# An Optimal Double Inequality between Seiffert and Geometric Means

Authors: Yu-Ming Chu; Miao-Kun Wang; Zi-Kui Wang
Journal: Journal of Applied Mathematics
 (2011)

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## Abstract

Forα,β∈(0,1/2) we prove that the double inequality G(αa+(1−α)b,αb+(1−α)a)<P(a,b)<G(βa+(1−β)b,βb+(1−β)a) holds for all a,b>0 with a≠b if and only if α≤(1−1−4/π2)/2 and β≥(3−3)/6. Here, G(a,b) and P(a,b) denote the geometric and Seiffert means of two positive numbers a and b, respectively.


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## Body

## 1. Introduction

Fora,b>0 with a≠b the Seiffert mean P(a,b) was introduced by Seiffert [1] as follows:(1.1)P(a,b)=a-b4arctana/b-π.Recently, the bivariate mean values have been the subject of intensive research. In particular, many remarkable inequalities for the Seiffert mean can be found in the literature [1–9].LetH(a,b)=2ab/(a+b), G(a,b)=ab, L(a,b)=(a-b)/(loga-logb), I(a,b)=1/e(bb/aa)1/(b-a), A(a,b)=(a+b)/2, C(a,b)=(a2+b2)/(a+b), and Mp(a,b)=[(ap+bp)/2]1/p(p≠0) and M0(a,b)=ab be the harmonic, geometric, logarithmic, identric, arithmetic, contraharmonic, and pth power means of two different positive numbers a and b, respectively. Then it is well known that (1.2)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)<C(a,b)<max⁡{a,b}
for all a,b>0 with a≠b.For alla,b>0 with a≠b, Seiffert [1] established that L(a,b)<P(a,b)<I(a,b); Jagers [4] proved that M1/2(a,b)<P(a,b)<M2/3(a,b) and M2/3(a,b) is the best possible upper power mean bound for the Seiffert mean P(a,b); Seiffert [7] established that P(a,b)>A(a,b)G(a,b)/L(a,b) and P(a,b)>2A(a,b)/π; Sándor [6] presented that (A(a,b)+G(a,b))/2<P(a,b)<A(a,b)(A(a,b)+G(a,b))/2 and A2(a,b)G(a,b)3<P(a,b)<(G(a,b)+2A(a,b))/3; Hästö [3] proved that P(a,b)>Mlog2/logπ(a,b) and Mlog2/logπ(a,b) is the best possible lower power mean bound for the Seiffert mean P(a,b).Very recently, Wang and Chu [8] found the greatest value α and the least value β such that the double inequality Aα(a,b)H1-α(a,b)<P(a,b)<Aβ(a,b)H1-β(a,b) holds for a,b>0 with a≠b; For any α∈(0,1), Chu et al. [10] presented the best possible bounds for Pα(a,b)G1-α(a,b) in terms of the power mean; In [2] the authors proved that the double inequality αA(a,b)+(1-α)H(a,b)<P(a,b)<βA(a,b)+(1-β)H(a,b) holds for all a,b>0 with a≠b if and only if α≤2/π and β≥5/6; Liu and Meng [5] proved that the inequalities (1.3)α1C(a,b)+(1-α1)G(a,b)<P(a,b)<β1C(a,b)+(1-β1)G(a,b),α2C(a,b)+(1-α2)H(a,b)<P(a,b)<β2C(a,b)+(1-β2)H(a,b)
hold for all a,b>0 with a≠b if and only if α1≤2/9, β1≥1/π, α2≤1/π and β2≥5/12.For fixeda,b>0 with a≠b and x∈[0,1/2], let (1.4)g(x)=G(xa+(1-x)b,xb+(1-x)a).Then it is not difficult to verify thatg(x) is continuous and strictly increasing in [0,1/2]. Note that g(0)=G(a,b)<P(a,b) and g(1/2)=A(a,b)>P(a,b). Therefore, it is natural to ask what are the greatest value α and least value β in (0,1/2) such that the double inequality G(αa+(1-α)b,αb+(1-α)a)<P(a,b)<G(βa+(1-β)b,βb+(1-β)a) holds for all a,b>0 with a≠b. The main purpose of this paper is to answer these questions. Our main result is the following Theorem 1.1.Theorem 1.1.
Ifα,β∈(0,1/2), then the double inequality
(1.5)G(αa+(1-α)b,αb+(1-α)a)<P(a,b)<G(βa+(1-β)b,βb+(1-β)a)
holds for all a,b>0 with a≠b if and only if α≤(1-1-4/π2)/2 and β≥(3-3)/6.


## 2. Proof of Theorem1.1

Proof of Theorem1.1.
Letλ=(1-1-4/π2)/2 and μ=(3-3)/6. We first prove that inequalities
(2.1)P(a,b)>G(λa+(1-λ)b,λb+(1-λ)a),(2.2)P(a,b)<G(μa+(1-μ)b,μb+(1-μ)a)
hold for all a,b>0 with a≠b.
Without loss of generality, we assume thata>b. Let t=a/b>1 and p∈(0,1/2), then from (1.1) one has
(2.3)log⁡G(pa+(1-p)b,pb+(1-p)a)-log⁡P(a,b)=12log⁡[(pt2+(1-p))((1-p)t2+p)]-log⁡t2-14arctant-π.
Let
(2.4)f(t)=12log⁡[(pt2+(1-p))((1-p)t2+p)]-log⁡t2-14arctant-π,
then simple computations lead to
(2.5)f(1)=0,(2.6)lim⁡t→+∞f(t)=12log⁡[p(1-p)]+log⁡π,(2.7)f′(t)=t(t2+1)(t2-1)(4arctant-π)(pt2+(1-p))((1-p)t2+p)f1(t),
where
(2.8)f1(t)=4(t2-1)(pt2+1-p)[(1-p)t2+p]t(t2+1)2-4arctant+π.(2.9)f1(1)=0,(2.10)lim⁡t→+∞f1(t)=+∞,(2.11)f1′(t)=4f2(t2)t2(t2+1)4,
where f2(t)=p(1-p)t5-(3p-2)(3p-1)t4+2(5p2-5p+1)t3+2(5p2-5p+1)t2-(3p-2)(3p-1)t+p(1-p).
Note that(2.12)f2(1)=0,(2.13)lim⁡t→+∞f2(t)=+∞,(2.14)f2′(t)=5p(1-p)t4-4(3p-2)(3p-1)t3+6(5p2-5p+1)t2+4(5p2-5p+1)t-(3p-2)(3p-1),(2.15)f2′(1)=0,(2.16)lim⁡t→+∞f2′(t)=+∞,(2.17)f2′′(t)=20p(1-p)t3-12(3p-2)(3p-1)t2+12(5p2-5p+1)t+4(5p2-5p+1),(2.18)f2′′(t)=-8(6p2-6p+1),(2.19)lim⁡t→+∞f2′′(t)=+∞,(2.20)f3′′′(t)=60p(1-p)t2-24(3p-2)(3p-1)t+12(5p2-5p+1),(2.21)f2′′′(1)=-36(6p2-6p+1),(2.22)lim⁡t→+∞f2′′′(t)=+∞,(2.23)f2(4)(t)=120p(1-p)t-24(3p-2)(3p-1),(2.24)f2(4)(1)=-48(7p2-7p+1),(2.25)lim⁡t→+∞f2(4)(t)=+∞.
We divide the proof into two cases.
Case 1 (p=λ=(1-1-4/π2)/2).
Then (2.6), (2.18), (2.21), and (2.24) become
(2.26)lim⁡t→+∞f(t)=0,(2.27)f2′′(1)=-8(π2-6)π2<0,(2.28)f2′′′(1)=-36(π2-6)π2<0,(2.29)f2(4)(1)=-48(π2-7)π2<0.
From (2.23) we clearly see that f2(4)(t) is strictly increasing in [1,+∞), then (2.25) and inequality (2.29) lead to the conclusion that there exists λ1>1 such that f2(4)(t)<0 for t∈[1,λ1) and f2(4)(t)>0 for t∈(λ1,+∞). Thus, f2′′′(t) is strictly decreasing in [1,λ1] and strictly increasing in [λ1,+∞).
It follows from (2.22) and inequality (2.28) together with the piecewise monotonicity of f2′′′(t) that there exists λ2>λ1>1 such that f2′′(t) is strictly decreasing in [1,λ2] and strictly increasing in [λ2,+∞). Then (2.19) and inequality (2.27) lead to the conclusion that there exists λ3>λ2>1 such that f2′(t) is strictly decreasing in [1,λ3] and strictly increasing in [λ3,+∞).
From (2.15) and (2.16) together with the piecewise monotonicity of f2′(t) we know that there exists λ4>λ3>1 such that f2(t) is strictly decreasing in [1,λ4] and strictly increasing in [λ4,+∞). Then (2.11)–(2.13) lead to the conclusion that there exists λ5>λ4>1 such that f1(t) is strictly decreasing in [1,λ5] and strictly increasing in [λ5,+∞).
It follows from (2.7)–(2.10) and the piecewise monotonicity of f1(t) that there exists λ6>λ5>1 such that f(t) is strictly decreasing in [1,λ6] and strictly increasing in [λ6,+∞).
Therefore, inequality (2.1) follows from (2.3)–(2.5) and the piecewise monotonicity of f(t).
Case 2 (p=μ=(3-3)/6).
Then (2.18), (2.21) and (2.24) become
(2.30)f2′′(1)=0,(2.31)f2′′′(1)=0,(2.32)f2(4)(1)=8>0.
From (2.23) we clearly see that f2(4)(t) is strictly increasing in [1,+∞), then inequality (2.32) leads to the conclusion that f2′′′(t) is strictly increasing in [1,+∞).
Therefore, inequality (2.2) follows from (2.3)–(2.5), (2.7)–(2.9), (2.11), (2.12), (2.15), and inequalities (2.30) and (2.31) together with the monotonicity of f2′′′(t).
Next, we prove thatλ=(1-1-4/π2)/2 is the best possible parameter such that inequality (2.1) holds for all a,b>0 with a≠b. In fact, if (1-1-4/π2)/2=λ<p<1/2, then (2.6) leads to
(2.33)lim⁡t→+∞f(t)=12log⁡[p(1-p)]+log⁡π>0.
Inequality (2.33) implies that there exists T=T(p)>1 such that
(2.34)f(t)>0
for t∈(T,+∞).
It follows from (2.3) and (2.4) together with inequality (2.34) that P(a,b)<G(pa+(1-p)b,pb+(1-p)a) for a/b∈(T2,+∞).
Finally, we prove thatμ=(3-3)/6 is the best possible parameter such that inequality (2.2) holds for all a,b>0 with a≠b. In fact, if 0<p<μ=(3-3)/6, then from (2.18) we get f2′′(1)<0, which implies that there exists δ>0 such that
(2.35)f2′′(t)<0
for t∈[1,1+δ).
Therefore,P(a,b)>G(pa+(1-p)b,pb+(1-p)a) for a/b∈(1,(1+δ)2) follows from (2.3)–(2.5), (2.7)–(2.9), (2.11), (2.12), and (2.15) together with inequality (2.35).



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*Source: 261237-2011-11-23.xml*