AAAAbstract and Applied Analysis1687-04091085-3375Hindawi Publishing Corporation52064810.1155/2011/520648520648Research ArticleOptimal Lower Power Mean Bound for the Convex Combination of Harmonic and Logarithmic MeansChuYu-Ming1WangShan-Shan2ZongCheng2LasieckaIrena1Department of MathematicsHuzhou Teachers CollegeHuzhou 313000Chinahutc.zj.cn2School of ScienceHangzhou Normal UniversityHangzhou 310012Chinahznu.edu.cn201114062011201101022011150520112011Copyright © 2011 Yu-Ming Chu et al.This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We find the least value λ(0,1) and the greatest value p=p(α) such that αH(a,b)+(1α)L(a,b)>Mp(a,b) for α[λ,1) and all a,b>0 with ab, 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.

1. Introduction

For p, the p-th power mean Mp(a,b) and logarithmic mean L(a,b) of two positive numbers a and b are defined byMp(a,b)={(ap+bp2)1/p,p0,ab,p=0,L(a,b)={b-alogb-loga,ab,a,a=b, respectively.

It is well known that Mp(a,b) is continuous and strictly increasing with respect to p for fixed a, b>0 with ab. 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 . It might be surprising that the logarithmic mean has applications in physics, economics, and even in meteorology . In , 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 . In [21, 22], it is shown that L can be expressed in terms of Gauss's hypergeometric function 2F1. And, in , 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 n1.

Let A(a,b)=1/2(a+b),I(a,b)=1/e(bb/aa)1/(b-a)(ba),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 thatmin{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 ab.

In , Alzer and Janous established the following best possible inequality: Mlog2/log3(a,b)<23A(a,b)+13G(a,b)<M2/3(a,b) for all a,b>0 with ab.

In [8, 11, 24], the authors presented bounds for L in terms of G and AG2/3(a,b)A1/3(a,b)<L(a,b)<23G(a,b)+13A(a,b) for all a,b>0 with ab.

The following companion of (1.3) provides inequalities for the geometric and arithmetic means of L and I. A proof can be found in  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 ab.

The following sharp bounds for L, I, (LI)1/2, and (L+I)/2 in terms of the power means are proved in [4, 5, 7, 9, 16, 25, 26]: M0(a,b)<L(a,b)<M1/3(a,b),M2/3(a,b)<I(a,b)<Mlog2(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 ab.

Alzer and Qiu  found the sharp bound of 1/2(L(a,b)+I(a,b)) in terms of the power mean as follows: Mc(a,b)<12(L(a,b)+I(a,b)) for all a,b>0 with ab, with the best possible parameter c=log2/(1+log2).

The main purpose of this paper is to find the least value λ(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 ab.

2. Lemmas

In order to establish our main result we need three lemmas, which we present in this section.

Lemma 2.1.

Let α(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 f(1)=6481(1-α)2(56α2+23α+11)>0,limt+f(t)=12(1-α)(1-p)=8(1-α)(1+2α)>0,f(t)=-2p(1-p)tp-4f1(t), where 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,limt+f1(t)=-,f1(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, Lemma 2.1 follows from (2.1) and (2.2) together with the piecewise monotonicity of f(t).

Lemma 2.2.

Let α(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.

Let g1(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 g(t)=tp-3g1(t),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, 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, 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-2g1′′′(1)=881(1-α)(576α4+3872α3+660α2+6612α-785)>0,g1(4)(t)=-4p(1-p)t-p-3g2(t), where 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,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, Lemma 2.2 follows from (2.7)–(2.11) and the monotonicity of g1(t).

Lemma 2.3.

Let α(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.

Let h1(t)=t-ph(t) and h2(t)=tp+2h1(t), then simple computations lead to 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], 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, 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, 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), h2(1)=0,h2′′(t)=f(t)logt+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, Lemma 2.3 follows from (2.15)–(2.18) and the monotonicity of h2(t).

3. Main Result

Theorem 3.1.

Inequality 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 ab, 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 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M164"><mml:mi>α</mml:mi><mml:mo>=</mml:mo><mml:mrow><mml:mrow><mml:mn>1</mml:mn></mml:mrow><mml:mo>/</mml:mo><mml:mrow><mml:mn>4</mml:mn></mml:mrow></mml:mrow></mml:math></inline-formula>).

Without loss of generality, we assume that a>b and put t=a/b>1, then from (1.1) and (1.2), we have 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)logt-38(t2+1)logtb. Let F(t)=3t4-4(2t3-t2+2t)logt-3, then simple computations lead to F(1)=0,F(t)=4(3t3-2t2+t-2)-8(3t2-t+1)logt,F(1)=0,F′′(t)=4tF1(t), where F1(t)=9t3-10t2+3t-2-2(6t-1)tlogt, F′′(1)=F1(1)=0,F1(t)=27t2-32t+5-2(12t-1)logt,F1(1)=0,F1′′(t)=2tF2(t), where F2(t)=27t2-12tlogt-28t+1, F1′′(1)=F2(1)=0,F2(t)=54t-12logt-40>0 for t>1.

Therefore, inequality (3.1) follows easily from (3.2)–(3.6).

Case 2 (<inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M175"><mml:mi>α</mml:mi><mml:mo>∈</mml:mo><mml:mo stretchy="false">(</mml:mo><mml:mrow><mml:mrow><mml:mn>1</mml:mn></mml:mrow><mml:mo>/</mml:mo><mml:mrow><mml:mn>4</mml:mn></mml:mrow></mml:mrow><mml:mo>,</mml:mo><mml:mn>1</mml:mn><mml:mo stretchy="false">)</mml:mo></mml:math></inline-formula>).

Without loss of generality, we assume that a>b. Let p=(1-4α)/3(-1,0) and t=a/b>1, then from (1.1) and (1.2), one has 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)logt-(tp+12)1/p].   Let G(t)=log[2αtt+1+(1-α)(t-1)logt]-1plogtp+12. Then simple computations lead to limt1G(t)=0,G(t)=h(t)t(t+1)(tp+1)logt[2αtlogt+(1-α)(t2-1)], where h(t) is defined as in Lemma 2.3.

From Lemma 2.3 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 that 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) if α[1/4,1).

For any α[1/4,1), p>(1-4α)/3, and x>0, one hasMp(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 x0 and making use of Taylor expansion, we haveJ(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.

If 0<α<1/4, then from (1.1) and (1.2), we have limx+M(1-4α)/3(1,x)αH(1,x)+(1-α)L(1,x)=23/(4α-1)×limx+(1+x(4α-1)/3)3/(1-4α)2α/(x+1)+((1-1/x)(1-α)/logx)=+.

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 ab.

Acknowledgments

This research is supported by the N. S. Foundation of China under Grant 11071069, N. S. Foundation of Zhejiang province under Grants nos. Y7080106 and Y6100170, and the Innovation Team Foundation of the Department of Education of Zhejiang Province under Grant no. T200924.

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