# Optimal Lehmer Mean Bounds for the Combinations of Identric and Logarithmic Means

Authors: Xu-Hui Shen; Wei-Ming Gong; Yu-Ming Chu
Journal: Chinese Journal of Mathematics
 (2013)

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## Abstract

For anyα∈0,1, we
answer the questions: what are the greatest values p and λ and the least values q and μ, such that the
inequalities Lpa,b<Iαa,bL1-αa,b<Lqa,b and Lλa,b<αIa,b+1-αLa,b<Lμa,b hold
for all a,b>0 with a≠b? Here, Ia,b, La,b, and Lpa,b denote the identric, logarithmic, and pth Lehmer
means of two positive numbers a and b, respectively.


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## Body

## 1. Introduction

Fora,b>0 with a≠b, the logarithmic mean L(a,b) and identric mean I(a,b) are defined by
(1)L(a,b)=a-bloga-logb,(2)I(a,b)=1e(bbaa)1/(b-a),
respectively. In the recent past, both mean values have been the subject of intensive research. In particular, many remarkable inequalities for L and I can be found in the literature [1–19]. In [14, 17, 20], inequalities between L, I, and the classical arithmetic-geometric mean of Gauss are proved. The ratio of identric means leads to the weighted geometric mean
(3)I(a2,b2)I(a,b)=(aabb)1/(a+b),
which has been investigated in [11, 13, 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 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 [5]. In [25, 26], it is shown that L 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,bj), where 0<a1<a2<⋯<an and 0<b1<b2<⋯<bn, is positive for all n≥1.Forp∈ℝ, the power mean Mp(a,b) and Lehmer mean Lp(a,b) of order p of two positive numbers a and b with a≠b are defined by
(4)Mp(a,b)=(ap+bp2)1/p,(5)Lp(a,b)=ap+1+bp+1ap+bp.It is well known thatMp(a,b) and Lp(a,b) are strictly increasing with respect to p∈ℝ for fixed a,b>0 with a≠b. The main properties for Mp and Lp are given in [27–32].LetA(a,b)=(a+b)/2, 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
(6)min{a,b}<H(a,b)=L-1(a,b)=M-1(a,b)<G(a,b)=L-1/2(a,b)=M0(a,b)<L(a,b)<I(a,b)<A(a,b)=L0(a,b)=M1(a,b)<max{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 power means are proved in [2–4, 6, 8, 9, 33]:
(7)M2/3(a,b)<I(a,b)<Mlog2(a,b),(8)M0(a,b)<L(a,b)<M1/3(a,b),M0(a,b)<I(a,b)L(a,b)<M1/2(a,b),12(I(a,b)+L(a,b))<M1/2(a,b)
for all a,b>0 with a≠b.Alzer and Qiu [19] proved that
(9)αA(a,b)+(1-α)G(a,b)<I(a,b)<βA(a,b)+(1-β)G(a,b),Mp(a,b)<12(I(a,b)+L(a,b))
for all a,b>0 with a≠b if and only if α≤2/3, β≥2/e=0.73575…, and p≤log2/(1+log2)=0.40938….The following sharp upper and lower Lehmer mean bounds forL, I, (LI)1/2, and (L+I)/2 are presented in [34]:
(10)L-1/3(a,b)<L(a,b)<L0(a,b),(11)M-1/6(a,b)<I(a,b)<L0(a,b),(12)L-1/4(a,b)<I1/2(a,b)L1/2(a,b)<L0(a,b),L-1/4(a,b)<12(I(a,b)+L(a,b))<L0(a,b)
for all a,b>0 with a≠b.The purpose of this paper is to present the best possible upper and lower Lehmer mean bounds of the productIα(a,b)L1-α(a,b) and the sum αI(a,b)+(1-α)L(a,b) for any α∈(0,1) and all a,b>0 with a≠b.


## 2. Lemmas

In order to establish our main result, we need several lemmas, which we present in this section.Lemma 1.
Ifp≥-1/3, then the following statements are true:(1)
18p2+21p+8>0;(2)
270p3+567p2+384p+104>0;(3)
1134p4+3429p3+3879p2+1950p+448>0;(4)
1458p5+4941p4+6903p3+4809p2+1714p+320>0;(5)
5346p6+18225p5+27297p4+21996p3+9672p2+2264p+320>0;(6)
9p3+33p2+51p+14>0.Proof.
(1) We clearly see that(13)18p2+21p+8>18p2+(-13)×21+8=18p2+1>0.
(2) Let(14)g1(p)=270p3+567p2+384p+104.
Then simple computations lead to
(15)g1(-13)=29>0,g1′(p)=6(135p2+189p+64)>6[135p2+189×(-13)+64]=6(135p2+1)>0.
Therefore, Lemma1(2) follows from (14) and (15).
(3) Let(16)g2(p)=1134p4+3429p3+3879p2+1950p+448.
Then simple computations yield
(17)g2(-13)=116>0,g2′(p)=4536p3+10287p2+7758p+1950,g2′(-13)=339>0,g2′′(p)=13608p2+20574p+7758>13608p2+20574×(-13)+7758=13608p2+900>0.
Therefore, Lemma1(3) follows from (16) and (17).
(4) Let(18)g3(p)=1458p5+4941p4+6903p3+4809p2+1714p+320.
Then simple computations lead to
(19)g3(-13)=2473>0,g3′(p)=7290p4+19764p3+20709p2+9618p+1714,g3′(-13)=167>0,g3′′(p)=29160p3+59292p2+41418p+9618,g3′′(-13)=1320>0,g3′′′(p)=87480p2+118584p+41418>87480p2+41418+118584×(-13)=87480p2+1890>0.
Therefore, Lemma1(4) follows from (18) and (19).
(5) Let(20)g4(p)=5346p6+18225p5+27297p4+21996p3+9672p2+2264p+320.
Then simple computations yield
(21)g4(-13)=2843>0,(22)g4′(p)=32076p5+91125p4+109188p3+65988p2+19344p+2264,(23)g4′(-13)=97>0,(24)g4′′(p)=160380p4+364500p3+327564p2+131976p+19344,(25)g4′′(-13)=228>0,(26)g4′′′(p)=108(5940p3+10125p2+6066p+1222),(27)g4′′′(-13)=11340>0,(28)g4(4)(p)=1944(990p2+1125p+337).
It follows from (28) and the discriminant Δ=(1125)2-4×990×337=-68895<0 of the quadratic function f(p)=990p2+1125p+337 that
(29)g4(4)(p)>0.
Therefore, Lemma1(5) follows from (20)–(27) and (29).
(6) Let(30)g5(p)=9p3+33p2+51p+14.
Then we have
(31)g5(-13)=13>0,g5′(p)=27p2+66p+51>27p2-66×(13)+51=27p2+29>0.
Therefore, Lemma1(6) follows from (30) and (31).Lemma 2.
Suppose thath(t)=(6p+1)t6p+6 -3t6p+4-2(3p+2)t6p+3 +(3p+1)t3p+6-3pt3p+5 -3(p+1)t3p+4+3(p+1)t3p+2 +3pt3p+1-(3p+1)t3p +2(3p+2)t3 +3t2-6p-1. If p∈(-1/3,-1/6), then h(t)>0 for t>1.Proof.
Leth1(t)=h′(t)/t, h2(t)=t4-3ph1′′(t), h3(t)=h2′′′(t)/t, and h4(t)=t4-3ph3′′′(t)/[9p(p+1)(3p+1)(3p+2)2(6p+1)]; then elaborated computations lead to
(32)h(1)=0,(33)h1(t)=6(p+1)(6p+1)t6p+4-6(3p+2)t6p+2-6(2p+1)(3p+2)t6p+1+3(p+2)(3p+1)t3p+4-3p(3p+5)t3p+3-3(p+1)(3p+4)t3p+2+3(p+1)(3p+2)t3p+3p(3p+1)t3p-1-3p(3p+1)t3p-2+6(3p+2)t+6,(34)h1(1)=0,(35)h1′(t)=12(p+1)(3p+2)(6p+1)t6p+3-12(3p+1)(3p+2)t6p+1-6(2p+1)(3p+2)(6p+1)t6p+3(p+2)(3p+1)(3p+4)t3p+3-9p(p+1)(3p+5)t3p+2-3(p+1)(3p+2)(3p+4)t3p+1+9p(p+1)(3p+2)t3p-1+3p(3p-1)(3p+1)t3p-2-3p(3p-2)(3p+1)t3p-3+6(3p+2),(36)h1′(1)=0,(37)h2(t)=36(p+1)(2p+1)(3p+2)(6p+1)t3p+6-12(3p+1)(3p+2)(6p+1)t3p+4-36p(2p+1)(3p+2)(6p+1)t3p+3+9(p+1)(p+2)(3p+1)(3p+4)t6-9p(p+1)(3p+2)(3p+5)t5-3(p+1)(3p+1)(3p+2)(3p+4)t4+9p(p+1)(3p-1)(3p+2)t2+3p(3p-2)(3p-1)(3p+1)t-9p(p-1)(3p-2)(3p+1),(38)h2(1)=12(3p+1)(18p2+21p+8),(39)h2′(t)=108(p+1)(p+2)(2p+1)(3p+2)(6p+1)t3p+5-12(3p+1)(3p+2)(3p+4)(6p+1)t3p+3-108p(p+1)(2p+1)(3p+2)(6p+1)t3p+2+54(p+1)(p+2)(3p+1)(3p+4)t5-45p(p+1)(3p+2)(3p+5)t4-12(p+1)(3p+1)(3p+2)(3p+4)t3+18p(p+1)(3p-1)(3p+2)t+3p(3p-1)(3p-2)(3p+1),(40)h2′(1)=12(3p+1)(9p+7)(18p2+21p+8),(41)h2′′(t)=108(p+1)(p+2)(2p+1)×(3p+2)(3p+5)(6p+1)t3p+4-36(p+1)(3p+1)(3p+2)×(3p+4)(6p+1)t3p+2-108p(p+1)(2p+1)(3p+2)2(6p+1)t3p+1+270(p+1)(p+2)(3p+1)(3p+4)t4-180p(p+1)(3p+2)(3p+5)t3-36(p+1)(3p+1)(3p+2)(3p+4)t2+18p(p+1)(3p-1)(3p+2),(42)h2′′(1)=36(3p+1)(p+1)(270p3+567p2+384p+104),(43)h3(t)=108(p+1)(p+2)(2p+1)(3p+2)×(3p+4)(3p+5)(6p+1)t3p+2-36(p+1)(3p+1)(3p+2)2×(3p+4)(6p+1)t3p-108p(p+1)(2p+1)(3p+1)(3p+2)2×(6p+1)t3p-1+1080(p+1)(p+2)×(3p+1)(3p+4)t2-540p(p+1)(3p+2)(3p+5)t-72(p+1)(3p+1)(3p+2)(3p+4),(44)h3(1)=36(3p+1)(p+1)×(1134p4+3429p3+3879p2+1950p+448),(45)h3′(t)=108(p+1)(p+2)(2p+1)(3p+2)2×(3p+4)(3p+5)(6p+1)t3p+1-108p(p+1)(3p+1)(3p+2)2×(3p+4)(6p+1)t3p-1-108p(p+1)(2p+1)(3p-1)(3p+1)(3p+2)2×(6p+1)t3p-2+2160(p+1)(p+2)×(3p+1)(3p+4)t-540p(p+1)(3p+2)(3p+5),(46)h3′(1)=108(p+1)(3p+1)×(1458p5+4941p4+6903p3+4809p2+1714p+320),(47)h3′′(t)=108(p+1)(p+2)(2p+1)(3p+1)×(3p+2)2(3p+4)(3p+5)(6p+1)t3p-108p(p+1)(3p-1)(3p+1)(3p+2)2(3p+4)×(6p+1)t3p-2-108p(p+1)(2p+1)(3p-2)×(3p-1)(3p+1)(3p+2)2(6p+1)t3p-3+2160(p+1)(p+2)(3p+1)(3p+4),(48)h3′′(1)=108(p+1)(3p+1)×(5346p6+18225p5+27297p4+21996p3+9672p2+2264p+320),(49)h4(t)=36(p+2)(2p+1)(3p+4)(3p+5)t3-12(1-3p)(2-3p)(3p+4)t+36(1-p)(1-3p)(2-3p)(2p+1)>36(p+2)(2p+1)(3p+4)(3p+5)t-12(1-3p)(2-3p)(3p+4)t+36(1-p)(1-3p)(2-3p)(2p+1)=24(3p+4)(9p3+33p2+51p+14)t+36(1-p)(1-3p)(2-3p)(2p+1).
It follows from Lemma1(6) and (49) that
(50)h4(t)>0
for t>1.
From Lemma1(1)–(5) and (38)–(48), we clearly see that
(51)h2(1)>0,h2′(1)>0,h2′′(1)>0,h3(1)>0,h3′(1)>0,h3′′(1)>0.
Therefore, Lemma2 follows from 9p(p+1)(3p+1)(3p+2)2(6p+1)>0 and (32)–(36) together with (50) and (51).Lemma 3.
Inequality[M2/3(a,b)]α[L-1/3(a,b)]1-α>L(α-2)/6(a,b) holds for any α∈(0,1) and all a,b>0 with a≠b.Proof.
Without loss of generality, we assume thata>b. Let t=(a/b)1/3>1 and p=(α-2)/6; then p∈(-1/3,-1/6), α=6p+2, and from (1) and (4) we have
(52)[M2/3(a,b)]α[L-1/3(a,b)]1-α-L(α-2)/6(a,b)=(b[(t3p+1)(t2+1)3p+2(t+1)6p+1-29p+3t6p+1(t3p+3+1)(t2+1)3p+2])×(29p+3t6p+1(t3p+1))-1.
Let
(53)f(t)=log(t3p+1)+(3p+2)log(t2+1)+(6p+1)log(t+1)-(6p+1)logt-log(t3p+3+1)-3(3p+1)log2.
Then simple computations lead to
(54)f(1)=0,(55)f′(t)=h(t)t(t+1)(t2+1)(t3p+1)(t3p+3+1),
where h(t) is defined as in Lemma 2.
From (54) and (55) together with Lemma 2, we clearly see that
(56)f(t)>0
for t>1.
Therefore, Lemma3 follows from (52) and (53) together with (56).


## 3. Main Results

Theorem 4.
InequalityL(α-2)/6(a,b)<Iα(a,b)L1-α(a,b)<L0(a,b) holds for any α∈(0,1) and all a,b>0 with a≠b, and L(α-2)/6(a,b) and L0(a,b) are the best possible lower and upper Lehmer mean bounds for the product Iα(a,b)L1-α(a,b).Proof.
InequalityIα(a,b)L1-α(a,b)<L0(a,b) follows directly from (10) and (11).
For the other inequality, we note that(57)Iα(a,b)L1-α(a,b)-L(α-2)/6(a,b)=Iα(a,b)L1-α(a,b)-[M2/3(a,b)]α[L-1/3(a,b)]1-α+[M2/3(a,b)]α[L-1/3(a,b)]1-α-L(α-2)/6(a,b).
Therefore,Iα(a,b)L1-α(a,b)>L(α-2)/6(a,b) follows from (7) and (10) together with Lemma 3 and (57).
Next, we prove thatL(α-2)/6(a,b) and L0(a,b) are the best possible lower and upper Lehmer mean bounds for the product Iα(a,b)L1-α(a,b).
For any0<ε<1 and x>0, from (1) and (2) we have
(58)limx→+∞Iα(1,x)L1-α(1,x)L-ε(1,x)=limx→+∞[(1-1x)1-αxε(logx)1-α(1-1x)1-αx-ε+1xε-1+1eα((logx)/(x-1)-1)×(1-1x)1-αxε(logx)1-α]=+∞,L(α-2)/6+ε(1,1+x)-Iα(1,1+x)L1-α(1,1+x)=(1+x)(α+4)/6+ε+1(1+x)(α-2)/6+ε+1-eα[(1+x)log(1+x)/x-1]×x1-α[log(1+x)]α.
Letting x→0 and making use of Taylor expansion, one has
(59)(1+x)(α+4)/6+ε+1(1+x)(α-2)/6+ε+1-eα[(1+x)log(1+x)/x-1]×x1-α[log(1+x)]α=1+12x+(α24-112+ε4)x2+o(x2)-[1+α2x+α(18α-16)x2+o(x2)]×[1+(12-α2)x+(18α2-124α-112)x2+o(x2)]=1+12x+(α24-112+ε4)x2-[1+12x+(α24-112)x2]+o(x2)=ε4x2+o(x2).
Equations (58) and (59) imply that for any 0<ε<1, there exist X1=X1(α,ε)>1 and δ1=δ1(α,ε)>0, such that Iα(1,x)L1-α(1,x)>L-ε(1,x) for x∈(X1,+∞) and L(α-2)/6+ε(1,1+x) >Iα(1,1+x)L1-α(1,1+x) for x∈(0,δ1).Theorem 5.
InequalityL(α-2)/6(a,b)<αI(a,b)+(1-α)L(a,b)<L0(a,b) holds for any α∈(0,1) and all a,b>0 with a≠b, and L(α-2)/6(a,b) and L0(a,b) are the best possible lower and upper Lehmer mean bounds for the sum αI(a,b)+(1-α)L(a,b).Proof.
InequalityαI(a,b)+(1-α)L(a,b)<L0(a,b) follows directly from (10) and (11), and inequality L(α-2)/6(a,b)<αI(a,b)+(1-α)L(a,b) follows from Theorem 4.
Next, we prove thatL(α-2)/6(a,b) and L0(a,b) are the best possible lower and upper Lehmer mean bounds for the sum αI(a,b)+(1-α)L(a,b).
For any0<ε<1 and x>0, from (1) and (2), we have
(60)limx→+∞αI(1,x)+(1-α)L(1,x)L-ε(1,x)=limx→+∞[x-ε+1xε-1+1×(αe(logx)/(x-1)xε+(1-α)(1-1/x)logxxε)]=+∞,(61)L(α-2)/6+ε(1,1+x)-αI(1,1+x)-(1-α)L(1,1+x)=(1+x)(α+4)/6+ε+1(1+x)(α-2)/6+ε+1-αe(1+x)log(1+x)/x-1-(1-α)xlog(1+x).
Letting x→0 and making use of Taylor expansion, one has
(62)(1+x)(α+4)/6+ε+1(1+x)(α-2)/6+ε+1-αe(1+x)log(1+x)/x-1-(1-α)xlog(1+x)=1+12x+(α24-112+ε4)x2+o(x2)-α[1+12x-124x2+o(x2)]-(1-α)[1+12x-112x2+o(x2)]=ε4x2+o(x2).
Equations (60)-(62) imply that for any 0<ε<1, there exist X2=X2(α,ε)>1 and δ2=δ2(α,ε)>0, such that αI(1,x)+(1-α)L(1,x)>L-ε(1,x) for x∈(X2,+∞) and L(α-2)/6+ε(1,1+x)>αI(1,1+x)+(1-α)L(1,1+x) for x∈(0,δ2).



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