# Optimal Inequalities among Various Means of Two Arguments

Authors: Ming-yu Shi; Yu-ming Chu; Yue-ping Jiang
Journal: Abstract and Applied Analysis
 (2009)

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## Abstract

We establish two optimal inequalities among power meanMp(a,b)=(ap/2+bp/2)1/p, arithmetic mean A(a,b)=(a+b)/2, logarithmic mean L(a,b)=(a−b)/(loga−logb), and geometric mean G(a,b)=ab.


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## Body

## 1. Introduction

Forp∈R, the power mean Mp(a,b) of order p of two positive numbers a and b is defined by(1.1)Mp(a,b)={(ap+bp2)1/p,p≠0,ab,p=0.In the recent past, the power meanMp(a,b) has been the subject of intensive research. In particular, many remarkable inequalities for the mean can be found in literature [1–11]. It is well-known that Mp(a,b) is continuous and strictly increasing with respect to p∈R for fixed a and b.If we denote by(1.2)I(a,b)={1e(aabb)1/(a-b),b≠a,a,b=a,L(a,b)={b-alogb-loga,b≠a,a,b=a,A(a,b)=(a+b)/2, G(a,b)=ab and H(a,b)=2ab/(a+b) the identric mean, logarithmic mean, arithmetic mean, geometric mean and the harmonic mean, of two positive real numbers a and b, respectively, then(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}
with equality if and only if b=a in each inequality.In [12], Alzer and Janous established the following sharp double inequality (see also [13, page 350]):(1.4)Mlog2/log3(a,b)≤23A(a,b)+13G(a,b)≤M2/3(a,b)
for all real numbers a,b>0.In [14], Mao proved(1.5)M1/3(a,b)≤13A(a,b)+23G(a,b)≤M1/2(a,b)
for all real numbers a,b>0, and the constant 1/3 in the left side inequality cannot be improved.In [15–17], the authors presented the bounds for L and I in terms of A and G as follows:(1.6)G2/3(a,b)A1/3(a,b)<L(a,b)<13A(a,b)+23G(a,b),13G(a,b)+23A(a,b)<I(a,b)
for all a,b>0 with a≠b.Alzer [18] proved(1.7)G(a,b)A(a,b)<L(a,b)I(a,b)<12(L(a,b)+I(a,b))<12(G(a,b)+A(a,b)).In [5], Alzer and Qiu established(1.8)αA(a,b)+(1-α)G(a,b)<I(a,b)<βA(a,b)+(1-β)G(a,b)
for α≤2/3, β≥2/e=0.73575… and a,b>0 with a≠b.The main purpose of this paper is to present the optimal bounds forAα(a,b)L1-α(a,b) and Gα(a,b)L1-α(a,b) for all α∈(0,1) in terms of the power mean Mp(a,b). Moreover, two optimal inequalities among A(a,b), G(a,b), L(a,b) and Mp(a,b) are proved.


## 2. Lemmas

In order to establish our main results we need two inequalities, which we present in this section.Lemma 2.1.
Ifr∈(0,1), then
(2.1)g(t)=-[t2+t(1+2r)/3+(1-2r)(t(1+2r)/3+1+t)]logt+(1-r)(t(1+2r)/3+2-t(1+2r)/3+t2-1)>0
for t∈(1,+∞).Proof.
Letp=(1+2r)/3, g1(t)=t1-pg′(t), g2(t)=tpg1′(t), g3(t)=t1-pg2′(t), g4(t)=tpg3′(t), g5(t)=t3-pg4′(t), g6(t)=tpg5′(t), g7(t)=t1-pg6′(t), and g8(t)=tpg7′(t), then simple computation yields
(2.2)g(1)=0,(2.3)g1(t)=-[2t2-p+(1-2r)t1-p+(1-2r)(p+1)t+p]logt+(1-r)(p+2)t2-(1-2r)t+(1-2r)t2-p-(1-2r)t1-p-p(1-r)-1,g1(1)=0,(2.4)g2(t)=-[2(2-p)t+(1-2r)(p+1)tp+(1-2r)(1-p)]logt+2(1-r)(2+p)t1+p-(1-2r)(p+2)tp-ptp-1+(2rp-4r-p)t-(1-2r)(2-p),g2(1)=0,(2.5)g3(t)=-[2(2-p)t1-p+p(p+1)(1-2r)]logt+p(1-p)t-1+(2rp-4r+p-4)t1-p-(1-2r)(1-p)t-p+2(1-r)(2+p)(1+p)t-(1-2r)(p2+3p+1),g3(1)=6p-2-4r=0,(2.6)g4(t)=-2(1-p)(2-p)logt+2(1-r)(2+p)(1+p)tp-p(1+p)(1-2r)tp-1+p(p-1)tp-2+p(1-p)(1-2r)t-1-p2-2rp2+6rp-4r+7p-8,g4(1)=12p-4-8r=0,(2.7)g5(t)=2p(1-r)(p+1)(p+2)t2+p(1-p2)(1-2r)t-2(2-p)(1-p)t2-p-p(1-p)(1-2r)t1-p+p(p-1)(p-2),g5(1)=2p3+2(1-4r)p2+4(3-r)p-4=827(-10r3-15r2+24r+1)>0,(2.8)g6(t)=4p(p+1)(p+2)(1-r)tp+1+p(1-p2)(1-2r)tp-2(2-p)2(1-p)t-p(1-p)2(1-2r),g6(1)=4p3+4(1-4r)p2+8(3-r)p-8=1627(-10r3-15r2+24r+1)>0,(2.9)g7(t)=4p(2+p)(1+p)2(1-r)t-2(1-p)(2-p)2t1-p+p2(1-p2)(1-2r),g7(1)=(3-2r)p4+2(9-8r)p3+11(1-2r)p2+8(3-r)p-8=181(-32r5-400r4-888r3-412r2+1576r+156)>0,(2.10)g8(t)=4p(1+p)2(2+p)(1-r)tp-2(1-p)2(2-p)2,(2.11)g8(1)=2(1-2r)p4+4(7-4r)p3-2(3+10r)p2+8(4-r)p-8=881(-8r5-60r4-82r3-79r2+198r+31)>0.
From (2.10) we clearly see that g8(t) is strictly increasing in (1,+∞). Therefore, Lemma 2.1 follows from (2.2)–(2.9) and (2.11) together with the monotonicity of g8(t).Lemma 2.2.
Ifr∈(0,1), then
(2.12)g(t)=[rt(1-r)/3+1+(r-2)t(1-r)/3+(r-2)t+r]logt+2(1-r)(t(1-r)/3+1-t(1-r)/3+t-1)>0
for t∈(1,+∞).Proof.
Letp=(1-r)/3, g1(t)=t1-pg′(t), g2(t)=tpg1′(t), g3(t)=t1-pg2′(t), g4(t)=tp+2g3′(t), and g5(t)=t1-pg4′(t), then simple computation leads to
(2.13)g(1)=0,(2.14)g1(t)=[r(p+1)t+(r-2)t1-p+p(r-2)]logt-rt1-p+rt-p+(2+2p-2pr-r)t+2pr-2p+r-2,g1(1)=0,(2.15)g2(t)=[r(p+1)tp+(1-p)(r-2)]logt+(2p+2-rp)tp+p(r-2)tp-1-rpt-1+pr-2,g2(1)=0,(2.16)g3(t)=pr(p+1)logt+(r-2)(1-p)t-p+rpt-p-1+p(p-1)(r-2)t-1+2p2-rp2+pr+2p+r,g3(1)=6p+2r-2=0,(2.17)g4(t)=p(p+1)rtp+1+p(1-p)(r-2)tp-p(1-p)(r-2)t-p(p+1)r,g4(1)=0,(2.18)g5(t)=rp(1+p)2t+p(1-p)(2-r)t1-p+p2(1-p)(r-2),(2.19)g5(1)=2p(p2+2rp-2p+1)=2(1-r)27(-5r2+10r+4)>0.
From (2.18) we clearly see that g5(t) is strictly increasing in (1,+∞). Therefore, Lemma 2.2 follows from (2.13)–(2.17) and (2.19) together with the monotonicity of g5(t).


## 3. Main Results

Theorem 3.1.
Ifα∈(0,1), then
(3.1)Aα(a,b)L1-α(a,b)≤M(1+2α)/3(a,b)
with equality if and only if a=b, and the parameter (1+2α)/3 cannot be improved.Proof.
Ifa=b, then we clearly see that
(3.2)a=Aα(a,b)L1-α(a,b)=M(1+2α)/3(a,b)=b.
Ifa≠b, then without loss of generality we assume that a>b and let t=a/b>1; hence elementary calculation yields
(3.3)M(1+2α)/3(a,b)-Aα(a,b)L1-α(a,b)=b[(t(1+2α)/3+12)3/(1+2α)-(1+t2)α(t-1logt)1-α].
Let
(3.4)f(t)=31+2αlog(t(1+2α)/3+1)-31+2αlog2-αlogt+12+αlog2-(1-α)log(t-1)+(1-α)log(logt),
then
(3.5)limt→1f(t)=0,(3.6)f′(t)=g(t)t(t+1)(t-1)(t(1+2α)/3+1)logt,
where
(3.7)g(t)=-[t2+t(1+2α)/3+(1-2α)(t(1+2α)/3+1+t)]logt+(1-α)(t(1+2α)/3+2-t(1+2α)/3+t2-1).
From Lemma2.1 and (3.6) we know that
(3.8)f′(t)>0
for t∈(1,+∞).
Therefore, we get
(3.9)M(1+2α)/3(a,b)>Aα(a,b)L1-α(a,b)
for a>b that follows from (3.3)–(3.5) and (3.8).
Next, we prove that the parameter(1+2α)/3 cannot be improved.
For any0<ε<(1+2α)/3, let 0<t<1, then (1.1) leads to
(3.10)Aα(t+1,1)L1-α(t+1,1)-M(1+2α)/3-ε(t+1,1)=(1+t2)α·[tlog(t+1)]1-α-[(t+1)(1+2α)/3-ε+12]1/((1+2α)/3-ε)=f1(t)[log(1+t)]1-α,
where
(3.11)f1(t)=(1+t2)α·t1-α-[(t+1)(1+2α-3ε)/3+12]3/(1+2α-3ε)[log(1+t)]1-α.
Making use of the Taylor expansion we get
(3.12)f1(t)=[ε8t2+o(t2)]·t1-α.
Equations (3.10) and (3.12) imply that for any α∈(0,1) and 0<ε<(1+2α)/3, there exists 0<δ1(ε,α)<1, such that
(3.13)Aα(1,1+t)L1-α(1,1+t)>M(1+2α)/3-ε(1,1+t)
for t∈(0,δ1).Remark 3.2.
For any0<α<1 we have
(3.14)M0(a,b)≤Aα(a,b)L1-α(a,b)
for all a,b>0, with equality if and only if a=b, and the parameter 0 in the lower bound cannot be improved.
In fact, ifa=b, then we clearly see that M0(a,b)=Aα(a,b)L1-α(a,b)=a. If a≠b, then M0(a,b)<Aα(a,b)L1-α(a,b) follows from M0(a,b)<L(a,b)<A(a,b).
Next, we prove that the parameter0 in the lower bound cannot be improved.
For anyε>0, we have
(3.15)limt→∞Mε(t,1)Aα(t,1)L1-α(t,1)=limt→∞((tε+1)/2)1/ε((t+1)/2)α((t-1)/logt)1-α=limt→∞((1+t-ε)/2)1/ε(logt)1-α((1+t-1)/2)α(1-t-1)1-α=+∞.
Equation (3.15) implies that for any ε>0, there exists T1=T1(ε)>1, such that
(3.16)Aα(t,1)L1-α(t,1)<Mε(t,1)
for t∈(T1,+∞).Theorem 3.3.
Ifα∈(0,1), then
(3.17)Gα(a,b)L1-α(a,b)≤M(1-α)/3(a,b)
for all a,b>0, with equality if and only if a=b, and the parameter (1-α)/3 cannot be improved.Proof.
Ifa=b, then we clearly see that
(3.18)a=Gα(a,b)L1-α(a,b)=M(1-α)/3(a,b)=b.
Ifa≠b, then without loss of generality, we assume that a>b, let t=a/b>1; hence simple computation leads to
(3.19)M(1-α)/3(a,b)-Gα(a,b)L1-α(a,b)=b[(t(1-α)/3+12)3/(1-α)-tα/2(t-1logt)1-α].
Let
(3.20)h(t)=31-αlog(t(1-α)/3+1)-31-αlog2-αlogt+12+αlog2-(1-α)log(t-1)+(1-α)log(logt),
then
(3.21)limt→1h(t)=0,(3.22)h′(t)=g(t)2t(t-1)(t(1-α)/3+1)logt,
where
(3.23)g(t)=[αt(1-α)/3+1+(α-2)t(1-α)/3+(α-2)t+α]logt+2(1-α)(t(1-α)/3+1-t(1-α)/3+t-1).
From Lemma2.2 and (3.22) we know that
(3.24)h′(t)>0
for t∈(1,+∞).
Therefore, we get
(3.25)M(1-α)/3(a,b)>Gα(a,b)L1-α(a,b)
for a>b that follows from (3.19)–(3.21) and (3.24).
Next, we prove that the constants(1-α)/3 cannot be improved.
For any0<ε<(1-α)/3, let 0<t<1, then (1.1) leads to
(3.26)Gα(t+1,1)L1-α(t+1,1)-M(1-α)/3-ε(t+1,1)=(1+t)α/2·[tlog(t+1)]1-α-[(t+1)(1-α)/3-ε+12]1/((1-α)/3-ε)=h1(t)[log(1+t)]1-α,
where
(3.27)h1(t)=(1+t)α/2·t1-α-[(t+1)(1-α-3ε)/3+12]3/(1-α-3ε)[log(1+t)]1-α.
Making use of the Taylor expansion we get
(3.28)h1(t)=[ε8t2+o(t2)]·t1-α.
Equations (3.26) and (3.28) imply that for any α∈(0,1) and 0<ε<(1-α)/3, there exists 0<δ2(ε,α)<1, such that
(3.29)Aα(1,1+t)L1-α(1,1+t)>M(1-α)/3-ε(1,1+t)
for t∈(0,δ2).Remark 3.4.
For anyε>0, we have
(3.30)limt→∞Mε(t,1)Gα(t,1)L1-α(t,1)=limt→∞((tε+1)/2)1/εtα/2((t-1)/logt)1-α=limt→∞((1+t-ε)/2)1/εtα/2(logt)1-α(1-t-1)1-α=+∞.
Therefore, (3.30) implies that inequality
(3.31)M0(a,b)≤Gα(a,b)L1-α(a,b)
holds for all α∈(0,1) and a,b>0, with equality if and only if a=b and the parameter 0 in the lower bound cannot be improved.



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*Source: 694394-2009-11-10.xml*