Vol. 17, No. 2 (2011), pp. 85–96.
OPTIMAL GENERALIZED LOGARITHMIC MEAN
BOUNDS FOR THE GEOMETRIC COMBINATION OF
ARITHMETIC AND HARMONIC MEANS
Bo-Yong Long
1and Yu-Ming Chu
21College of Mathematics Science, Anhui University, Hefei 230039,
P. R. China, lbymxy@163.com
2Department of Mathematics, Huzhou Teachers College, Huzhou 313000,
P. R. China, chuyuming@hutc.zj.cn
Abstract. In this paper, we answer the question: forα ∈ (0,1), what are the greatest valuep=p(α) and least valueq=q(α), such that the double inequality Lp(a, b) ≤Aα(a, b)H1−α(a, b)≤ Lq(a, b) holds for all a, b > 0? where Lp(a, b),
A(a, b), andH(a, b) are thep-th generalized logarithmic, arithmetic, and harmonic means ofaandb, respectively.
Key words: Generalized logarithmic mean, arithmetic mean, harmonic mean.
Abstrak. Dalam paper ini, kami menjawab pertanyaan: untukα∈(0,1), berapa nilai terbesarp=p(α) dan nilai terkecilq=q(α), sehingga ketidaksamaan ganda Lp(a, b)≤Aα(a, b)H1−α(a, b)≤Lq(a, b) dipenuhi untuk semuaa, b >0? dengan
Lp(a, b),A(a, b), danH(a, b) secara berturut-turut adalah rata-rata logaritmik yang diperumum, aritmatik, dan harmonik ke-pdariaandb.
Kata kunci: Rata-rata logaritmik yang diperumum, rata-rata aritmatik, rata-rata harmonik.
2000 Mathematics Subject Classification: 26E60.
86
1. Introduction
Forp∈Rthe generalized logarithmic meanLp(a, b) of two positive numbers
aandbis defined by
It is well-known thatLp(a, b) is continuous and strictly increasing with respect
top∈Rfor fixeda, b >0 witha6=b. In the recent past, the generalized logarithmic mean has been the subject of intensive research. In particular, many remarkable inequalities for Lp can be found in the literature [1, 8, 9, 13, 14, 19, 20, 22, 23].
It might be surprising that the generalized logarithmic mean has applications in economics, physics and even in meteorology [10, 17, 18]. In [10] the authors study a variant of Jensen’s functional equation involving Lp, which appear in a heat
conduction problem.
LetA(a, b) = (a+b)/2,I(a, b) = 1/e(bb/aa)1/(b−a), L(a, b) = (b−a)(lnb− lna),G(a, b) =√ab, andH(a, b) = 2ab/(a+b) be the arithmetic, identric, logarith-mic, geometric, and harmonic means of two positive numbersaand b witha6=b, respectively. Then
min{a, b}< H(a, b)< G(a, b) =L−2(a, b)< L(a, b) =L−1(a, b)
< I(a, b) =L0(a, b)< A(a, b) =L1(a, b)<max{a, b}.
In [7, 11, 21] the authors present bounds forL(a, b) and I(a, b) in terms of
G(a, b) andA(a, b).
Theorem 1.1. For all positive real numbersa andbwith a6=bwe have
A13(a, b)G
The proof of the following Theorem 1.2 can be found in [5].
Theorem 1.2. For all positive real numbersa andbwith a6=bwe have
The main properties of these means are given in [5]. Several authors discussed the relationship of certain means toMp(a, b). The following sharp bounds forL,I,
(IL)1/2and (I+L)/2 in terms of power means are proved in [2, 3, 6, 12, 15, 16].
Theorem 1.3. For all positive real numbersa andbwith a6=bwe have
M0(a, b)< L(a, b)< M1/3(a, b), M2/3(a, b)< I(a, b)< Mln 2(a, b),
M0(a, b)< I1/2(a, b)L1/2(a, b)< M1/2(a, b)
and
1
2[I(a, b) +L(a, b)]< M1/2(a, b).
The following Theorems 1.4-1.6 were established by Alzer and Qiu in [4].
Theorem 1.4. The inequalities
αA(a, b) + (1−α)G(a, b)< I(a, b)< βA(a, b) + (1−β)G(a, b)
hold for all positive real numbersaandb with a6=b if and only if
α≤2/3 and β≥2/e= 0.73575· · · .
Theorem 1.5. Let aandb be real numbers with a6=b. If0< a, b≤e, then
[G(a, b)]A(a,b)<[L(a, b)]I(a,b)<[A(a, b)]G(a,b).
And, ifa, b≥e, then
[A(a, b)]G(a,b)<[I(a, b)]L(a,b)<[G(a, b)]A(a,b).
Theorem 1.6. For all positive real numbersa andbwith a6=bwe have
Mc(a, b)<
1
2(L(a, b) +I(a, b))
with the best possible parameterc= ln 2/(1 + ln 2) = 0.40938· · ·.
It is the aim of this paper to answer the question: forα∈(0,1), what are the greatest valuep=p(α) and least valueq=q(α), such that the double inequality
Lp(a, b)≤Aα(a, b)H1
−α(a, b)
≤Lq(a, b)
88
2. Main Results
Theorem 2.1. Forα∈(0,1) and alla, b >0 we have (1) L6α−5(a, b) =Aα(a, b)H1
−α(a, b) =L
−1/α(a, b)forα= 1/3orα= 1/2; (2) L−1/α(a, b)≤Aα(a, b)H1
−α(a, b)
≤L6α−5(a, b)forα∈(0,1/3)
S
(1/2,1)
and L−1/α(a, b) ≥Aα(a, b)H1
−α(a, b)
≥ L6α−5(a, b) for α ∈(1/3,1/2), with
equality if and only if a=b, and the parameters−1/α and 6α−5 in either case are best possible.
Proof. (1) Ifα= 1/3 orα= 1/2, anda=b, then we clearly see thatL6α−5(a, b) =
Aα(a, b)H1−α(a, b) =L
−1/α(a, b) =a.
Ifα= 1/3 anda6=b, then we have
L6α−5(a, b) =L−1/α(a, b) =L−3(a, b)
=2
1/3(ab)2/3 (a+b)1/3 =A
1/3(a, b)H2/3(a, b) =Aα(a, b)H1−α(a, b).
Ifα= 1/2 anda6=b, then we get
L6α−5(a, b) =L−1/α(a, b) =L−2(a, b)
= (ab)1/2=A1/2(a, b)H1/2(a, b) =Aα(a, b)H1−α(a, b).
(2) Ifα∈(0,1) anda=b, then we clearly see thatL−1/α(a, b) =Aα(a, b)H1−α(a, b) =
L6α−5(a, b). Without loss of generality, we assume thatt=a/b >1 in the following discussion.
Ifα= 2/3, then one has
lnL6α−5(a, b)−ln[Aα(a, b)H1
−α(a, b)]
= lnL−1(a, b)−ln[A2/3(a, b)H1/3(a, b)]
= ln(t−1 lnt )−ln[2
−1/3(1 +t)1/3t1/3]. (1)
Letf1(t) = ln(t−1 lnt)−ln[2
−1/3(1 +t)1/3t1/3], then simple computations lead to
lim
t→1+f1(t) = 0, (2) f′
1(t) =
g1(t)
3t(t−1)(t+ 1) lnt, (3)
whereg1(t) = (t2+ 4t+ 1) lnt−3(t2−1).
g1(1) = 0, (4)
g′ 1(t) =
h1(t)
92
fort >1. Therefore,L−1α(a, b)> Aα(a, b)H1−α(a, b) forα∈(1/3,1/2) and a6=b
follows from (31)-(35) and (38).
Finally, we prove that the parameters−1/αand 6α−5 in either case are best possible.
Firstly, we show that the parameter 6α−5 in either case is best possible. We divide the proof into seven cases.
Case 1. α= 2/3. For anyǫ >0 andx >0 one has
[Aα(1,1 +x)H1−α(1,1 +x)]1+ǫ
−[L6α−5−ǫ(1,1 +x)]1+ǫ
= [A2/3(1,1 +x)H1/3(1,1 +x)]1+ǫ−[L−1−ǫ(1,1 +x)]1+ǫ
= f1(x)
(1 +x)ǫ−1, (39)
wheref1(x) = [(1 +x)ǫ−1](1 +x)(1+ǫ)/3(1 +x/2)(1+ǫ)/3−ǫx(1 +x)ǫ. Lettingx→0 and making use of Taylor expansion we get
f1(x) = ǫ 2(1 +ǫ)
24 x
3+o(x3). (40)
Equations (39) and (40) imply that forǫ >0 there existsδ1=δ1(ǫ)>0, such that L6α−5−ǫ(1,1 +x)< Aα(1,1 +x)H(1−α)(1,1 +x) forα= 2/3 andx∈(0, δ1).
Case 2. α= 5/6. For anyǫ∈(0,1) andx >0 we have
[Aα(1,1 +x)H1−α(1,1 +x)]ǫ−[L
6α−5−ǫ(1,1 +x)]ǫ
= [A5/6(1,1 +x)H1/6(1,1 +x)]ǫ−[L−ǫ(1,1 +x)]ǫ
= f2(x)
(1 +x)1−ǫ−1, (41)
wheref2(x) = [(1 +x)1−ǫ−1](1 +x)ǫ/6(1 +x/2)2ǫ/3−(1−ǫ)x. Lettingx→0 and making using of Taylor expansion we get
f2(x) =ǫ(1−ǫ) 24 x
3+o(x3). (42)
Equations (41) and (42) show that forǫ∈(0,1) there exists δ2 =δ2(ǫ)>0, such that L6α−5−ǫ(1,1 +x)< Aα(1,1 +x)H(1−α)(1,1 +x) for α= 5/6 and x∈
(0, δ2).
Case 3. α∈(0,1/3). For anyǫ >0 andx >0 one has
[Aα(1,1 +x)H1−α(1,1 +x)]5+ǫ−6α
−[L6α−5−ǫ(1,1 +x)]5+ǫ−6α
= f3(x)
[(1 +x)4+ǫ−6α−1](1 +x
2)(1
−2α)(5+ǫ−6α), (43) wheref3(x) = [(1+x)4+ǫ−6α−1](1+x)(1−α)(5+ǫ−6α)−(4+ǫ−6α)x(1+x)4+ǫ−6α(1+
Lettingx→0 and making use of Taylor expansion we get
f3(x) =ǫ(4 +ǫ−6α)(5 +ǫ−6α)
24 x
3+o(x3). (44)
Equations (43) and (44) imply that for anyα∈(0,1/3) andǫ >0 there exists
δ3 =δ3(ǫ, α)>0, such that L6α−5−ǫ(1,1 +x)< Aα(1,1 +x)H(1−α)(1,1 +x) for
x∈(0, δ3).
Case 4. α∈(1/3,1/2). For anyǫ∈(0,4−6α) andx >0 we get
[L6α−5+ǫ(1,1 +x)]5
−6α−ǫ
−[Aα(1,1 +x)H1−α(1,1 +x)]5−6α−ǫ
= f4(x)
[(1 +x)4−6α−ǫ−1](1 +x
2)(1
−2α)(5−6α−ǫ), (45) wheref4(x) = (4−6α−ǫ)x(1 +x)4−6α−ǫ(1 +x/2)(1−2α)(5−6α−ǫ)−[(1 +x)4−6α−ǫ−
1](1 +x)(1−α)(5−6α−ǫ)
Lettingx→0 and making using Taylor expansion one has
f4(x) =ǫ(4 +ǫ−6α)(5 +ǫ−6α)
24 x
3+o(x3). (46)
Equations (45) and (46) imply that for anyα ∈ (1/3,1/2) and ǫ ∈ (0,4− 6α) there exists δ4 = δ4(ǫ, α) > 0, such that L6α−5+ǫ(1,1 +x) > Aα(1,1 +
x)H(1−α)(1,1 +x) forx∈(0, δ4).
Case 5. α∈(1/2,2/3). For anyǫ >0 andx >0 we have
[Aα(1,1 +x)H1−α(1,1 +x)]5−6α+ǫ
−[L6α−5−ǫ(1,1 +x)]5−6α+ǫ
= f5(x)
(1 +x)4−6α+ǫ−1, (47)
wheref5(x) = [(1 +x)4−6α+ǫ−1](1 +x/2)(2α−1)(5−6α+ǫ)(1 +x)(1−α)(5−6α+ǫ)−(4− 6α+ǫ)x(1 +x)4−6α+ǫ.
Lettingx→0 and making use of Taylor expansion we get
f5(x) = ǫ(4 +ǫ−6α)
2(5 +ǫ−6α)
24 x
3+o(x3). (48)
Equations (47) and (48) imply that for anyα∈ (1/2,2/3) and ǫ > 0 there existsδ5=δ5(ǫ, α)>0, such thatL6α−5−ǫ(1,1 +x)< Aα(1,1 +x)H(1−α)(1,1 +x)
forx∈(0, δ5).
Case 6. α∈(2/3,5/6). For anyǫ∈(0,6α−4) andx >0 one has
[Aα(1,1 +x)H1−α(1,1 +x)]5−6α+ǫ
−[L6α−5−ǫ(1,1 +x)]5
−6α+ǫ
= f6(x)
(1 +x)6α−4−ǫ−1, (49)
94
Lettingx→0 and making using Taylor expansion we obtain
f6(x) =ǫ(5 +ǫ−6α)(6α−4−ǫ)
Lettingx→0 and making use of Taylor expansion we get
f7(x) =ǫ(6α−5−ǫ)(6α−4−ǫ)
Secondly, we prove that the parameter−1/αin either case is the best possible. The proof is divided into two cases.
lim
t→+∞{2
1−2α(1 +1
t)
2α−1
−t−ǫα2
1+ǫα[(
1
α−1 +ǫ)(1−
1
t)
1−t−(1 α−1+ǫ)
]1+αǫα}
= 21−α>0. (56)
From (55) and (56) we clearly see that for any α ∈ (1/3,1/2) and ǫ > 0 there exists T2 = T2(ǫ, α) > 1, such that L−1/α−ǫ(1, t) < Aα(1, t)H1−α(1, t) for
t∈(T2,∞).
Acknowledgement The authors wish to thank the anonymous referees for their careful reading of the manuscript and their fruitful comments and suggestions. This research was supported by the Natural Science Foundation of China under Grants 11071069 and 11171307, and the Innovation Team Foundation of the Department of Education of Zhejiang Province under Grant T200924.
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