# Optimal Inequalities for Power Means

Authors: Yong-Min Li; Bo-Yong Long; Yu-Ming Chu; Wei-Ming Gong
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We present the best possible power mean bounds for the productMpα(a,b)M-p1-α(a,b) for any p>0, α∈(0,1), and all a,b>0 with a≠b. Here, Mp(a,b) is the pth power mean of two positive numbers a and b.


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## Body

## 1. Introduction

Forp∈ℝ, the pth power mean Mp(a,b) of two positive numbers a and b is defined by (1.1)Mp(a,b)={(ap+bp2)1/p,p≠0,ab,p=0.It is well known thatMp(a,b) is continuous and strictly increasing with respect to p∈ℝ for fixed a,b>0 with a≠b. Many classical means are special cases of the power mean, for example, M-1(a,b)=H(a,b)=2ab/(a+b), M0(a,b)=G(a,b)=ab and M1(a,b)=A(a,b)=(a+b)/2 are the harmonic, geometric and arithmetic means of a and b, respectively. Recently, the power mean has been the subject of intensive research. In particular, many remarkable inequalities and properties for the power mean can be found in literature [1–22].LetL(a,b)=(a-b)/(loga-logb), P(a,b)=(a-b)/[4arctan(a/b)-π] and I(a,b)=1/e(aa/bb)1/(a-b) be the logarithmic, Seiffert and identric means of two positive numbers a and b with a≠b, respectively. Then it is well known that (1.2)min⁡{a,b}<H(a,b)<G(a,b)<L(a,b)<P(a,b)<I(a,b)<A(a,b)<max⁡{a,b},
for all a,b>0 with a≠b.In [23–29], the authors presented the sharp power mean bounds for L, I, (IL)1/2 and (L+I)/2 as follows: (1.3)M0(a,b)<L(a,b)<M1/3(a,b),M2/3(a,b)<I(a,b)<Mlog⁡2(a,b),M0(a,b)<L(a,b)I(a,b)<M1/2(a,b),12(L(a,b)+I(a,b))<M1/2(a,b),
for all a,b>0 with a≠b.Alzer and Qiu [12] proved that the inequality (1.4)12(L(a,b)+I(a,b))>Mp(a,b)
holds for all a,b>0 with a≠b if and only if p≤(log2)/(1+log2)=0.40938….The following sharp bounds for the sumαA(a,b)+(1-α)L(a,b), and the products Aα(a,b)L1-α(a,b) and Gα(a,b)L1-α(a,b) in terms of power means were proved in [5, 8]: (1.5)Mlog⁡2/(log⁡2-log⁡α)(a,b)<αA(a,b)+(1-α)L(a,b)<M(1+2α)/3(a,b),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),
for any α∈(0,1) and all a,b>0 with a≠b.In [2, 7] the authors answered the questions: for any α∈(0,1), what are the greatest values p1=p1(α), p2=p2(α), p3=p3(α), and p4=p4(α), and the least values q1=q1(α), q2=q2(α), q3=q3(α), and q4=q4(α), such that the inequalities (1.6)Mp1(a,b)<Pα(a,b)L1-α(a,b)<Mq1(a,b),Mp2(a,b)<Aα(a,b)G1-α(a,b)<Mq2(a,b),Mp3(a,b)<Gα(a,b)H1-α(a,b)<Mq3(a,b),Mp4(a,b)<Aα(a,b)H1-α(a,b)<Mq4(a,b),
hold for all a,b>0 with a≠b?It is the aim of this paper to present the best possible power mean bounds for the productMpα(a,b)M-p1-α(a,b) for any p>0, α∈(0,1) and all a,b>0 with a≠b.


## 2. Main Result

Theorem 2.1.
Letp>0, α∈(0,1) and a,b>0 with a≠b. Then(1)
M(2α-1)p(a,b)=Mpα(a,b)M-p1-α(a,b)=M0(a,b) for α=1/2,(2)
M(2α-1)p(a,b)>Mpα(a,b)M-p1-α(a,b)>M0(a,b) for α>1/2 and M(2α-1)p(a,b)<Mpα(a,b)M-p1-α(a,b)<M0(a,b) for α<1/2, and the bounds M(2α-1)p(a,b) and M0(a,b) for the product Mpα(a,b)M-p1-α(a,b) in either case are best possible.Proof.
From (1.1) we clearly see that Mp(a,b) is symmetric and homogenous of degree 1. Without loss of generality, we assume that b=1, a=x>1.(1)
Ifα=1/2, then (1.1) leads to(2.1)Mpα(x,1)M-p1-α(x,1)=(1+xp2)1/p(1+x-p2)-1/p=(1+xp2)1/p(2xp1+xp)1/p=x=M02(x,1)=M(2α-1)p2(x,1).(2)
Firstly, we compare the value ofM(2α-1)p(x,1) to the value of Mpα(x,1)M-p1-α(x,1) for α∈(0,1/2)∪(1/2,1). From (1.1) we have(2.2)log⁡[Mpα(x,1)M-p1-α(x,1)]-log⁡M(2α-1)p(x,1)=αplog⁡1+xp2-1-αplog⁡1+x-p2-1(2α-1)plog⁡1+x(2α-1)p2.
Let(2.3)f(x)=αplog⁡1+xp2-1-αplog⁡1+x-p2-1(2α-1)plog⁡1+x(2α-1)p2,
then simple computations lead to
(2.4)f(1)=0,(2.5)f′(x)=g(x)x(1+xp)(1+x(2α-1)p),
where
(2.6)g(x)=(α-1)x2αp+αxp-αx(2α-1)p+1-α,g(1)=0,(2.7)g′(x)=αpxp-1h(x),
where
(2.8)h(x)=2(α-1)x(2α-1)p-(2α-1)x2(α-1)p+1,h(1)=0,(2.9)h′(x)=-2p(1-α)(2α-1)x2(α-1)p-1(xp-1).
Ifα∈(1/2,1), then (2.9) implies that h(x) is strictly decreasing in [1,+∞). Therefore, M(2α-1)p(x,1)>Mpα(x,1)M-p1-α(x,1) follows easily from (2.2)–(2.8) and the monotonicity of h(x).
Ifα∈(0,1/2), then (2.9) leads to the conclusion that h(x) is strictly increasing in [1,+∞). Therefore, M(2α-1)p(x,1)<Mpα(x,1)M-p1-α(x,1) follows easily from (2.2)–(2.8) and the monotonicity of h(x).
Secondly, we compare the value ofM0(x,1) to the value of Mpα(x,1)M-p1-α(x,1). It follows from (1.1) that(2.10)log⁡[Mpα(x,1)M-p1-α(x,1)]-log⁡M0(x,1)=αplog⁡1+xp2-1-αplog⁡1+x-p2-12log⁡x.
Let(2.11)F(x)=αplog⁡1+xp2-1-αplog⁡1+x-p2-12log⁡x,
then simple computations lead to
(2.12)F(1)=0,(2.13)F′(x)=(2α-1)(xp-1)x(1+xp)(1+x(2α-1)p).
Ifα∈(1/2,1), then (2.13) implies that F(x) is strictly increasing in [1,+∞). Therefore, Mpα(x,1)M-p1-α(x,1)>M0(x,1) follows easily from (2.10)–(2.12) and the monotonicity of F(x).
Ifα∈(0,1/2), then (2.13) leads to the conclusion that F(x) is strictly decreasing in [1,+∞). Therefore, Mpα(x,1)M-p1-α(x,1)<M0(x,1) follows easily from (2.10)–(2.12) and the monotonicity of F(x).
Next, we prove that the boundM(2α-1)p(a,b) for the product Mpα(a,b)M-p1-α(a,b) in either case is best possible.
Ifα∈(0,1/2), then for any ϵ∈(0,(1-2α)p) and x>0 we have(2.14)Mpα(1+x,1)M-p1-α(1+x,1)-M(2α-1)p+ϵ(1+x,1)=[1+(1+x)p2]α/p[1+(1+x)-p2](α-1)/p-[1+(1+x)(2α-1)p+ϵ2]1/[(2α-1)p+ϵ].
Lettingx→0 and making use of Taylor’s expansion, one has(2.15)[1+(1+x)p2]α/p[1+(1+x)-p2](α-1)/p-[1+(1+x)(2α-1)p+ϵ2]1/[(2α-1)p+ϵ]=[1+α2x+α(p+α-2)8x2+o(x2)]×[1+1-α2x-(1-α)(p+α+1)8x2+o(x2)]-[1+12x+(2α-1)p+ϵ-18x2+o(x2)]=[1+12x+(2α-1)p-18x2+o(x2)]-[1+12x+(2α-1)p+ϵ-18x2+o(x2)]=-ϵ8x2+o(x2).
Equations (2.14) and (2.15) imply that for any α∈(0,1/2) and ϵ∈(0,(1-2α)p) there exists δ1=δ1(ϵ)>0, such that Mpα(1+x,1)M-p1-α(1+x,1)<M(2α-1)p+ϵ(1+x,1) for x∈(0,δ1).
Ifα∈(1/2,1), then for any ϵ∈(0,(2α-1)p) and x>0 we have(2.16)Mpα(1+x,1)M-p1-α(1+x,1)-M(2α-1)p-ϵ(1+x,1)=[1+(1+x)p2]α/p[1+(1+x)-p2](α-1)/p-[1+(1+x)(2α-1)p-ϵ2]1/[(2α-1)p-ϵ].
Lettingx→0 and making use of Taylor’s expansion, one has(2.17)[1+(1+x)p2]α/p[1+(1+x)-p2](α-1)/p-[1+(1+x)(2α-1)p-ϵ2]1/[(2α-1)p-ϵ]=[1+α2x+α(p+α-2)8x2+o(x2)]×[1+1-α2x-(1-α)(p+α+1)8x2+o(x2)]-[1+12x+(2α-1)p-ϵ-18x2+o(x2)]=[1+12x+(2α-1)p-18x2+o(x2)]-[1+12x+(2α-1)p-ϵ-18x2+o(x2)]=ϵ8x2+o(x2).
Equations (2.16) and (2.17) imply that for any α∈(1/2,1) and ϵ∈(0,(2α-1)p) there exists δ2=δ2(ϵ)>0, such that Mpα(1+x,1)M-p1-α(1+x,1)>M(2α-1)p-ϵ(1+x,1) for x∈(0,δ2).
Finally, we prove that the boundM0(a,b) for the product Mpα(a,b)M-p1-α(a,b) in either case is best possible.
Ifα∈(0,1/2), then for any ϵ>0 we clearly see that(2.18)lim⁡x→+∞Mpα(x,1)M-p1-α(x,1)M-ϵ(x,1)=+∞.
Equation (2.18) implies that for any α∈(0,1/2) and ϵ>0 there exists T1=T1(ϵ)>1, such that Mpα(x,1)M-p1-α(x,1)>M-ϵ(x,1) for x∈(T1,+∞).
Ifα∈(1/2,1), then for any ϵ>0 we have(2.19)lim⁡x→+∞Mpα(x,1)M-p1-α(x,1)Mϵ(x,1)=0.
Equation (2.19) implies that for any α∈(1/2,1) and ϵ>0 there exists T2=T2(ϵ)>1, such that Mpα(x,1)M-p1-α(x,1)<Mϵ(x,1) for x∈(T2,+∞).



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*Source: 182905-2012-04-08.xml*