# A Best Possible Double Inequality for Power Mean

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

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## Abstract

We answer the question: for anyp,q∈ℝ with p≠q and p≠-q, what are the greatest value λ=λ(p,q) and the least value μ=μ(p,q), such that the double inequality Mλ(a,b)<Mp(a,b)Mq(a,b)<Mμ(a,b) holds for all a,b>0 with a≠b? Where 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 case 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 [1–15].LetL(a,b)=(a-b)/(loga-logb) and I(a,b)=1/e(aa/bb)1/(a-b) be the logarithmic and identric means of two positive numbers a and b with a≠b, respectively. Then it is well known that
(1.2)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)
for all a,b>0 with a≠b.In [16–22], 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)<Mlog2(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] as follows:
(1.5)Mlog2/(log2-logα)(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 question: 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?In [4], the authors presented the greatest value p=p(α,β) and the least value q=q(α,β) such that the double inequality
(1.7)Mp(a,b)<Aα(a,b)Gβ(a,b)H1-α-β(a,b)<Mq(a,b)
holds for all a,b>0 with a≠b and α,β>0 with α+β<1.It is the aim of this paper to answer the question: for anyp, q∈ℝ with p≠q and p≠-q, what are the greatest value λ=λ(p,q) and the least value μ=μ(p,q), such that the double inequality
(1.8)Mλ(a,b)<Mp(a,b)Mq(a,b)<Mμ(a,b)
holds for all a,b>0 with a≠b?


## 2. Main Result

In order to establish our main result, we need a lemma which we present in this section.Lemma 2.1.
Letp, q≠0,  p≠q and x>1. Then
(2.1)Mp(x,1)Mq(x,1)<M(p+q)/22(x,1)
for p+q>0, and
(2.2)Mp(x,1)Mq(x,1)>M(p+q)/22(x,1)
for p+q<0.Proof.
From (1.1), we have
(2.3)log[Mp(x,1)Mq(x,1)]-logM(p+q)/22(x,1)=1plog1+xp2+1qlog1+xq2-4p+qlog1+x(p+q)/22.
Let(2.4)f(x)=1plog1+xp2+1qlog1+xq2-4p+qlog1+x(p+q)/22,
then simple computations lead to
(2.5)f(1)=0,(2.6)f′(x)=(1-x(p+q)/2)(xp/2-xq/2)2x(1+xp)(1+xq)(1+x(p+q)/2).
Equation (2.6) implies that
(2.7)f′(x)<0
for p+q>0, and
(2.8)f′(x)>0
for p+q<0.
Therefore, inequality (2.1) follows from (2.3)–(2.5) and inequality (2.7), and inequality (2.2) follows from (2.3)–(2.5) and inequality (2.8).
Let(2.9)E0={(p,q)∈ℝ2:p=q},E0′={(p,q)∈ℝ2:p=-q},E1={(p,q)∈ℝ2:p,q>0,p>q},E1′={(p,q)∈ℝ2:p,q>0,p<q},E2={(p,q)∈ℝ2:p,q<0,p>q},E2′={(p,q)∈ℝ2:p,q<0,p<q},E3={(p,q)∈ℝ2:p>0,q=0},E3′={(p,q)∈ℝ2:p=0,q>0},E4={(p,q)∈ℝ2:p>0,q<0,p+q>0},E4′={(p,q)∈ℝ2:p<0,q>0,p+q>0},E5={(p,q)∈ℝ2:p=0,q<0},E5′={(p,q)∈ℝ2:p<0,q=0},E6={(p,q)∈ℝ2:p>0,q<0,p+q<0},E6′={(p,q)∈ℝ2:p<0,q>0,p+q<0}.
Then we clearly see thatℝ2=⋃i=06Ei⋃i=06Ei′, and it is not difficult to verify that the identity Mp(a,b)Mq(a,b)=M(p+q)/2(a,b) holds for all a, b>0 if (p,q)∈E0⋃E0′. Let
(2.10)λ={2pq(p+q),(p,q)∈E1⋃E1′,(p+q)2,(p,q)∈E2⋃E2′⋃E5⋃E5′⋃E6⋃E6′,0,(p,q)∈E3⋃E3′⋃E4⋃E4′,μ={2pq(p+q),(p,q)∈E2⋃E2′,(p+q)2,(p,q)∈E1⋃E1′⋃E3⋃E3′⋃E4⋃E4′,0,(p,q)∈E5⋃E5′⋃E6⋃E6′.Then we have Theorem2.2 as follows.Theorem 2.2.
The double inequality(2.11)Mλ(a,b)<Mp(a,b)Mq(a,b)<Mμ(a,b)
holds for all a, b>0 with a≠b, and Mλ(a,b) and Mμ(a,b) are the best possible lower and upper power mean bounds for the geometric mean of Mp(a,b) and Mq(a,b).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 and p>q. We divide the proof of inequality (2.11) into three cases.
Case
1. (p,q)∈E1⋃‍E2. Then from Lemma 2.1, we clearly see that
(2.12)Mp(x,1)Mq(x,1)<M(p+q)/2(x,1)
for (p,q)∈E1, and
(2.13)Mp(x,1)Mq(x,1)>M(p+q)/2(x,1)
for (p,q)∈E2.
From (1.1), we get
(2.14)log[Mp(x,1)Mq(x,1)]-logM2pq/(p+q)2(x,1)=1plog1+xp2+1qlog1+xq2-p+qpqlog1+x2pq/(p+q)2.
Let(2.15)F(x)=1plog1+xp2+1qlog1+xq2-p+qpqlog1+x2pq/(p+q)2,
then simple computations lead to
(2.16)F(1)=0,(2.17)F′(x)=xqG(x)x(1+xp)(1+xq)(1+x2pq/(p+q)),
where
(2.18)G(x)=xp-q-x(2pq+p2-q2)/(p+q)+2xp-x2pq/(p+q)-2xq(p-q)/(p+q)+1,(2.19)G(1)=0,(2.20)G′(x)=x(pq-q2-p-q)/(p+q)H(x),
where
(2.21)H(x)=(p-q)xp(p-q)/(p+q)-2pq+p2-q2p+qxp+2px(p2+q2)/(p+q)-2pqp+qxq-2q(p-q)p+q,(2.22)H(1)=2(p-q)2p+q,(2.23)H′(x)=pp+qxp-1I(x),
where
(2.24)I(x)=(p-q)2x-2pq/(p+q)+2(p2+q2)x-q(p-q)/(p+q)-2q2x-(p-q)-2pq-p2+q2,(2.25)I(1)=2(p-q)2,(2.26)I′(x)=2q(p-q)p+qx(q2-pq-p-q)/(p+q)J(x),
where
(2.27)J(x)=-p(p-q)x-q+q(p+q)x-p(p-q)/(p+q)-p2-q2,(2.28)J(1)=-2p(p-q),(2.29)J′(x)=pq(p-q)x-q-1(1-x(q2-p2+2pq)/(p+q)).
If(p,q)∈E1, then (2.15), (2.18), (2.21), (2.22), (2.24), (2.25), (2.27), and (2.28) lead to
(2.30)limx→+∞F(x)=0,(2.31)limx→+∞G(x)=-∞,(2.32)limx→+∞H(x)=-∞,(2.33)H(1)>0,(2.34)limx→+∞I(x)=-2pq-p2+q2<0,(2.35)I(1)>0,(2.36)limx→+∞J(x)=-(p2+q2)<0,(2.37)J(1)<0.
We divide the discussion into two subcases.
Subcase 1.1. (p,q)∈E1. Then (2.26) and (2.29) together with inequalities (2.36) and (2.37) imply that I(x) is strictly decreasing in [1,+∞). In fact, if (q2-p2+2pq)/(p+q)≥0, then (2.29) and inequality (2.37) imply that J(x)<0 for x∈[1,+∞). If (q2-p2+2pq)/(p+q)<0, then (2.29) and inequality (2.36) lead to the conclusion that J(x)<0 for x∈[1,+∞).
From inequalities (2.34) and (2.35) together with the monotonicity of I(x), we know that there exists λ1>1 such that I(x)>0 for x∈[1,λ1) and I(x)<0 for x∈(λ1,+∞). Then (2.23) leads to the conclusion that H(x) is strictly increasing in [1,λ1] and strictly decreasing in [λ1,+∞).
It follows from (2.32) and (2.33) together with the piecewise monotonicity of H(x) that there exists λ2>λ1>1 such that H(x)>0 for [1,λ2] and H(x)<0 for (λ2,+∞). Then (2.20) leads to the conclusion that G(x) is strictly increasing in [1,λ2] and strictly decreasing in [λ2,+∞).
From (2.17), (2.19) and (2.31) together with the piecewise monotonicity of G(x), we clearly see that there exists λ3>λ2>1 such that F(x) is strictly increasing in [1,λ3] and strictly decreasing in [λ3,+∞).
Therefore,Mp(x,1)Mq(x,1)>M2pq/(p+q)(x,1) follows from (2.14)–(2.16) and (2.30) together with the piecewise monotonicity of F(x).
Subcase
1.2.
(
p
,
q
)
∈
E
2. Then (2.30) and (2.35) again hold, and (2.18), (2.21), (2.22), and (2.28) lead to
(2.38)limx→+∞G(x)=+∞,(2.39)limx→+∞H(x)=+∞,(2.40)H(1)<0,(2.41)J(1)>0.
It follows from (2.29) and inequalities (q2-p2+2pq)/(p+q)<0 and (2.41) that J(x)>0 for x∈[1,+∞). Then (2.26) and inequality (2.35) lead to the conclusion that I(x)>0 for x∈[1,+∞). Therefore, H(x) is strictly increasing in [1,+∞) follows from (2.23).
It follows from (2.20) and (2.39) together with inequality (2.40) and the monotonicity of H(x) that there exists μ1>1 such that G(x) is strictly decreasing in [1,μ1] and strictly increasing in [μ1,+∞).
From (2.17), (2.19) and (2.38) together with the piecewise monotonicity of G(x), we clearly see that there exists μ2>μ1>1 such that F(x) is strictly decreasing in [1,μ2] and strictly increasing in [μ2,+∞).
Therefore,Mp(x,1)Mq(x,1)<M2pq/(p+q)(x,1) follows from (2.14)–(2.16) and (2.30) together with the piecewise monotonicity of F(x).
Case
2. (p,q)∈E3⋃‍E5. Clearly, we have M0(x,1)<Mp(x,1)Mq(x,1) for (p,q)∈E3 and M0(x,1)>Mp(x,1)Mq(x,1) for (p,q)∈E5. Therefore, we need only to prove that
(2.42)M0(x,1)Mr(x,1)<Mr/2(x,1)
for r>0, and
(2.43)M0(x,1)Mr(x,1)>Mr/2(x,1)
for r<0.
From (1.1), one has
(2.44)log[M0(x,1)Mr(x,1)]-logMr/22(x,1)=12logx+1rlog1+xr2-4rlog1+xr/22.
Let(2.45)f(x)=12logx+1rlog1+xr2-4rlog1+xr/22,
then simple computations lead to
(2.46)f(1)=0,(2.47)f′(x)=-(xr/2-1)32x(1+xr)(1+xr/2).
Ifr>0 (or r<0, resp.), then (2.47) leads to the conclusion that f(x) is strictly decreasing (or increasing, resp.) in [1,+∞). Therefore, inequalities (2.42) and (2.43) follow from (2.44)–(2.46) and the monotonicity of f(x).
Case
3.(p,q)∈E4⋃‍E6. Then from Lemma 2.1, we clearly see that M(p+q)/2(x,1)>Mp(x,1)Mq(x,1) for (p,q)∈E4 and Mp(x,1)Mq(x,1)>M(p+q)/2(x,1) for (p,q)∈E6. Therefore, we need only to prove that
(2.48)Mp(x,1)Mq(x,1)>M0(x,1)
for (p,q)∈E4, and
(2.49)Mp(x,1)Mq(x,1)<M0(x,1)
for (p,q)∈E6.
From (1.1), we get
(2.50)log[Mp(x,1)Mq(x,1)]-logM02(x,1)=1plog1+xp2+1qlog1+xq2-logx.
Let(2.51)f(x)=1plog1+xp2+1qlog1+xq2-logx,
then simple computations lead to
(2.52)f(1)=0,(2.53)f′(x)=xp+q-1x(1+xp)(1+xq).
If(p,q)∈E4 (or E6, resp.), then (2.53) implies that f(x) is strictly increasing (or decreasing, resp.) in [1,+∞). Therefore, inequalities (2.48) and (2.49) follow from (2.50)–(2.52) and the monotonicity of f(x).
Next, we prove thatMλ(a,b) and Mμ(a,b) are the best possible lower and upper power mean bounds for the geometric mean of Mp(a,b) and Mq(a,b). We divide the proof into six cases.
Case
A. (p,q)∈E1. Then for any ϵ∈(0,(p+q)/2) and x>0, from (1.1), one has
(2.54)Mp(1+x,1)Mq(1+x,1)-M(p+q)/2-ϵ2(1+x,1)=[1+(1+x)p2]1/p[1+(1+x)q2]1/q-[1+(1+x)(p+q)/2-ϵ2]4/(p+q-2ϵ),(2.55)limx→+∞M2pq/(p+q)+ϵ2(x,1)Mp(x,1)Mq(x,1)=2ϵ(p+q)2/pq[2pq+ϵ(p+q)]>1.
Lettingx→0 and making use of Taylor expansion, we get
(2.56)[1+(1+x)p2]1/p[1+(1+x)q2]1/q-[1+(1+x)(p+q)/2-ϵ2]4/(p+q-2ϵ)=ϵ4x2+o(x2).
Equations (2.54) and (2.56) together with inequality (2.55) imply that for any ϵ∈(0,p+q/2), there exist δ1=δ1(ϵ)>0 and X1=X1(p,q,ϵ)>1 such that Mp(1+x,1)Mq(1+x,1)>M(p+q)/2-ϵ(1+x,1) for x∈(0,δ1) and Mp(x,1)Mq(x,1)<M2pq/(p+q)+ϵ(x,1) for x∈(X1,+∞).
Case
B. (p,q)∈E2. Then for ϵ∈(0,-(p+q)/2) and x>0, making use of (1.1) and Taylor expansion, we have
(2.57)M(p+q)/2+ϵ2(1+x,1)-Mp(1+x,1)Mq(1+x,1)=ϵ4x2+o(x2)(x→0),(2.58)limx→+∞Mp(x,1)Mq(x,1)M2pq/(p+q)-ϵ2(x,1)=2ϵ(p+q)2/pq[2pq-ϵ(p+q)]>1.
Equation (2.57) and inequality (2.58) imply that for any ϵ∈(0,-(p+q)/2), there exist δ2=δ2(ϵ)>0 and X2=X2(p,q,ϵ)>1 such that M(p+q)/2+ϵ(1+x,1)>Mp(1+x,1)Mq(1+x,1) for x∈(0,δ2) and Mp(x,1)Mq(x,1)>M2pq/(p+q)-ϵ(x,1) for x∈(X2,+∞).
Case
C. (p,q)∈E3. Then for ϵ∈(0,p/2) and x>0, making use of (1.1) and Taylor expansion, we have
(2.59)Mp(1+x,1)M0(1+x,1)-Mp/2-ϵ2(1+x,1)=ϵ4x2+o(x2)(x→0),limx→+∞Mϵ2(x,1)Mp(x,1)M0(x,1)=+∞.
Equation (2.59) leads to the conclusion that for any ϵ∈(0,p/2), there exist δ3=δ3(ϵ)>0 and X3=X3(p,ϵ)>1 such that Mp(1+x,1)M0(1+x,1)>Mp/2-ϵ(1+x,1) for x∈(0,δ3) and Mϵ(x,1)>Mp(x,1)M0(x,1) for x∈(X3,+∞).
Case
D. (p,q)∈E4. Then for ϵ∈(0,(p+q)/2) and x>0, making use of (1.1) and Taylor expansion, we have
(2.60)Mp(1+x,1)Mq(1+x,1)-M(p+q)/2-ϵ2(1+x,1)=ϵ4x2+o(x2)(x→0),limx→+∞Mϵ2(x,1)Mp(x,1)Mq(x,1)=+∞.
Equation (2.60) implies that for any ϵ∈(0,(p+q)/2), there exist δ4=δ4(ϵ)>0 and X4=X4(p,q,ϵ)>1 such that M(p+q)/2-ϵ(1+x,1)<Mp(1+x,1)Mq(1+x,1) for x∈(0,δ4) and Mϵ(x,1)>Mp(x,1)Mq(x,1) for x∈(X4,+∞).
Case
E. (p,q)∈E5. Then for any ϵ∈(0,-q/2) and x>0, making use of (1.1) and Taylor expansion, one has
(2.61)Mq/2+ϵ2(1+x,1)-M0(1+x,1)Mq(1+x,1)=ϵ4x2+o(x2)(x→0),limx→+∞M0(x,1)Mq(x,1)M-ϵ2(x,1)=+∞.
Equation (2.61) leads to the conclusion that for any ϵ∈(0,-q/2), there exist δ5=δ5(ϵ)>0 and X5=X5(q,ϵ)>1 such that Mq/2+ϵ(1+x,1)>M0(1+x,1)Mq(1+x,1) for x∈(0,δ5) and M-ϵ(x,1)<M0(x,1)Mq(x,1) for x∈(X5,+∞).
Case
F. (p,q)∈E6. Then for any ϵ∈(0,-(p+q)/2) and x>0, making use of (1.1) and Taylor expansion, one has
(2.62)M(p+q)/2+ϵ2(1+x,1)-Mp(1+x,1)Mq(1+x,1)=ϵ4x2+o(x2)(x→0),limx→+∞Mp(x,1)Mq(x,1)M-ϵ2(x,1)=+∞.
Equation (2.62) shows that for any ϵ∈(0,-(p+q)/2), there exist δ6=δ6(ϵ)>0 and X6=X6(p,q,ϵ)>1 such that M(p+q)/2+ϵ(1+x,1)>Mp(1+x,1)Mq(1+x,1) for x∈(0,δ6) and Mp(x,1)Mq(x,1)>M-ϵ2(x,1) for x∈(X6,+∞).



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