# Optimal Bounds for Seiffert Mean in terms of One-Parameter Means

Authors: Hua-Nan Hu; Guo-Yan Tu; Yu-Ming Chu
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

The authors present the greatest valuer1 and the least value r2 such that the double inequality Jr1(a,b)<T(a,b)<Jr2(a,b) holds for all a,b>0 with a≠b, where T(a,b) and Jp(a,b) denote the Seiffert and pth one-parameter means of two positive numbers a and b, respectively.


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## Body

## 1. Introduction

Forp∈ℝ the pth one-parameter mean Jp(a,b) and the Seiffert mean T(a,b) of two positive real numbers a and b are defined by
(1.1)Jp(a,b)={p(ap+1-bp+1)(p+1)(ap-bp),a≠b,p≠0,-1,a-bloga-logb,a≠b,p=0,ab(loga-logb)a-b,a≠b,p=-1,a,a=b,(1.2)T(a,b)={a-b2arctan((a-b)/(a+b)),a≠b,a,a=b,
respectively. Recently, both mean values have been the subject of intensive research. In particular, many remarkable inequalities and properties for Jp and T can be found in the literature [1–14].It is well known that the one-parameter meanJp(a,b) is continuous and strictly increasing with respect to p∈ℝ for fixed a, b>0 with a≠b. Many mean values are the special case of the one-parameter mean, for example:
(1.3)J1(a,b)=(a+b)2,the arithmetic mean,J1/2(a,b)=(a+ab+b)3,the Heronian mean,J-1/2(a,b)=ab,the geometric mean,J-2(a,b)=2ab(a+b),the harmonic mean.Seiffert [4] proved that the double inequality
(1.4)M1(a,b)<T(a,b)<M2(a,b)
holds for all a,b>0 with a≠b, where Mr(a,b)=[(ar+br)/2]1/r(r≠0) and M0(a,b)=ab is the rth power mean of a and b.In [15–17], the authors presented the best possible bounds for the Seiffert mean in terms of the Lehmer, power-type Heron, and one-parameter Gini means as follows:(1.5)L0(a,b)<T(a,b)<L1/3(a,b),Hlog3/log(π/2)(a,b)<T(a,b)<H5/2(a,b),S1(a,b)<T(a,b)<S5/3(a,b).
for all a,b>0 with a≠b, where Lr(a,b)=(ar+1+br+1)/(ar+br), Hk(a,b)=[(ak+(ab)k/2+bk)/3]1/k(k≠0) and H0(a,b)=ab, and Sα(a,b)=[(aα-1+bα-1)/(a+b)]1/(α-2)(α≠2) and S2(a,b)=(aabb)1/(a+b) denote the Lehmer, power-type Heron, and one-parameter Gini means of a and b, respectively.The purpose of this paper is to answer the question: what are the greatest valuer1 and the least value r2 such that the double inequality
(1.6)Jr1(a,b)<T(a,b)<Jr2(a,b)
holds for all a,b>0 with a≠b?


## 2. Lemma

In order to establish our main result we need the following lemma.Lemma 2.1.
Ifp=2/(π-2)=1.75⋯, t≥1 and g(t)=-(p-1)t2p+2+(p+1)t2p+p(p+1)tp+3-2(p+1)2tp+2+2p(p+3)tp+1-2(p+1)2tp+p(p+1)tp-1+(p+1)t2-(p-1), then there exists λ∈(1,∞) such that g(t)>0 for t∈(1,λ) and g(t)<0 for t∈(λ,∞).Proof.
Letg1(t)=g′(t)/t, g2(t)=t4-pg1′(t) and g3(t)=t4-pg2(5)(t)/[4p2(p-1)2(p+1)2]. Then simple computations lead to
(2.1)g(1)=0,(2.2)limt→+∞g(t)=-∞,(2.3)g1(t)=-2(p-1)(p+1)t2p+2p(p+1)t2p-2+p(p+1)(p+3)×tp+1-2(p+1)2(p+2)tp+2p(p+1)(p+3)tp-1-2p(p+1)2tp-2+p(p+1)(p-1)tp-3+2(p+1),(2.4)g1(1)=0,(2.5)limt→+∞g1(t)=-∞,(2.6)g2(t)=-4p(p-1)(p+1)tp+3+4p(p+1)(p-1)tp+1+p(p+1)2×(p+3)t4-2p(p+1)2(p+2)t3+2p(p-1)(p+1)(p+3)×t2-2p(p+1)2(p-2)t+p(p+1)(p-1)(p-3),(2.7)g2(1)=0,(2.8)limt→+∞g2(t)=-∞,(2.9)g2′(t)=-4p(p-1)(p+1)(p+3)tp+2+4p(p+1)2(p-1)tp+4p(p+1)2(p+3)t3-6p(p+1)2(p+2)t2+4p(p-1)(p+1)(p+3)t-2p(p+1)2(p-2),(2.10)g2′(1)=0,(2.11)limt→+∞g2′(t)=-∞,(2.12)g2′′(t)=-4p(p-1)(p+1)(p+2)(p+3)tp+1+4p2(p+1)2×(p-1)tp-1+12p(p+1)2(p+3)t2-12p(p+1)2×(p+2)t+4p(p-1)(p+1)(p+3),(2.13)g2′′(1)=12p(2-p)(p+1)2>0,(2.14)limt→+∞g2′′(t)=-∞,(2.15)g2′′′(t)=-4p(p-1)(p+1)2(p+2)(p+3)tp+4p2(p+1)2×(p-1)2tp-2+24p(p+1)2(p+3)t-12p(p+1)2(p+2),(2.16)g2′′′(1)=12p(2-p)(2p+3)(p+1)2>0,(2.17)limt→+∞g2′′′(t)=-∞,(2.18)g2(4)(t)=-4p2(p-1)(p+1)2(p+2)(p+3)tp-1+4p2(p+1)2×(p-1)2(p-2)tp-3+24p(p+1)2(p+3),(2.19)g2(4)(1)=8p(p+1)2(-4p3+2p2+5p+9)>8p(p+1)2[-4×1.83+2×1.752+5×1.75+9]=4.376p(p+1)2>0,(2.20)limt→+∞g2(4)(t)=-∞,(2.21)g3(t)=-(p+2)(p+3)t2+(p-2)(p-3)≤-(p+2)(p+3)+(p-2)(p-3)=-10p<0
for t∈[1,∞).
From the inequality (2.21) we clearly see that g2(4)(t) is strictly decreasing in [1,∞). Then (2.19) and (2.20) lead to the conclusion that there exists λ1>1 such that g2(4)(t)>0 for t∈[1,λ1) and g2(4)(t)<0 for t∈(λ1,∞). Hence, g2′′′(t) is strictly increasing in [1,λ1] and strictly decreasing in [λ1,∞).
It follows from (2.16) and (2.17) together with the monotonicity of g2′′′(t) that there exists λ2>1 such that g2′′′(t)>0 for t∈[1,λ2) and g2′′′(t)<0 for t∈(λ2,∞). Therefore, g2′′(t) is strictly increasing in [1,λ2] and strictly decreasing in [λ2,∞).
From (2.13) and (2.14) together with the monotonicity of g2′′(t) we know that there exists λ3>1 such that g2′′(t)>0 for t∈[1,λ3) and g2′′(t)<0 for t∈(λ3,∞). So, g2′(t) is strictly increasing in [1,λ3] and strictly decreasing in [λ3,∞).
Equations (2.10) and (2.11) together with the monotonicity of g2′(t) imply that there exists λ4>1 such that g2′(t)>0 for t∈(1,λ4) and g2′(t)<0 for t∈(λ4,∞). Hence, g2(t) is strictly increasing in [1,λ4] and strictly decreasing in [λ4,∞).
It follows from (2.7) and (2.8) together with the monotonicity of g2(t) that there exists λ5>1 such that g2(t)>0 for t∈(1,λ5) and g2(t)<0 for t∈(λ5,∞). Therefore, g1(t) is strictly increasing in [1,λ5] and strictly decreasing in [λ5,∞).
From (2.4) and (2.5) together with the monotonicity of g1(t) we clearly see that there exists λ6>1 such that g1(t)>0 for t∈(1,λ6) and g1(t)<0 for t∈(λ6,∞). So, g(t) is strictly increasing in [1,λ6] and strictly decreasing in [λ6,∞).
Therefore, Lemma2.1 follows from (2.1) and (2.2) together with the monotonicity of g(t).


## 3. Main Result

Theorem 3.1.
The double inequality(3.1)J2/(π-2)(a,b)<T(a,b)<J2(a,b)
holds for all a,b>0 with a≠b, and J2/(π-2)(a,b) and J2(a,b) are the best possible lower and upper one-parameter mean bounds for the Seiffert mean T(a,b), respectively.Proof.
Without loss of generality, we assume thata>b. Let t=a/b>1. Then from (1.1) and (1.2) we have
(3.2)J2(a,b)-T(a,b)=b(t2+t+1)6(t+1)arctan((t-1)/(t+1))[4arctant-1t+1-3(t2-1)t2+t+1].
Let
(3.3)f(t)=4arctant-1t+1-3(t2-1)t2+t+1.
Then simple computations lead to
(3.4)f(1)=0,f′(t)=(t-1)4(t2+1)(t2+t+1)2>0,
for t>1.
Therefore,T(a,b)<J2(a,b) for all a,b>0 with a≠b follows from (3.2)–(3.4).
Next, we prove that(3.5)T(a,b)>J2/(π-2)(a,b)
for all a,b>0 with a≠b.
Letp=2/(π-2)=1.75⋯. Then (1.1) and (1.2) lead to
(3.6)T(a,b)-Jp(a,b)=bp(tp+1-1)2(p+1)(tp-1)arctan((t-1)/(t+1))[(p+1)(t-1)(tp-1)p(tp+1-1)-2arctant-1t+1].
Let
(3.7)G(t)=(p+1)(t-1)(tp-1)p(tp+1-1)-2arctant-1t+1.
Then simple computations lead to
(3.8)limt→1G(t)=limt→+∞G(t)=0,(3.9)G′(t)=g(t)p(tp+1-1)2(t2+1),
where g(t) is defined as in Lemma 2.1.
From Lemma2.1 and (3.9) we know that there exists λ>1 such that G(t) is strictly increasing in [1,λ] and strictly decreasing in [λ,∞). Then (3.8) leads to that
(3.10)G(t)>0,
for t>1.
Therefore, the inequality (3.5) follows from (3.6), (3.7), and (3.10).
Finally, we prove thatJ2/(π-2)(a,b) and J2(a,b) are the best possible lower and upper one-parameter mean bounds for the Seiffert mean T(a,b), respectively.
Letp=2/(π-2), 0<ε<2 and x>0. Then from (1.1) and (1.2) one has
(3.11)limx→+∞Jp+ε(x,1)T(x,1)=p+εp+ε+1×π2>pp+1×π2=1,(3.12)T(1+x,1)-J2-ε(1+x,1)=h(x)2(3-ε)[(1+x)2-ε-1]arctan(x/(2+x)),
where
(3.13)h(x)=(3-ε)x[(1+x)2-ε-1]-2(2-ε)[(1+x)3-ε-1]arctanx2+x.
Lettingx→0 and making use of Taylor expansion we get
(3.14)h(x)=(3-ε)x[(2-ε)x+(2-ε)(1-ε)2x2-ε(1-ε)(2-ε)6x3+o(x3)]-2(2-ε)[(3-ε)x+(3-ε)(2-ε)2x2+(1-ε)(2-ε)(3-ε)6x3+o(x3)]-2(2-ε)2×[x2-x24+x312+o(x3)]=112ε(2-ε)(3-ε)x4+o(x4).
The inequality (3.11) implies that for any 0<ε<2, there exists X=X(ε)>1, such that T(x,1)<J2/(π-2)+ε(x,1) for x∈(X,+∞).
Equations (3.12)–(3.14) imply that for any 0<ε<2, there exists δ=δ(ε)>0 such that T(1+x,1)>J2-ε(1+x,1) for x∈(0,δ).



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