# Optimal Lower Power Mean Bound for the Convex Combination of Harmonic and Logarithmic Means

Authors: Yu-Ming Chu; Shan-Shan Wang; Cheng Zong
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

We find the least valueλ∈(0,1) and the greatest value p=p(α) such that αH(a,b)+(1−α)L(a,b)>Mp(a,b) for α∈[λ,1) and all a,b>0 with a≠b, where H(a,b), L(a,b), and Mp(a,b) are the harmonic,
logarithmic, and p-th power means of two positive numbers a and b, respectively.


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## Body

## 1. Introduction

Forp∈ℝ, the p-th power mean Mp(a,b) and logarithmic mean L(a,b) of two positive numbers 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-alog⁡⁡b-log⁡⁡a,a≠b,a,a=b,
respectively.It is well known thatMp(a,b) is continuous and strictly increasing with respect to p∈ℝ for fixed a, b>0 with a≠b. In the recent past, both mean values have been the subject of intensive research. In particular, many remarkable inequalities for Mp(a,b) and L(a,b) can be found in the literature [1–17]. It might be surprising that the logarithmic mean has applications in physics, economics, and even in meteorology [18–20]. In [18], 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 [8]. In [21, 22], it is shown that L can be expressed in terms of Gauss's hypergeometric function 2F1. And, in [21], 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),I(a,b)=1/e(bb/aa)1/(b-a)(b≠a),I(a,b)=a (b=a), G(a,b)=ab, and H(a,b)=2ab/(a+b) be the arithmetic, identric, 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)<I(a,b)<A(a,b)=M1(a,b)<max⁡⁡{a,b}
for all a, b>0 with a≠b.In [23], Alzer and Janous established the following best possible inequality: (1.4)Mlog⁡⁡2/log⁡⁡3(a,b)<23A(a,b)+13G(a,b)<M2/3(a,b)
for all a,b>0 with a≠b.In [8, 11, 24], the authors presented 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 companion of (1.3) provides inequalities for the geometric and arithmetic means of L and I. A proof can be found in [25] (1.6)G1/2(a,b)A1/2(a,b)<L1/2(a,b)I1/2(a,b)<12L(a,b)+12I(a,b)<12G(a,b)+12A(a,b)
for all a,b>0 with a≠b.The following sharp bounds forL, I, (LI)1/2, and (L+I)/2 in terms of the power means are proved in [4, 5, 7, 9, 16, 25, 26]: (1.7)M0(a,b)<L(a,b)<M1/3(a,b),M2/3(a,b)<I(a,b)<Mlog⁡⁡2(a,b),M0(a,b)<L1/2(a,b)I1/2(a,b)<M1/2(a,b),12L(a,b)+12I(a,b)<M1/2(a,b)
for all a,b>0 with a≠b.Alzer and Qiu [27] found the sharp bound of 1/2(L(a,b)+I(a,b)) in terms of the power mean as follows: (1.8)Mc(a,b)<12(L(a,b)+I(a,b))
for all a,b>0 with a≠b, with the best possible parameter c=log2/(1+log2).The main purpose of this paper is to find the least valueλ∈(0,1) and the greatest value p=p(α) such that αH(a,b)+(1-α)L(a,b)>Mp(a,b) for α∈[λ,1) and all a,b>0 with a≠b.


## 2. Lemmas

In order to establish our main result we need three lemmas, which we present in this section.Lemma 2.1.
Letα∈(1/4,1), p=(1-4α)/3∈(-1,0), and f(t)=-4αp(p+1)2(p+2)tp-1+2(1-α)p2(1-p2)tp-2+2(1-α)p(1-p)2(2-p)tp-3+12(1-α)(1-p). Then f(t)>0 for t∈[1,+∞).Proof.
Simple computations lead to(2.1)f(1)=6481(1-α)2(56α2+23α+11)>0,(2.2)lim⁡t→+∞f(t)=12(1-α)(1-p)=8(1-α)(1+2α)>0,(2.3)f′(t)=-2p(1-p)tp-4f1(t),
where
(2.4)f1(t)=-2α(p+1)2(p+2)t2+(1-α)p(p+1)(2-p)t+(1-α)(1-p)(2-p)(3-p),f1(1)=427(1-α)(148α2-11α+25)>0,(2.5)lim⁡t→+∞f1(t)=-∞,(2.6)f′1(t)=-4α(p+1)2(p+2)t+(1-α)p(p+1)(2-p)=-427(1-α)2[16α(7-4α)t+(4α-1)(4α+5)]<0
for t∈[1,+∞).
Inequality (2.6) implies that f1(t) is strictly decreasing in [1,+∞), then from (2.4) and (2.5) we know that λ1>1 exists such that f1(t)>0 for t∈[1,λ1) and f1(t)<0 for t∈(λ1,+∞). Hence, equation (2.3) leads to the conclusion that f(t) is strictly increasing in [1,λ1] and strictly decreasing in [λ1,+∞).Therefore, Lemma2.1 follows from (2.1) and (2.2) together with the piecewise monotonicity of f(t).Lemma 2.2.
Letα∈(1/4,1), p=(1-4α)/3∈(-1,0), and g(t)=-(1-α)(p+1)(p+2)2(p+3)tp+(p+1)(p3-αp3-19αp2+3p2-34αp+2p-8α)tp-1+(1-α)p(p3-8p2-p+4)tp-2+(1-α)(1-p)(p3+5p2-14p+4)tp-3+4(1-α)(7-4p)-4p(1-α)t-1+4α(1+p)t-2, then g(t)>0 for t∈[1,+∞).Proof.
Letg1(t)=-(1-α)(p+1)(p+2)2(p+3)t3+(p+1)(p3-αp3-19αp2+3p2-34αp+2p-8α)t2+(1-α)p(p3-8p2-p+4)t+(1-α)(1-p)(p3+5p2-14p+4)+4(1-α)(7-4p)t3-p-4p(1-α)t2-p+4α(1+p)t1-p. Then simple computations lead to
(2.7)g(t)=tp-3g1(t),(2.8)g1(1)=1627(1-α)(80α2+110α-1)>0,g1′(t)=-3(1-α)(p+1)(p+2)2(p+3)t2+2(p+1)(p3-αp3-19αp2+3p2-34αp+2p-8α)t+(1-α)p(p3-8p2-p+4)+4(1-α)(7-4p)(3-p)t2-p-4p(1-α)(2-p)t1-p+4α(1-p2)t-p,
(2.9)g1′(1)=3227(1-α)(-16α3+38α2+176α-9)>0,g1′′(t)=-6(1-α)(p+1)(p+2)2(p+3)t+2(p+1)(p3-αp3-19αp2+3p2-34αp+2p-8α)+4(1-α)(7-4p)(3-p)(2-p)t1-p-4p(1-α)(2-p)(1-p)t-p-4αp(1-p2)t-p-1,
(2.10)g1′′(1)=881(1-α)(-128α4+896α3+288α2+5294α-437)>0,g1′′′(t)=-6(1-α)(p+1)(p+2)2(p+3)+4(1-α)(7-4p)(3-p)(2-p)(1-p)t-p+4p2(1-α)(2-p)(1-p)t-p-1+4αp(1+p)2(1-p)t-p-2(2.11)g1′′′(1)=881(1-α)(576α4+3872α3+660α2+6612α-785)>0,(2.12)g1(4)(t)=-4p(1-p)t-p-3g2(t),
where
(2.13)g2(t)=(1-α)(7-4p)(3-p)(2-p)t2+(1-α)p(2-p)(p+1)t+α(p+1)2(p+2),g2(1)=427(1-α)(96α3+232α2+388α+175)>0,(2.14)g2′(t)=2(1-α)(7-4p)(3-p)(2-p)t+(1-α)p(2-p)(p+1)≥g2′(1)=49(1-α)(5+4α)(12α2+31α+23)>0
for t∈[1,+∞).From (2.13) and (2.14), we clearly see that g2(t)>0 for t∈[1,+∞), then (2.12) leads to the conclusion that g1′′′(t) is strictly in [1,+∞).Therefore, Lemma2.2 follows from (2.7)–(2.11) and the monotonicity of g1′′′(t).Lemma 2.3.
Letα∈(1/4,1), p=(1-4α)/3∈(-1,0), and h(t)=2α(1-tp+1)tlog2t+(1-α)(1+tp-1)(1+t)2tlogt+(1-α)(1+t)2(1-t)(tp+1), then h(t)>0 for t∈(1,+∞).Proof.
Leth1(t)=t-ph′′(t) and h2(t)=tp+2h1′(t), then simple computations lead to
(2.15)h(1)=0,h′(t)=2α[1-(p+2)tp+1]log2t+[(p+2-αp-6α)tp+1+2(1-α)(p+1)tp+(1-α)ptp-1+3(1-α)t2+4(1-α)t+3α+1]logt-(1-α)[(p+3)tp+2+(p+1)tp+1-(p+3)tp-(p+1)tp-1+2t2-2],
(2.16)h′(1)=0,h1(t)=-2α(p+1)(p+2)log2t+[(p2-αp2+3p-11αp-14α+2)+2(1-α)p(p+1)t-1-(1-α)p(1-p)t-2+6(1-α)t1-p+4(1-α)t-p+4αt-1-p]logt-(1-α)(p+2)(p+3)t+(1-α)(p2+5p+2)t-1+(1-α)(p2+p-1)t-2-(1-α)t1-p+4(1-α)t-p+(1+3α)t-1-p-(1-α)p2-(1-α)p-5α+1,
(2.17)h1(1)=0,h2(t)=-[4α(p+1)(p+2)tp+1+2(1-α)p(p+1)tp-2(1-α)p(1-p)tp-1-6(1-α)(1-p)t2+4(1-α)pt+4α(1+p)]logt-(1-α)(p+2)(p+3)tp+2+(p2-αp2+3p-11αp-14α+2)tp+1+(1-α)(p2-3p-2)tp-(1-α)(p2+3p-2)tp-1+(1-α)(p+5)t2+4(1-α)(1-p)t+α-3αp-p-1,
(2.18)h2(1)=0,h2′(t)=-[4α(p+1)2(p+2)tp+2(1-α)p2(p+1)tp-1+2(1-α)p(1-p)2tp-2-12(1-α)(1-p)t+4(1-α)p]logt-(1-α)(p+2)2(p+3)tp+1+(p+1)(p2-αp2-15αp+3p-22α+2)tp+(1-α)p(p2-5p-4)tp-1+(1-α)(1-p)(p2+5p-2)tp-2+4(1-α)(4-p)t-4α(p+1)t-1+4(1-α)(1-2p),
(2.19)h2′(1)=0,h2′′(t)=f(t)log⁡⁡t+g(t),
where f(t) and g(t) are defined as in Lemmas 2.1 and 2.2, respectively.
From (2.19) and (2.10) together with Lemmas 2.1 and 2.2, we clearly see that h2(t) is strictly increasing in [1,+∞).Therefore, Lemma2.3 follows from (2.15)–(2.18) and the monotonicity of h2(t).


## 3. Main Result

Theorem 3.1.
Inequality(3.1)αH(a,b)+(1-α)L(a,b)>M(1-4α)/3(a,b)
holds for α∈[1/4,1) and all a,b>0 with a≠b, and M(1-4α)/3(a,b) is the best possible lower power mean bound for the sum αH(a,b)+(1-α)L(a,b).Proof.
We divide the proof of inequality (3.1) into two cases.
Case 1 (α=1/4).
Without loss of generality, we assume thata>b and put t=a/b>1, then from (1.1) and (1.2), we have
(3.2)αH(a,b)+(1-α)L(a,b)-M(1-4α)/3(a,b)=14[H(a,b)+3L(a,b)]-ab=3t4-4(2t3-t2+2t)log⁡⁡t-38(t2+1)log⁡⁡tb.
Let
(3.3)F(t)=3t4-4(2t3-t2+2t)log⁡⁡t-3,
then simple computations lead to
(3.4)F(1)=0,F′(t)=4(3t3-2t2+t-2)-8(3t2-t+1)log⁡⁡t,F′(1)=0,F′′(t)=4tF1(t),
where F1(t)=9t3-10t2+3t-2-2(6t-1)tlogt,
(3.5)F′′(1)=F1(1)=0,F1′(t)=27t2-32t+5-2(12t-1)log⁡⁡t,F1′(1)=0,F1′′(t)=2tF2(t),
where F2(t)=27t2-12tlogt-28t+1,
(3.6)F1′′(1)=F2(1)=0,F2′(t)=54t-12log⁡⁡t-40>0
for t>1.
Therefore, inequality (3.1) follows easily from (3.2)–(3.6).
Case 2 (α∈(1/4,1)).
Without loss of generality, we assume thata>b. Let p=(1-4α)/3∈(-1,0) and t=a/b>1, then from (1.1) and (1.2), one has
(3.7)αH(a,b)+(1-α)L(a,b)-M(1-4α)/3(a,b)=αH(a,b)+(1-α)L(a,b)-Mp(a,b)=b[2αtt+1+(1-α)(t-1)log⁡⁡t-(tp+12)1/p].
Let
(3.8)G(t)=log⁡[2αtt+1+(1-α)(t-1)log⁡⁡t]-1plog⁡⁡tp+12.
Then simple computations lead to
(3.9)lim⁡t→1G(t)=0,(3.10)G′(t)=h(t)t(t+1)(tp+1)log⁡⁡t[2αtlog⁡⁡t+(1-α)(t2-1)],
where h(t) is defined as in Lemma 2.3.
From Lemma2.3 and (3.10), we clearly see that G(t) is strictly increasing in (1,+∞).Therefore, inequality (3.1) follows from (3.7)–(3.9) and the monotonicity of G(t).Next, we prove thatM(1-4α)/3(a,b) is the best possible lower power mean bound for the sum αH(a,b)+(1-α)L(a,b) if α∈[1/4,1).For anyα∈[1/4,1), p>(1-4α)/3, and x>0, one has(3.11)Mp(1+x,1)-αH(1+x,1)-(1-α)L(1+x,1)=J(x)21/p(1+x2)log⁡⁡(1+x),
where J(x)=(1+x/2)[1+(1+x)p]1/plog(1+x)-21/p[α(1+x)log(1+x)+(1-α)x(1+x/2)]. Letting x→0 and making use of Taylor expansion, we have(3.12)J(x)=21/p8(p-1-4α3)x3+o(x3).Equations (3.11) and (3.12) imply that for any α∈[1/4,1) and p>(1-4α)/3 there exists δ>0, such that αH(1+x,1)+(1-α)L(1+x,1)<Mp(1+x,1) for x∈(0,δ).Remark 3.2.
If0<α<1/4, then from (1.1) and (1.2), we have
(3.13)lim⁡x→+∞M(1-4α)/3(1,x)αH(1,x)+(1-α)L(1,x)=23/(4α-1)×lim⁡x→+∞(1+x(4α-1)/3)3/(1-4α)2α/(x+1)+((1-1/x)(1-α)/log⁡⁡x)=+∞.
Equation (3.13) implies that for any 0<α<1/4, there exists X>1, such that M(1-4α)/3(1,x)>αH(1,x)+(1-α)L(1,x) for x∈(X,+∞). Therefore, λ=1/4 is the least value of λ in (0,1) such that inequality (3.1) holds for all a,b>0 with a≠b.



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*Source: 520648-2011-07-07.xml*