f′(α)=2(log2-1)-2logα-1α,(2.2)f′′(α)=1α2(1-2α).
From (2.2) we clearly see that f′′(α)>0 for α∈(0,1/2), and f′′(α)<0 for α∈(1/2,1). Then from (2.1) we get
(2.3)f′(α)≤f′(12)=4(log2-1)<0
for α∈(0,1).
Thereforef(α)>f(1)=3log2 for α∈(0,1) follows from (2.3).Lemma 2.2.
Letα∈(0,1), if p=log2/(log2-logα), then
(2.4)-p3+(4α-1)p2-3αp+α<0.Proof.
Forα∈(0,1), let t=-logα, then t∈(0,+∞) and
(2.5)-p3+(4α-1)p2-3αp+α=f(t)(t+log2)3et,
where f(t)=(t+log2)3-3log2(t+log2)2+(log2)2(t+log2)(4-et)-(log2)3et.
To prove Lemma2.2 we need only to prove that f(t)<0 for t∈(0,+∞). Elementary calculations yield that
(2.6)f(0)=0,(2.7)f'(t)=3(t+log2)2-6log2(t+log2)-(log2)2tet-(1+2log2)(log2)2et+4(log2)2,(2.8)f'(0)=-2(log2)3<0,(2.9)limt→+∞f′(t)=-∞,(2.10)f′′(t)=6t-(log2)2tet-2(log2)2(1+log2)et,(2.11)f′′(0)=-2(1+log2)(log2)2<0,(2.12)limt→+∞f′′(t)=-∞,(2.13)f′′′(t)=6-(log2)2tet-(log2)2(3+2log2)et,(2.14)f′′′(0)=6-3(log2)2-2(log2)3>0,(2.15)limt→+∞f′′′(t)=-∞,(2.16)f(4)(t)=-(log2)2tet-2(log2)2(2+log2)et<0
for t∈(0,+∞).
Making use of a computer and the mathematica software, from (2.10) we get
(2.17)f′′(1.15)=0.01679⋯,(2.18)f′′(1.16)=-0.0077⋯.
From (2.14)–(2.16) we clearly see that there exists a unique t0∈(0,+∞), such that f′′′(t)>0 for t∈[0,t0) and f′′′(t)<0 for t∈(t0,+∞). Hence we know that f′′(t) is strictly increasing in [0,t0] and strictly decreasing in [t0,+∞).
From (2.11), (2.12), (2.17), (2.18) and the monotonicity of f′′(t) in [0,t0] and in [t0,+∞) we know that there exist exactly two numbers t1,t2∈(0,+∞) with t1<t2, such that f′′(t)<0 for t∈[0,t1)∪(t2,+∞) and f′′(t)>0 for t∈(t1,t2), and t2 satisfies
(2.19)1.15<t2<1.16.
Hence, we know thatf′(t) is strictly decreasing in [0,t1]∪[t2,+∞) and strictly increasing in [t1,t2].
Making use of a computer and the mathematica software, from (2.7) and (2.19), we get
(2.20)f′(t2)<3(1.16+log2)2-6log2(1.15+log2)-1.15×e1.15×(log2)2-(1+2log2)×(log2)2×e1.15+4(log2)2=-0.807⋯<0.
Now, (2.8), (2.9), (2.20) and the monotonicity of f′(t) in [0,t1]∪[t2,+∞) and in [t1,t2] imply that
(2.21)f′(t)<0
for t∈(0,+∞).
Therefore,f(t)<0 for t∈(0,+∞) follows from (2.6) and (2.21).Lemma 2.3.
Forα∈(0,1) and g(t)=α(t-tp)(logt)2+2(1-α)(t+tp)logt-2(1-α)(t-1)(1+tp), one has the following.(1)
Ifp=log2/(log2-logα), then there exists λ∈(1,+∞) such that g(t)>0 for t∈(1,λ) and g(t)<0 for t∈(λ,+∞).(2)
Ifp=(1+2α)/3, then g(t)<0 for t∈(1,+∞).Proof.
Letg1(t)=t1-pg′(t), g2(t)=tpg1′(t), g3(t)=tg2′(t), g4(t)=t2-pg3′(t), g5(t)=tg4′(t), and p∈{log2/(log2-logα),(1+2α)/3}, then simple computations lead to
(2.22)g(1)=0,(2.23)limt→+∞g(t)=-∞,(2.24)g1(t)=α(t1-p-p)(logt)2+2(t1-p+p-αp-α)logt+2(1-α)(1+p)(1-t),(2.25)g1(1)=0,(2.26)limt→+∞g1(t)=-∞,(2.27)g2(t)=α(1-p)(logt)2+2(1+α-p-αptp-1)logt+2(p-αp-α)tp-1-2(1-α)(1+p)tp+2,(2.28)g2(1)=0,(2.29)limt→+∞g2(t)=-∞,(2.30)g3(t)=2α(1-p)(1+ptp-1)logt+2[(1-α)p2-(1+α)p+α]tp-1-2p(1-α)(1+p)tp+2(1+α-p),(2.31)g3(1)=2(1+2α-3p),(2.32)limt→+∞g3(t)=-∞,(2.33)g4(t)=2α(1-p)t1-p-2αp(1-p)2logt-2p2(1-α)(1+p)t+2(p-1)[(1-α)p2-(1+2α)p+α],(2.34)g4(1)=2p(1+2α-3p),(2.35)limt→+∞g4(t)=-∞,(2.36)g5(t)=2α(1-p)2t1-p-2p2(1-α)(1+p)t-2αp(1-p)2,(2.37)g5(1)=-2[p3-(4α-1)p2+3αp-α],(2.38)g5′(t)=2α(1-p)3t-p-2p2(1-α)(1+p),(2.39)g5′(1)=-2[p3-(4α-1)p2+3αp-α].
(1) If p=log2/(log2-logα), then from (2.31), (2.34), (2.37)–(2.39), and Lemmas 2.1-2.2 we clearly see that
(2.40)g3(1)>0,(2.41)g4(1)>0,(2.42)g5(1)<0,(2.43)g5′(1)<0,
and g5′(t) is strictly decreasing in [1,+∞).
From (2.43) and the monotonicity of g5′(t) we know that g5(t) is strictly decreasing in [1,+∞).
The monotonicity ofg5(t) and (2.42) implies that g5(t)<0 for t∈[1,+∞), then we conclude that g4(t) is strictly decreasing in [1,+∞).
From the monotonicity ofg4(t) and (2.35) together with (2.41) we clearly see that there exists t1∈(1,+∞), such that g4(t)>0 for t∈[1,t1) and g4(t)<0 for t∈(t1,+∞). Hence we know that g3(t) is strictly increasing in [1,t1] and strictly decreasing in [t1,+∞).
The monotonicity ofg3(t) in [1,t1] and in [t1,+∞) together with (2.32) and (2.40) imply that there exists t2∈(1,+∞), such that g3(t)>0 for t∈[1,t2) and g3(t)<0 for t∈(t2,+∞). Then we know that g2(t) is strictly increasing in [1,t2] and strictly decreasing in [t2,+∞).
From (2.28) and (2.29) together with the monotonicity of g2(t) in [1,t2] and in [t2,+∞) we clearly see that there exists t3∈(1,+∞), such that g2(t)>0 for t∈[1,t3) and g2(t)<0 for t∈(t3,+∞). Hence we know that g1(t) is strictly increasing in [1,t3] and strictly decreasing in [t3,+∞).
Equations (2.25) and (2.26) together with the monotonicity of g1(t) in [1,t3] and in [t3,+∞) imply that there exists t4∈(1,+∞), such that g1(t)>0 for t∈[1,t4) and g1(t)<0 for t∈(t4,+∞). Then we conclude that g(t) is strictly increasing in [1,t4] and strictly decreasing in [t4,+∞).
Now (2.22), (2.23) and the monotonicity of g(t) in [1,t4] and in [t4,+∞) imply that there exists λ∈(1,+∞), such that g(t)>0 for t∈[1,λ) and g(t)<0 for t∈(λ,+∞).
(2) Ifp=(1+2α)/3, then (2.31), (2.34), and (2.37)–(2.39) lead to
(2.44)g4(1)=g3(1)=0,(2.45)g5′(1)=g5(1)=-127[8(1-α3)+12α(1-α2)+60α2(1-α)]<0,
and g5′(t) is strictly decreasing in [1,+∞).
Therefore, Lemma2.3(2) follows from (2.22), (2.25), (2.28), (2.44), (2.45), and the monotonicity of g5′(t).


## 3. Main Result

Theorem 3.1.
Forα∈(0,1), the double inequality Mlog2/(log2-logα)(a,b)≤αA(a,b)+(1-α)L(a,b)≤M(1+2α)/3(a,b) holds for all a,b>0, each inequality becomes an equality if and only if a=b, and the given parameters log2/(log2-logα) and (1+2α)/3 in each inequality are best possible.Proof.
Ifa=b, then from (1.1) and (1.2) we clearly see that Mlog2/(log2-logα)(a,b)=αA(a,b)+(1-α)L(a,b)=M(1+2α)/3(a,b)=a for α∈(0,1). Next, we assume that a≠b.
Firstly, we prove thatMlog2/(log2-logα)(a,b)<αA(a,b)+(1-α)L(a,b)<M(1+2α)/3(a,b) for a,b>0 with a≠b.
Without loss of generality, we assume thata>b. Let t=a/b>1 and p∈{log2/(log2-logα),(1+2α)/3}, then (1.1) and (1.2) leads to
(3.1)αA(a,b)+(1-α)L(a,b)-Mp(a,b)=b[α(t+1)logt+2(1-α)(t-1)2logt-(tp+12)1/p].
Let
(3.2)f(t)=log[α(t+1)logt+2(1-α)(t-1)2logt]-1plog(tp+12),
then
(3.3)limt→1f(t)=0,(3.4)f′(t)=g(t)t[α(t+1)logt+2(1-α)(t-1)](1+tp)logt,
where g(t)=α(t-tp)(logt)2+2(1-α)(t+tp)logt-2(1-α)(t-1)(1+tp).
Ifp=log2/(log2-logα), then it is not difficult to verify that
(3.5)limt→+∞f(t)=0.
From (3.4) and Lemma 2.3(1) we know that there exists λ∈(1,+∞), such that f(t) is strictly increasing in [1,λ] and strictly decreasing in [λ,+∞). Then (3.3) and (3.5) together with the monotonicity of f(t) in [1,λ] and in [λ,+∞) imply that f(t)>0 for t∈(1,+∞), and from (3.1) and (3.2) we know that αA(a,b)+(1-α)L(a,b)>Mlog2/(log2-logα)(a,b) for all a,b>0 with a≠b.
Ifp=(1+2α)/3, then from Lemma 2.3(2) and (3.1)–(3.4) we clearly see that αA(a,b)+(1-α)L(a,b)<M(1+2α)/3(a,b) for all a,b>0 with a≠b.
Secondly, we prove that the parameterslog2/(log2-logα) and (1+2α)/3 cannot be improved in each inequality.
For anyɛ>0 and x>1, from (1.1) and (1.2) we get
(3.6)limx→+∞Mlog2/(log2-logα)+ɛ(1,x)αA(1,x)+(1-α)L(1,x)=2α×(12)(log2-logα)/(log2+ɛ(log2-logα))>2α×(12)(log2-logα)/log2=1.
Inequality (3.6) implies that for any ɛ>0 there exists X=X(ɛ)>1, such that Mlog2/(log2-logα)+ɛ(1,x)>αA(1,x)+(1-α)L(1,x) for x∈(X,+∞). Hence the parameter log2/(log2-logα) cannot be improved in the left-side inequality.
Next for0<ɛ<(1+2α)/3, let 0<x<1, then (1.1) and (1.2) leads to
(3.7)[αA(1,1+x)+(1-α)L(1,1+x)](1+2α-3ɛ)/3-[M(1+2α)/3-ɛ(1,1+x)](1+2α-3ɛ)/3=[(1-α)x+α(1+x/2)log(1+x)log(1+x)](1+2α-3ɛ)/3-1+(1+x)(1+2α-3ɛ)/32=f(x)[log(1+x)](1+2α-3ɛ)/3,
where f(x)=[(1-α)x+α(1+α/2)log(1+x)](1+2α-3ɛ)/3-((1+(1+x)(1+2α-3ɛ)/3)/2)[log(1+x)](1+2α-3ɛ)/3.
Letx→0, making use of the Taylor expansion we get
(3.8)f(x)=124ɛ(1+2α-3ɛ)x(1+2α-3ɛ)/3[x2+o(x2)].
Equations (3.7) and (3.8) imply that for any 0<ɛ<(1+2α)/3 there exists 0<δ=δ(ɛ,α)<1, such that αA(1,1+x)+(1-α)L(1,1+x)>M(1+2α)/3-ɛ(1,1+x) for x∈(0,δ). Hence the parameter (1+2α)/3 cannot be improved in the right-side inequality.



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