AAAAbstract and Applied Analysis1687-04091085-3375Hindawi Publishing Corporation69439410.1155/2009/694394694394Research ArticleOptimal Inequalities among Various Means of Two ArgumentsShiMing-yu1ChuYu-ming2JiangYue-ping1BaleanuDumitru1College of Mathematics and EconometricsHunan UniversityChangsha 410082Chinahnu.cn2Department of MathematicsHuzhou Teachers CollegeHuzhou 313000Chinahutc.zj.cn200910112009200910092009191020092009Copyright © 2009This 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 establish two optimal inequalities among power mean Mp(a,b)=(ap/2+bp/2)1/p, arithmetic mean A(a,b)=(a+b)/2, logarithmic mean L(a,b)=(ab)/(logalogb), and geometric mean G(a,b)=ab.

1. Introduction

For pR, the power mean Mp(a,b) of order p of two positive numbers a and b is defined by

Mp(a,b)={(ap+bp2)1/p,p0,ab,p=0.

In the recent past, the power mean Mp(a,b) has been the subject of intensive research. In particular, many remarkable inequalities for the mean can be found in literature . It is well-known that Mp(a,b) is continuous and strictly increasing with respect to pR for fixed a and b.

If we denote by

I(a,b)={1e(aabb)1/(a-b),ba,a,b=a,L(a,b)={b-alogb-loga,ba,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

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 , Alzer and Janous established the following sharp double inequality (see also [13, page 350]):

Mlog2/log3(a,b)23A(a,b)+13G(a,b)M2/3(a,b) for all real numbers a,b>0.

In , Mao proved

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 , the authors presented the bounds for L and I in terms of A and G as follows:

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 ab.

Alzer  proved

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 , Alzer and Qiu established

αA(a,b)+(1-α)G(a,b)<I(a,b)<βA(a,b)+(1-β)G(a,b) for α2/3, β2/e=0.73575 and a,b>0 with ab.

The main purpose of this paper is to present the optimal bounds for Aα(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.

If r(0,1), then 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.

Let p=(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 g(1)=0,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,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,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,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,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,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,  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,g8(t)=4p(1+p)2(2+p)(1-r)tp-2(1-p)2(2-p)2,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.

If r(0,1), then 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.

Let p=(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 g(1)=0,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,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,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,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,g5(t)=rp(1+p)2t+p(1-p)(2-r)t1-p+p2(1-p)(r-2),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 ResultsTheorem 3.1.

If α(0,1), then 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.

If a=b, then we clearly see that a=Aα(a,b)L1-α(a,b)=M(1+2α)/3(a,b)=b.

If ab, then without loss of generality we assume that a>b and let t=a/b>1; hence elementary calculation yields

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

f(t)=31+2αlog(t(1+2α)/3+1)-31+2αlog2-αlogt+12+αlog2-(1-α)log(t-1)+(1-α)log(logt), then limt1f(t)=0,f(t)=g(t)t(t+1)(t-1)(t(1+2α)/3+1)logt, where 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 Lemma 2.1 and (3.6) we know that

f(t)>0 for t(1,+).

Therefore, we get

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 any 0<ε<(1+2α)/3, let 0<t<1, then (1.1) leads to

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 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

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

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 any 0<α<1 we have 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, if a=b, then we clearly see that M0(a,b)=Aα(a,b)L1-α(a,b)=a. If ab, 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 parameter 0 in the lower bound cannot be improved.

For any ε>0, we have

limtMε(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

Aα(t,1)L1-α(t,1)<Mε(t,1) for t(T1,+).

Theorem 3.3.

If α(0,1), then 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.

If a=b, then we clearly see that a=Gα(a,b)L1-α(a,b)=M(1-α)/3(a,b)=b.

If ab, then without loss of generality, we assume that a>b, let t=a/b>1; hence simple computation leads to

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 h(t)=31-αlog(t(1-α)/3+1)-31-αlog2-αlogt+12+αlog2-(1-α)log(t-1)+(1-α)log(logt), then limt1h(t)=0,h(t)=g(t)2t(t-1)(t(1-α)/3+1)logt, where 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 Lemma 2.2 and (3.22) we know that

h(t)>0 for t(1,+).

Therefore, we get

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 any 0<ε<(1-α)/3, let 0<t<1, then (1.1) leads to

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 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

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

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 any ε>0, we have limtMε(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 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.

Acknowledgments

This work was supported by the National Natural Science Foundation of China (Grant no. 60850005) and the Natural Science Foundation of Zhejiang Province (Grant no. D7080080 and no. Y607128).

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