# Sharp Power Mean Bounds for the Combination of Seiffert and Geometric Means

Authors: Yu-Ming Chu; Ye-Fang Qiu; Miao-Kun Wang
Journal: Abstract and Applied Analysis
 (2010)

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## Abstract

We answer the question: forα∈(0,1), what are the greatest value p
and the least value q such that the double inequality Mp(a,b)<Pα(a,b)G1−α(a,b)<Mq(a,b) holds for all a,b>0 with a≠b. Here, Mp(a,b), P(a,b), and G(a,b) denote the power of order p, Seiffert, and geometric means of two positive numbers a and b, respectively.


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## Body

## 1. Introduction

Forp∈ℝ, the power mean Mp(a,b) of order p and the Seiffert mean P(a,b) of two positive numbers a and b are defined by (1.1)Mp(a,b)={(ap+bp2)1/p,p≠0,ab,p=0,P(a,b)={a-b4arctan(a/b)-π,a≠b,a,a=b.
The main properties of the power mean are given in [1]. It is well known that Mp(a,b) is strictly increasing with respect to p for fixed a,b>0 with a≠b. Recently, the power mean has been the subject of intensive research. In particular, many remarkable inequalities for the power mean can be found in the literature [2–16].The Seiffert meanP was introduced by Seiffert in [17], it can be rewritten in the following symmetric form (see [18, (2.4)]): (1.2)P(a,b)={a-b2arcsin((a-b)/(a+b)),a≠b,a,a=b.LetA(a,b)=(1/2)(a+b), G(a,b)=ab, L(a,b)={(b-a)/(logb-loga),b≠a,a,b=a,H(a,b)=2ab/(a+b) and I(a,b)={1/e(bb/aa)1/(b-a),b≠a,a,b=a be the arithmetic, geometric, logarithmic, harmonic, and identric means of two positive numbers a and b, respectively. Then it is well known that (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}
for all a,b>0 with a≠b.In [9], Alzer and Janous presented the sharp power mean bounds for the sum (2/3)A(a,b)+(1/3)G(a,b) as follows: (1.4)Mlog2/log3(a,b)<23A(a,b)+13G(a,b)<M2/3(a,b)
for all a,b>0 with a≠b.In [17], Seiffert proved that (1.5)L(a,b)<P(a,b)<I(a,b)
for all a,b>0 with a≠b.The following power mean bounds for the Seiffert mean was given by Jagers [19]: (1.6)M1/2(a,b)<P(a,b)<M2/3(a,b)
for all a,b>0 with a≠b.In [20, 21], the authors presented the bounds for the Seiffert mean P in terms of A and G as follows: (1.7)P(a,b)>3A(a,b)G(a,b)A(a,b)+2G(a,b),12A(a,b)+12G(a,b)<P(a,b)<23A(a,b)+13G(a,b),P(a,b)>A1/3(a,b)G2/3(a,b)
for all a,b>0 with a≠b.The following sharp lower power mean bounds for(1/3)G(a,b)+(2/3)H(a,b), (2/3)G(a,b)+(1/3)H(a,b), and P(a,b) can be found in [4, 6]: (1.8)13G(a,b)+23H(a,b)>M-2/3(a,b),23G(a,b)+13H(a,b)>M-1/3(a,b),P(a,b)>Mlog2/logπ(a,b)
for all a,b>0 with a≠b.The purpose of this paper is to answer the question: forα∈(0,1), what are the greatest value p and the least value q such that the double inequality Mp(a,b)<Pα(a,b)G1-α(a,b)<Mq(a,b) holds for all a,b>0 with a≠b.


## 2. Lemmas

In order to prove our main result, we need several lemmas which we present in this section.Lemma 2.1.
Letλ∈(0,1/3), x∈[1,∞) and h(x)=(1-3λ)x2λ+1-(1+3λ)x2λ-(1+3λ)x+(1-3λ). Then there exists x0>1 such that h(x)<0 for x∈[1,x0), h(x)>0 for x∈(x0,∞) and h(x0)=0.Proof.
Simple computations lead to(2.1)h(1)=-12λ<0,(2.2)limx→+∞h(x)=+∞,(2.3)h′(x)=(1-3λ)(2λ+1)x2λ-2λ(1+3λ)x2λ-1-(1+3λ),h′(1)=-6λ(1+2λ)<0,(2.4)limx→+∞h′(x)=+∞,(2.5)h′′(x)=2λx2λ-2[(1+2λ)(1-3λ)x+(1-2λ)(1+3λ)]>0.
Inequality (2.5) implies that h'(x) is strictly increasing in [1,∞). Then (2.3) and (2.4) lead to that there exists x1>1 such that h'(x)<0 for x∈[1,x1) and h'(x)>0 for x∈(x1,∞). Hence, h(x) is strictly decreasing in [1,x1] and strictly increasing in [x1,∞).
Therefore, Lemma2.1 follows from (2.1) and (2.2) together with the monotonicity of h(x).Lemma 2.2.
Ifλ∈(1/10,1/3), then the following statements are true:(1)
300λ4+324λ3+29λ2-45λ-8<0;(2)
-1176λ5-24λ4+114λ3+10λ2-3λ-1<0;(3)
-24λ5+72λ4+178λ3+6λ2-25λ-3<0.Proof.
Simple computations lead to(1)
300λ4+324λ3+29λ2-45λ-8<300×(1/3)4 + 324×(1/3)3+29×(1/3)2-45/10-8 = (590-24310)/54<0;(2)
-1176λ5-24λ4+114λ3 + 10λ2-3λ-1<-1176×(1/10)5-24×(1/10)4 + 114×(1/3)3+10×(1/3)2-3/10-1 = (3070-110710)/750<0;(3)
-24λ5+72λ4+178λ3 + 6λ2-25λ-3<-24×(1/10)5+72×(1/3)4 + 178×(1/3)3+6×(1/3)2-25/10-3 = (34750-1703710)/6750<0.Lemma 2.3.
Ifα∈(0,3/10], then
(2.6)Iα(a,b)G1-α(a,b)<M2α/3(a,b)
holds for all  a,b>0 with a≠b.Proof.
Without loss of generality, we assume thata>b. Let t=a/b>1 and β=α/3∈(0,1/10]. Then
(2.7)log[M2α/3(a,b)]-log[Iα(a,b)G1-α(a,b)]=log[M2β(a,b)]-log[I3β(a,b)G1-3β(a,b)]=12βlog1+t2β2-3βtt-1logt-1-3β2logt+3β.
Let
(2.8)f(t)=12βlog1+t2β2-3βtt-1logt-1-3β2logt+3β.
Then simple computations lead to
(2.9)limt→1f(t)=0,(2.10)f′(t)=g(t)(t-1)2,
where g(t)=(t2β+1-2t2β+t2β-1) / (1+t2β)-3β(t-1) + 3βlogt-(1-3β)(t-2+1/t)/2,
(2.11)g(1)=0,g′(t)=g1(t)2t2(1+t2β)2,
where g1(t)=(1-3β)t4β+2 + 6βt4β+1-(1+3β)t4β-2βt2β+2 + 4βt2β+1-2βt2β-(1+3β)t2+6βt+(1-3β),
(2.12)g1(1)=0,(2.13)g1′(1)=0,(2.14)g1′′(1)=0.
Let g2(t)=g1′′′(t)/(8βt2β-3) and g3(t)=g2′′′(t)/(2βt2β-3). Then
(2.15)g2(t)=(1-3β)(2β+1)(4β+1)t2β+2+3β(4β+1)(4β-1)t2β+1-(1+3β)(4β-1)(2β-1)t2β-β(β+1)(2β+1)t2+β(2β+1)(2β-1)t-β(β-1)(2β-1),(2.16)g2(1)=0,(2.17)g2′(1)=2(1-10β)(1+10β)≥0,(2.18)g2′′(1)=2(6β+1)(1-10β)(1+10β)≥0,(2.19)g3(t)=2(1-3β)(2β+1)2(4β+1)(β+1)t2+3β(4β+1)(4β-1)×(2β+1)(2β-1)t-2(1+3β)(4β-1)(2β-1)2(β-1),(2.20)g3(1)=9β(3-28β2)>0,(2.21)g3′(t)=(2β+1)(4β+1)[4(1-3β)(2β+1)(β+1)t+3β(4β-1)(2β-1)],(2.22)g3′(1)=(2β+1)(4β+1)(4+3β-46β2),(2.23)minβ∈[0,1/10](4+3β-46β2)=310-610>0.
From (2.21)–(2.23) we clearly see that g3′(t)>g3′(1)≥(310-6)/10(2β+1)(4β+1)>0 for t∈(1,∞), hence g3(t) is strictly increasing in [1,∞). Then (2.20) implies that g3(t)>0 for t∈(1,∞), hence g2′′(t) is strictly increasing in [1,∞).
It follows from (2.18) and the monotonicity of g2′′(t) that g2′′(t)>0 for t∈(1,∞), hence g2′(t) is strictly increasing in [1,∞). Then (2.17) implies that g2′(t)>0 for t∈(1,∞), therefore g2(t) is strictly increasing in [1,∞).
Equation (2.16) and the monotonicity of g2(t) lead to that g2(t)>0 for t∈(1,∞), so g1′′(t) is strictly increasing in [1,∞).
From (2.9)–(2.14) and the monotonicity of g1′′(t) we can deduce that (2.24)f(t)>0
for t∈(1,∞).
Therefore, Lemma2.3 follows from (2.7) and (2.8) together with (2.24).Remark 2.4.
In [22, Theorem3.1], the authors proved that(2.25)I1/2(a,b)G1/2(a,b)<M1/3(a,b),(2.26)I2/3(a,b)G1/3(a,b)<M4/9(a,b),(2.27)I1/3(a,b)G2/3(a,b)<M2/9(a,b)
for all a,b>0 with a≠b.
Obviously, (2.6) is the generalization of (2.25)–(2.27).Remark 2.5.
Ifα∈(3/10,1), then β=α/3∈(1/10,1/3) and (2.17) leads to(2.28)g2′(1)=2(1-10β)(1+10β)<0.
Inequality (2.28) and the continuity of g2′(t) imply that there exists δ>0 such that (2.29)g′2(t)<0
for t∈[1,1+δ).
From (2.29) and (2.9)–(2.16) we can deduce that(2.30)f(t)<0
for t∈(1,1+δ).
Equations (2.7) and (2.8) together with (2.30) lead to(2.31)Iα(a,b)G1-α(a,b)>M2α/3(a,b)
for all a,b>0 with a/b∈(1,1+δ)∪(1/(1+δ),1).
Therefore,α0=3/10 is the largest value in (0,1) such that inequality (2.6) holds for α∈(0,α0].


## 3. Main Result

Theorem 3.1.
Ifα∈(0,1), then
(3.1)M0(a,b)<Pα(a,b)G1-α(a,b)<M2α/3(a,b)
holds for all a,b>0 with a≠b, and M0(a,b) and M2α/3(a,b) are the best possible lower and upper power mean bounds for the product Pα(a,b)G1-α(a,b).Proof.
For alla,b>0 with a≠b, from (1.3), (1.5) and Lemma 2.1 we clearly see that(3.2)Pα(a,b)G1-α(a,b)>M0(a,b)
for all  α∈(0,1), and
(3.3)Pα(a,b)G1-α(a,b)<M2α/3(a,b)
for α∈(0,3/10].
Next, we prove that  (3.3) is also true for α∈(3/10,1) and all a,b>0 with a≠b.
Without loss of generality, we assume thata>b. Let t=a/b>1 and β=α/3∈(1/10,1/3). Then (1.1) leads to(3.4)log[M2α/3(a,b)]-log[Pα(a,b)G1-α(a,b)]=log[M2β(a,b)]-log[P3β(a,b)G1-3β(a,b)]=12βlog1+t4β2-3βlogt2-14arctant-π-(1-3β)logt.
Let
(3.5)F(t)=12βlog1+t4β2-3βlogt2-14arctant-π-(1-3β)logt.
Then simple computations lead to
(3.6)limt→1F(t)=0,(3.7)F′(t)=(1-3β)t4β+2-(1+3β)t4β-(1+3β)t2+(1-3β)t(t2-1)(t4β+1)+12β(t2+1)(4arctant-π).
From Lemma 2.1 we know that there exists λ0>1 such that
(3.8)(1-3β)λ04β+2-(1+3β)λ04β-(1+3β)λ02+(1-3β)=0,(3.9)(1-3β)t4β+2-(1+3β)t4β-(1+3β)t2+(1-3β)<0
for t∈[1,λ0), and
(3.10)(1-3β)t4β+2-(1+3β)t4β-(1+3β)t2+(1-3β)>0
for t∈(λ0,∞).
We divide two cases to prove that(3.11)F′(t)>0
for t>1.Case 1.
t∈[λ0,∞). Then (3.11) follows from (3.7) and (3.8) together with (3.10).Case 2.
t∈(1,λ0). Then (3.7) can be written as(3.12)F′(t)=(1-3β)t4β+2-(1+3β)t4β-(1+3β)t2+(1-3β)t(t2-1)(t4β+1)(4arctant-π)F1(t),
where
(3.13)F1(t)=4arctant+12βt(t2-1)(t4β+1)(t2+1)[(1-3β)t4β+2-(1+3β)t4β-(1+3β)t2+(1-3β)]-π.
Let x=t2∈(1,λ02), then (3.13) leads to
(3.14)F1(1)=0,F1′(t)=4F2(x)(t2+1)2[(1-3β)t4β+2-(1+3β)t4β-(1+3β)t2+(1-3β)]2,
where
(3.15)F2(x)=(18β2-9β+1)x4β+3-(18β2-3β+1)x4β+2-(18β2+3β+1)x4β+1+(18β2+9β+1)x4β+2(6β2-1)x2β+3-2(6β2-1)x2β+2-2(6β2-1)x2β+1+2(6β2-1)x2β+(18β2+9β+1)x3-(18β2+3β+1)x2-(18β2-3β+1)x+(18β2-9β+1).
Let F3(x)=(x4-2β/8β)F2(4)(x), then (3.15) leads to
(3.16)F2(1)=0,F2′(1)=0,F2′′(1)=0,F2′′′(1)=0,(3.17)F3(x)=(18β2-9β+1)(4β+1)(4β+3)(2β+1)x2β+3-(18β2-3β+1)(2β+1)(4β+1)(4β-1)x2β+2-(18β2+3β+1)(4β+1)(4β-1)(2β-1)x2β+1+(18β2+9β+1)(4β-1)(2β-1)(4β-3)x2β+(6β2-1)(2β+3)(β+1)(2β+1)x3-(6β2-1)(β+1)(2β+1)(2β-1)x2-(6β2-1)(2β+1)(2β-1)(β-1)x+(6β2-1)(2β-1)(β-1)(2β-3).
Fromβ∈(1/10,1/3) and x>1, we clearly see that (18β2-9β+1)(4β+1)(4β+3)(2β+1)x2β+3<(18β2-9β+1)(4β+1)(4β+3)(2β+1)x3, -(18β2-3β+1)(2β+1)(4β+1)(4β-1)x2β+2<-(18β2-3β+1)(2β+1)(4β+1)(4β-1)x2, -(18β2+3β+1)(4β+1)(4β-1)(2β-1)x2β+1<-(18β2+3β+1)(4β+1)(4β-1)(2β-1)x2 and (18β2+9β+1)(4β-1)(2β-1)(4β-3)x2β<(18β2+9β+1)(4β-1)(2β-1)(4β-3)x. These inequalities and Lemma 2.2 lead to(3.18)F3(x)<(18β2-9β+1)(4β+1)(4β+3)(2β+1)x3-(18β2-3β+1)(2β+1)(4β+1)(4β-1)x2-(18β2+3β+1)(4β+1)(4β-1)(2β-1)x2+(18β2+9β+1)(4β-1)(2β-1)(4β-3)x+(6β2-1)(2β+3)(β+1)(2β+1)x3-(6β2-1)(β+1)(2β+1)(2β-1)x2-(6β2-1)(2β+1)(2β-1)(β-1)x+(6β2-1)(2β-1)(β-1)(2β-3)=2β(300β4+324β3+29β2-45β-8)x3+(-1176β5-24β4+114β3+10β2-3β-1)x2+2(276β5-276β4+3β3+43β2-3β-1)x+(24β5-72β4+62β3-6β2-11β+3)<2β(300β4+324β3+29β2-45β-8)x+(-1176β5-24β4+114β3+10β2-3β-1)x+2(276β5-276β4+3β3+43β2-3β-1)x+24β5-72β4+62β3-6β2-11β+3=(-24β5+72β4+178β3+6β2-25β-3)x+24β5-72β4+62β3-6β2-11β+3<-24β5+72β4+178β3+6β2-25β-3+24β5-72β4+62β3-6β2-11β+3=12β(20β2-3)<0.
From (3.16)–(3.18) we can deduce that
(3.19)F2(x)<0
for x∈(1,λ02).
Equation (3.14) together with (3.19) imply that(3.20)F1(t)<0
for t∈(1,λ0).
Therefore, (3.11) follows from (3.9) and (3.12) together with (3.20).
It follows from (3.4)–(3.6) and (3.11) that(3.21)Pα(a,b)G1-α(a,b)<M2α/3(a,b)
for α∈(3/10,1) and all a,b>0 with a≠b.
At last, we prove thatM0(a,b) and M2α/3(a,b) are the best possible lower- and upper-power mean bounds for the product Pα(a,b)G1-α(a,b), respectively.
For any0<ɛ<(2/3)α and x>0, from (1.1) one has(3.22)limx→+∞Mɛ(1,x)Pα(1,x)G1-α(1,x)=limx→+∞[((1+x-ɛ)/2)1/ɛ((1-x-1)/(4arctanx-π))αx(1-α)/2]=+∞,(3.23)[Pα(1,1+x)G1-α(1,1+x)](2α/3)-ɛ-[M(2α/3)-ɛ(1,1+x)](2α/3)-ɛ=J(x)(4arctan1+x-π)α(2α-3ɛ)/3,
where J(x)=xα(2α-3ɛ)/3(1+x)(1-α)(2α-3ɛ)/6- (1/2) (4arctan1+x-π)α(2α-3ɛ)/3[1+(1+x)(2α-3ɛ)/3].
Letx→0, making use of the Taylor expansion we get(3.24)J(x)=124ɛ(2α-3ɛ)x(1/3)α(2α-3ɛ)+2+o(x(1/3)α(2α-3ɛ)+2).
Equation (3.22) implies that for any 0<ɛ<(2/3)α there exists X=X(ɛ,α)>1 such that Mɛ(1,x)>Pα(1,x)G1-α(1,x) for x∈(X,∞).
Equations (3.23) and (3.24) imply that for any 0<ɛ<(2/3)α there exists δ=δ(ɛ,α)>0 such that Pα(1,1+x)G1-α(1,1+x)>M2α/3-ɛ(1,1+x) for x∈(0,δ).



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