# Inequalities between Power Means and Convex Combinations of the Harmonic and Logarithmic Means

Authors: Wei-Mao Qian; Zhong-Hua Shen
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We prove thatαH(a,b)+(1−α)L(a,b)>M(1−4α)/3(a,b) for α∈(0,1) and all a,b>0 with a≠b if and only if α∈[1/4,1) and αH(a,b)+(1−α)L(a,b)<M(1−4α)/3(a,b) if and only if α∈(0,3345/80−11/16), and the parameter (1−4α)/3 is the best possible in either case. Here, H(a,b)=2ab/(a+b),  L(a,b)=(a−b)/(loga−logb), and  Mp(a,b)=((ap+bp)/2)1/p(p≠0)
and M0(a,b)=ab are the harmonic, logarithmic, and pth power means of a and b, respectively.


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## Body

## 1. Introduction

The classical logarithmic meanL(a,b) of two positive real numbers a and b with a≠b is defined by (1.1)L(a,b)=a-blog⁡⁡a-log⁡⁡b.In the recent past, the bivariate means have been the subject of intensive research. In particular, many remarkable inequalities forL(a,b) can be found in the literature [1–21]. It might be surprising that the logarithmic mean has applications in physics, economics, and even in meteorology [22–24]. In [22] the authors study a variant of Jensen's functional equation involving the logarithmic mean, which appears in a heat conduction problem. A representation of L(a,b) as an infinite product and an iterative algorithm for computing it as the common limit of two sequences of special geometric and arithmetic means are given in [4]. In [25, 26] it is shown that L(a,b) can be expressed in terms of Gauss hypergeometric function F12. And, in [26] 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.LetG(a,b)=ab, H(a,b)=2ab/(a+b), I(a,b)=1/e(aa/bb)1/(a-b), A(a,b)=(a+b)/2, Mp(a,b)=((ap+bp)/2)1/p(p≠0) and M0(a,b)=ab, and Lp(a,b)=(ap+1+bp+1)/(ap+bp) be the geometric, harmonic, identric, arithmetic, pth power, and pth Lehmer means of two positive numbers a and b, respectively. Then it is well known that both Mp(a,b) and Lp(a,b) are continuous and strictly increasing with respect to p∈ℝ for fixed a,b>0 with a≠b, and the inequalities(1.2)min⁡{a,b}<H(a,b)=M-1(a,b)=L-1(a,b)<G(a,b)=M0(a,b)=L-1/2(a,b)<L(a,b)<I(a,b)<A(a,b)=M1(a,b)=L0(a,b)<max⁡⁡{a,b}
hold for all a,b>0 with a≠b.In [4], Carlson proves that the double inequality (1.3)G(a,b)(A(a,b)+G(a,b))2<L(a,b)<12(A(a,b)+G(a,b))
holds for all a,b>0 with a≠b.In [5], Lin finds the best possible upper and lower power bounds for the logarithmic mean as follows: (1.4)M0(a,b)<L(a,b)<M1/3(a,b)
for all a,b>0 with a≠b.In [9], Sándor establishes that (1.5)G(a,b)I(a,b)<L(a,b)<A(a,b)+G(a,b)-I(a,b)
for all a,b>0 with a≠b.In [27], Alzer gives the optimal Lehmer mean bounds for L, (LI)1/2, and (L+I)/2 as follows: (1.6)L-1/3(a,b)<L(a,b)<L0(a,b),L-1/4(a,b)<L(a,b)I(a,b)<L0(a,b),L-1/4(a,b)<12(L(a,b)+I(a,b))<L0(a,b)
for all a,b>0 with a≠b.The following sharp bounds for(LI)1/2 and (L+I)/2 in terms of power mean are presented in [28]: (1.7)M0(a,b)<L(a,b)I(a,b)<M1/2(a,b),Mlog⁡⁡2/(1+log⁡⁡2)(a,b)<12(L(a,b)+I(a,b))<M1/2(a,b)
for all a,b>0 with a≠b.In [29, 30], the authors obtain the sharp bounds for the products Aα(a,b)L1-α(a,b) and Gα(a,b)L1-α(a,b) and the sum αA(a,b)+(1-α)L(a,b) in terms of power mean as follows: (1.8)M0(a,b)<Aα(a,b)L1-α(a,b)<M(1+2α)/3(a,b),M0(a,b)<Gα(a,b)L1-α(a,b)<M(1-α)/3(a,b),Mlog⁡⁡2/(log⁡⁡2-log⁡⁡α)(a,b)<αA(a,b)+(1-α)L(a,b)<M(1+2α)/3(a,b)
for any α∈(0,1) and all a,b>0 with a≠b.In [31], Zhu presents some bounds for I(a,b) in terms of A(a,b) and L(a,b) and L(a,b) in terms of G(a,b) and I(a,b).In [32],  Chu et al.  prove that the double inequality αA(a,b)+(1-α)H(a,b)<P(a,b)<βA(a,b)+(1-β)H(a,b) holds for all a,b>0 with a≠b if and only if α≤2/π and β≥5/6.It is the aim of this paper to give the optimal power mean bounds for the convex combination of harmonic and logarithmic means. Our main result is the following theorem.Theorem 1.1.
Forα∈(0,1) and all a,b>0 with a≠b, one has(1)
αH(a,b)+(1-α)L(a,b)>M(1-4α)/3(a,b) if and only if α∈[1/4,1);(2)
αH(a,b)+(1-α)L(a,b)<M(1-4α)/3(a,b) if and only if α∈(0,3345/80-11/16).
In particular, the parameter(1-4α)/3 is the best possible in either case.


## 2. Lemmas

In order to establish our main result we need to establish four lemmas, which we present in this section.Lemma 2.1.
Letα∈(1/4,1),p=(1-4α)/3∈(-1,0), and g(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 g(t)>0 for t∈[1,+∞).Proof.
Simple computations lead to(2.1)g(1)=6481(1-α)2(56α2+23α+11)>0,(2.2)lim⁡t→+∞⁡g(t)=12(1-α)(1-p)=8(1-α)(1+2α)>0,(2.3)g′(t)=-2p(1-p)tp-4g1(t),
where
(2.4)g1(t)=-2α(p+1)2(p+2)t2+(1-α)p(p+1)(2-p)t+(1-α)(1-p)(2-p)(3-p),g1(1)=427(1-α)(148α2-11α+25)>0,(2.5)lim⁡t→+∞⁡g1(t)=-∞,(2.6)g′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 g1(t) is strictly decreasing in [1,+∞). Then (2.4) and (2.5) lead to the conclusion that there exists λ1>1 such that g1(t)>0 for t∈[1,λ1) and g1(t)<0 for t∈(λ1,+∞). It follows from (2.3) that g(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 g(t).Lemma 2.2.
Letα∈(1/4,1),p=(1-4α)/3∈(-1,0), and h(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 h(t)>0 for t∈[1,+∞).Proof.
Let(2.7)h1(t)=t3-ph(t).
Then simple computations lead to
(2.8)h1(1)=1627(1-α)(80α2+110α-1)>0,(2.9)h1′(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,h1′(1)=3227(1-α)(-16α3+38α2+176α-9)>0,(2.10)h1′′(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,h1′′(1)=881(1-α)(-128α4+896α3+288α2+5294α-437)>0,(2.11)h1′′′(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,h1′′′(1)=881(1-α)(576α4+3872α3+660α2+6612α-785)>0,(2.12)h1(4)(t)=-4p(1-p)t-p-3h2(t),
where
(2.13)h2(t)=(1-α)(7-4p)(3-p)(2-p)t2+(1-α)p(2-p)(p+1)t+α(p+1)2(p+2),h2(1)=427(1-α)(96α3+232α2+388α+175)>0,h2′(t)=2(1-α)(7-4p)(3-p)(2-p)t+(1-α)p(2-p)(p+1)≥h2′(1)=49(1-α)(5+4α)(12α2+31α+23)>0
for t∈[1,+∞).
From inequalities (2.13) we clearly see that h2(t)>0 for t∈[1,+∞). Then (2.12) leads to the conclusion that h1′′′(t) is strictly increasing in [1,+∞).
Therefore, Lemma2.2 follows from (2.7)–(2.11) and the monotonicity of h1′′′(t).Lemma 2.3.
Letα∈(0,1),p=(1-4α)/3,λ0=3345/80-11/16=0.00903…, and f(t)=2α(1-tp+1)tlog2t+(1-α)(1+tp-1)(1+t)2tlogt+(1-α)(1+t)2(1-t)(tp+1). Then the following two statements are true:(1)
ifα∈(1/4,1), then f(t)>0 for t∈(1,+∞);(2)
ifα∈(0,λ0], then f(t)<0 for t∈(1,+∞).Proof.
Letf1(t)=t-pf′′(t), f2(t)=tp+2f1′(t), f3(t)=t4-pf2′′′(t), f4(t)=tp+2f3′′′(t) and f5(t)=t4-pf4′′′(t). Then simple computations lead to
(2.14)f(1)=0,(2.15)f′(t)=2α[1-(p+2)tp+1]log⁡2⁡t+[(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]log⁡⁡t-(1-α)[(p+3)tp+2+(p+1)tp+1-(p+3)tp-(p+1)tp-1+2t2-2],f′(1)=0,(2.16)f1(t)=-2α(p+1)(p+2)log⁡2⁡t+[(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+4αpt-1-p]log⁡⁡t-(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,f1(1)=0,(2.17)f2(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)]log⁡⁡t-(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,f2(1)=0,(2.18)f2′(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]log⁡⁡t-(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(1+p)t-1+4(1-α)(1-2p),f2′(1)=0,(2.19)f2′′(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)]log⁡⁡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)pt-1+4α(1+p)t-2+4(1-α)(7-4p).
(1) Ifα∈(1/4,1), then from (2.19) we note that
(2.20)f2′′(t)=g(t)log⁡⁡t+h(t),
where g(t) and h(t) are defined as in Lemmas 2.1 and 2.2, respectively.
Lemmas2.1 and 2.2 together with (2.20) imply that f2′(t) is strictly increasing in [1,+∞). Therefore, f(t)>0 for t∈(1,+∞) follows from (2.14)–(2.18) and the monotonicity of f2′(t).
(2) Ifα∈(0,λ0], then from (2.19) we have
(2.21)f2′′(1)=1627(1-α)(80α2+110α-1)=128027(1-α)(α-λ0)(α+334580+1116)≤0,(2.22)f3(t)=[4αp(1-p)(p+1)2(p+2)t2-2(1-α)p2(1-p2)(2-p)t-2(1-α)p(1-p)2(2-p)(3-p)]log⁡⁡t-(1-α)p(p+1)(p+2)2(p+3)t3+(p+1)(p4-αp4-22αp3+2p3-27αp2-p2-2p+18αp+8α)t2+(1-α)p(p4-12p3+15p2+8p-8)t+(1-α)(1-p)(p4+4p3-35p2+50p-12)+12(1-α)(1-p)t3-p+4p(1-α)t2-p-8α(1+p)t1-p,f3(1)=3281(1-α)(1-4α)(80α2+110α-1)=1024081(1-α)(14-α)(α-λ0)(α+334580+1116)≤0,(2.23)f3′(t)=[8αp(1-p)(p+1)2(p+2)t-2(1-α)p2(1-p2)(2-p)]log⁡⁡t-3(1-α)p(p+1)(p+2)2(p+3)t-2(p+1)×(3αp4-p4+26αp3-2p3+25αp2+p2-22αp+2p-8α)t-(1-α)p(p4+8p3-17p2-4p+8)-2(1-α)p(1-p)2(2-p)(3-p)t-1+12(1-α)(1-p)(3-p)t2-p+4(1-α)p(2-p)t1-p-8α(1-p2)t-p,f3′(1)=8243(1-α)(3328α4+128α3-7248α2+7118α-167)<8243(1-α)(3328λ04+128λ03+7118λ0-167)<8243(1-α)[3328×(0.01)4+128×(0.01)3+7118×0.01-167(0.01)4]<0,(2.24)f3′′(t)=8αp(1-p)(p+1)2(p+2)log⁡⁡t-6(1-α)p(p+1)(p+2)2(p+3)t-2(1-α)p2(1-p2)(2-p)t-1+2(1-α)p(1-p)2(2-p)(3-p)t-2+12(1-α)(1-p)(2-p)(3-p)t1-p+4(1-α)p(1-p)(2-p)t-p+8αp(1-p2)t-1-p-2(p+1)×(7αp4-p4+34αp3-2p3+21αp2+p2-30αp+2p-8α),f3′′(1)=8243(1-α)(7-4α)(256α4-64α3-1152α2+2066α-53)<8243(1-α)(7-4α)(256λ04+2066λ0-53)<8243(1-α)[7-4α)(256×(0.01)4+2066×0.01-53]<0,(2.25)f4(t)=-6(1-α)p(p+1)(p+2)2(p+3)tp+2+8αp(1-p)(p+1)2(p+2)tp+1+2(1-α)p2(1-p2)(2-p)tp-4(1-α)p(1-p)2(2-p)(3-p)tp-1+12(1-α)(1-p)2(2-p)(3-p)t2-4(1-α)p2(1-p)(2-p)t-8αp(1-p)(1+p)2,f4(1)=8243(1-α)(-1024α4+21952α3-10968α2+13474α-835)<8243(1-α)(21952λ03+13474λ0-835)<8243(1-α)[21952×(0.01)3+13474×0.01-835]<0,(2.26)f4′(t)=-6(1-α)p(p+1)(p+2)3(p+3)tp+1+8αp(1-p)(p+1)3(p+2)tp+2(1-α)p3(1-p2)(2-p)tp-1+4(1-α)p(1-p)3(2-p)(3-p)tp-2+24(1-α)(1-p)2(2-p)(3-p)t-4(1-α)p2(1-p)(2-p),f4′(1)=8729(1-α)(7-4α)(-1024α4+21952α3-10968α2+13474α-835)<8729(1-α)(7-4α)(21952λ03+13474λ0-835)<8729(1-α)(7-4α)[21952×(0.01)3+13474×0.01-835]<0,(2.27)f4′′(t)=-6(1-α)p(p+1)2(p+2)3(p+3)tp+8αp2(1-p)(p+1)3(p+2)tp-1-2(1-α)p3(1+p)(1-p)2(2-p)tp-2-4(1-α)p(1-p)3(2-p)2(3-p)tp-3+24(1-α)(1-p)2(2-p)(3-p),f4′′(1)=322187(1-α)(-4096α6+136320α5-241440α4+383672α3-209850α2+100113α-7255)<322187(1-α)(136320λ05+383672λ03+100113λ0-7255)<322187(1-α)[136320×(0.01)5+383672×(0.01)3+100113×0.01-7255]<0,(2.28)f5(t)=-6(1-α)p2(p+1)2(p+2)3(p+3)t3-8αp2(1-p)2(p+1)3(p+2)t2+2(1-α)p3(1+p)(1-p)2(2-p)2t+4(1-α)p(1-p)3(2-p)2(3-p)2,f5(1)=326561(1-α)(1-4α)(-4096α6+173568α5-190368α4+439136α3-191370α2+96723α-8665)<326561(1-α)(1-4α)(173568λ05+439136λ03+96723λ0-8665)<326561(1-α)(1-4α)[173568×(0.01)5+439136×(0.01)3+96723×0.01-8665]<0,(2.29)f5′(t)=-18(1-α)p2(p+1)2(p+2)3(p+3)t2-16αp2(1-p)2(p+1)3(p+2)t+2(1-α)p3(1+p)(1-p)2(2-p)2,f5′(1)=326561(1-α)2(1-4α)2(-17408α4+69920α3-119136α2+95282α-30845)<326561(1-α)(1-4α)2(69920λ03+95282λ0-30845)<326561(1-α)(1-4α)2(69920×(0.01)3+95282×0.01-30845)<0,(2.30)f5′′(t)=-36(1-α)p2(p+1)2(p+2)3(p+3)t-16αp2(1-p)2(p+1)3(p+2),f5′′(1)=1286561(1-α)3(1-4α)2(7-4α)(160α3-1856α2+3370α-2205)<1286561(1-α)3(1-4α)2(7-4α)(160λ03+3370λ0-2205)<1286561(1-α)3(1-4α)2(7-4α)(160×(0.01)3+3370×0.01-2205)<0,f5′′′(t)=-36(1-α)p2(p+1)2(p+2)3(p+3)=-128729(5-2α)(1-4α)2(1-α)3(7-4α)3<0.
Inequalities (2.30) imply that f5′(t) is strictly decreasing in [1,+∞). Then (2.29) leads to the conclusion that f5(t) is strictly decreasing in [1,+∞).
It follows from (2.28) and the monotonicity of f5(t) that f4′′(t) is strictly decreasing in [1,+∞). Then inequalities (2.25)–(2.27) lead to the conclusion that f4(t)<0 for t∈[1,+∞). Thus, f3′′(t) is strictly decreasing in [1,+∞).
From inequalities (2.22)–(2.24) and the monotonicity of f3′′(t) we clearly see that f3(t)<0 for t∈(1,+∞). Thus, f2′′(t) is strictly decreasing in [1,+∞).
It follows from (2.17) and (2.18) and inequality (2.21) together with the monotonicity of f2′′(t) that f2(t)<0 for t∈(1,+∞), which implies that f1(t) is strictly decreasing in [1,+∞).
Therefore,f(t)<0 for t∈(1,+∞) follows from (2.14)–(2.16) and the monotonicity of f1(t).Lemma 2.4.
3t4-4t(2t2-t+2)logt-3>0 for t>1.Proof.
Let(2.31)J(t)=3t4-4t(2t2-t+2)log⁡⁡t-3.
Then simple computations lead to
(2.32)J(1)=0,J′(t)=4(3t3-2t2+t-2)-8(3t2-t+1)log⁡⁡t,J′(1)=0,J′′(t)=4tJ1(t),
where J1(t)=9t3-10t2+3t-2-2(6t-1)tlogt,
(2.33)J1(1)=0,(2.34)J1′(t)=27t2-32t+5-2(12t-1)log⁡⁡t,J1′(1)=0,(2.35)J1′′(t)=2tJ2(t),
where J2(t)=27t2-12tlogt-28t+1,
(2.36)J2(1)=0,J2′(t)=54t-12log⁡⁡t-40>0
for t>1.Therefore, Lemma2.4 follows from (2.31)–(2.36).


## 3. Proof of Theorem1.1

Proof of Theorem1.1.
For alla,b>0 with a≠b, we first prove that
(3.1)αH(a,b)+(1-α)L(a,b)>M(1-4α)/3(a,b)
for α∈[1/4,1),
(3.2)αH(a,b)+(1-α)L(a,b)<M(1-4α)/3(a,b)
for α∈(0,3345/80-11/16).
Without loss of generality, we assume thata>b,t=a/b>1 and p=(1-4α)/3. We divide the proof into two cases.
Case 1 (α=1/4).
Letx=t=a/b>1. Then we clearly see that
(3.3)αH(a,b)+(1-α)L(a,b)-M(1-4α)/3(a,b)=14[H(a,b)+3L(a,b)]-ab=b[3x4-4x(2x2-x+2)log⁡⁡x-3]8(x2+1)log⁡⁡x.
Therefore, inequality (3.1) follows from (3.3) and Lemma 2.4.
Case 2 (α∈(0,3345/80-11/16)∪(1/4,1)).
Then we have(3.4)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(3.5)F(t)=log⁡⁡[2αtt+1+(1-α)(t-1)log⁡t]-1plog⁡⁡(tp+12).
Then simple computations lead to
(3.6)lim⁡t→1⁡F(t)=0,F′(t)=f(t)t(t+1)(tp+1)[2αtlog⁡⁡t+(1-α)(t2-1)]log⁡⁡t,
where f(t) is defined as in Lemma 2.3.
Ifα∈(1/4,1), then inequality (3.1) follows from (3.4)–(3.6) and Lemma 2.3(1). If α∈(0,3345/80-11/16), then inequality (3.2) follows from (3.4)–(3.6) and Lemma 2.3(2).
Next, we prove that the parameter(1-4α)/3 in inequalities (3.1) and (3.2) is the best possible.
For anyα∈(0,3345/80-11/16)∪(1/4,1), p≠0, and x>0, one has
(3.7)αH(1,1+x)+(1-α)L(1,1+x)-Mp(1,1+x)=Q(x)21/p(1+x/2)log⁡⁡(1+x),
where Q(x)=21/pα(1+x)log(1+x)+21/p(1-α)x(1+x/2)-(1+x/2)log(1+x)[1+(1+x)p]1/p.
Lettingx→0 and making use of Taylor expansion, we get
(3.8)Q(x)=21/p8(1-4α3-p)x3+o(x3).
Ifα∈[1/4,1) and p>(1-4α)/3, then (3.7) and (3.8) imply that there exists δ1=δ1(α,p)>0 such that αH(1,1+x)+(1-α)L(1,1+x)<Mp(1,1+x) for x∈(0,δ1). If α∈(0,3345/80-11/16) and p<(1-4α)/3, then (3.7) and (3.8) imply that there exists δ2=δ2(α,p)>0 such that αH(1,1+x)+(1-α)L(1,1+x)>Mp(1,1+x) for x∈(0,δ2).
Finally, we prove that there exista1,b1,a2,b2>0 with a1≠b1 and a2≠b2 such that αH(a1,b1)+(1-α)L(a1,b1)<M(1-4α)/3(a1,b1) and αH(a2,b2)+(1-α)L(a2,b2)>M(1-4α)/3(a2,b2) for any 3345/80-11/16<α<1/4.
If3345/80-11/16<α<1/4, then from the expression of f2′′(1) in (2.21) we clearly see that f2′′(1)>0, which leads to the conclusion that there exists λ>1 such that
(3.9)f2′′(t)>0
for t∈[1,λ).
From (2.14)–(2.18) and inequality (3.9) we know that
(3.10)f(t)>0
for t∈(1,λ). Equations (3.4)–(3.6) and inequality (3.10) lead to the conclusion that αH(a,b)+(1-α)L(a,b)>M(1-4α)/3(a,b) for all a/b∈(1,λ).
On the other hand, simple computations lead to(3.11)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-α)(1-1/x)/log⁡⁡x=+∞.
Equation (3.11) implies that there exists X=X(α)>1 such that αH(a,b)+(1-α)L(a,b)<M(1-4α)/3(a,b) for all a/b∈(X,+∞).



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*Source: 471096-2012-02-22.xml*