# Optimal Lower Generalized Logarithmic Mean Bound for the Seiffert Mean

Authors: Ying-Qing Song; Wei-Mao Qian; Yun-Liang Jiang; Yu-Ming Chu
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

We present the greatest valuep such that the inequality P(a,b)>Lp(a,b) holds for all a,b>0 with a≠b, where P(a,b) and Lp(a,b) denote the Seiffert and pth generalized logarithmic means of a and b, respectively.


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## Body

## 1. Introduction

Forp∈ℝ, the pth generalized logarithmic mean Lp:(0,∞)2→(0,∞) is defined by
(1)Lp(a,b)={[bp+1-ap+1(p+1)(b-a)]1/p,a≠b,p≠0,-1,1e(bbaa)1/(b-a),a≠b,p=0,b-alogb-loga,a≠b,p=-1,a,a=b,
and the Seiffert mean P:(0,∞)2→(0,∞) [1] is defined by
(2)P(a,b)={a-b4arctan(a/b)-π,a≠b,a,a=b.It is well known that the generalized logarithmic meanLp(a,b) is continuous and strictly increasing with respect to p∈ℝ for fixed a,b>0 with a≠b. The special cases of the generalized logarithmic mean are, for example, G(a,b)=ab=L-2(a,b) is the geometric mean, L(a,b)=(b-a)/ (logb-loga)=L-1(a,b) is the logarithmic mean, I(a,b)=1/e(bb/aa)1/(b-a)=L0(a,b) is the identric mean, and A(a,b)=(a+b)/2=L1(a,b) is the arithmetic mean. The Seiffert mean P(a,b) can be rewritten as (see [2, equation (2.4)])
(3)P(a,b)={a-b2arcsin[(a-b)/(a+b)],a≠b,a,a=b.Recently, the bivariate means have been the subject of intensive research. In particular, many remarkable inequalities and properties for the generalized logarithmic and the Seiffert means can be found in the literature [3–13].In [1, 11], Seiffert proved that the inequalities
(4)L-1(a,b)<P(a,b)<L0(a,b)=I(a,b),2πA(a,b)<P(a,b)<A(a,b),P(a,b)>G(a,b)A(a,b)L(a,b),P(a,b)>3A(a,b)G(a,b)A(a,b)+2G(a,b)
hold for all a,b>0 with a≠b.Sándor [14] presented the bounds for the Seiffert mean P(a,b) in terms of the arithmetic mean A(a,b) and geometric mean G(a,b) as follows:
(5)12A(a,b)+12G(a,b)<P(a,b)<23A(a,b)+13G(a,b)
for all a,b>0 with a≠b.Hästö [15] proved that the double inequality Mα(a,b)<P(a,b)<Mβ(a,b) holds for all a,b>0 with a≠b if and only if α≤log2/logπ and β≥2/3, where Mp(a,b)=((ap+bp)/2)1/p is the pth power mean of a and b.In [16], the authors found the greatest value α and least value β such that the double inequality αC(a,b)+(1-α)G(a,b)<P(a,b)<βC(a,b)+(1-β)G(a,b) holds for all a,b>0 with a≠b, where C(a,b)=(a2+b2)/(a+b) is the contraharmonic mean of a and b.Motivated by the first inequality in (4), Gao [17] gave the best possible constants λ and μ such that the double inequality λI(a,b)<P(a,b)<μI(a,b) holds for all a,b>0 with a≠b.In [18], the author solved the following open problem proposed by Long and Chu [19]: what is the smallest p (largest q) such that the inequality αA(a,b)+(1-α)G(a,b)<Lp(a,b)(>Lq(a,b)) holds for α∈(0,1/2)(α∈(1/2,1)) and all a,b>0 with a≠b?Chu et al. [20] proved that the double inequality Lp(a,b)<T(a,b)<Lq(a,b) holds for all a,b>0 with a≠b if and only if p≤3 and q≥q0=3.152…, where T(a,b)=(a-b)/[2arctan((a-b)/(a+b))] is the second Seiffert mean of a and b, and p0=3.152… is the unique solution of the equation (x+1)-1/x=2/π.In [21], the authors answered the question: what are the greatest value p1=p1(q) and the least value p2=p2(q) such that the double inequality Lp1(a,b)<[L(aq,bq)]1/q<Lp2(a,b) holds for any q>0 with q≠1 and all a,b>0 with a≠b?Motivated by the first inequality in (4), it is natural to ask what are the best possible generalized logarithmic mean bounds for the Seiffert mean P(a,b)? It is the aim of this paper to answer this question.


## 2. Preliminaries

In order to prove our main results, we need two lemmas, which we present in this section.Lemma 1.
Let the functionE:(-1,+∞)→ℝ+ be defined with
(6)(x+1)-1/x,x∈(-1,0)∪(0,+∞),1e,x=0.
Then, E is a continuous and strictly increasing function.Proof.
From (6), we clearly see that
(7)limx→0E(x)=1e=E(0),(8)limx→-1+E(x)=0,limx→+∞E(x)=1.
If x∈(-1,0)∪(0,+∞), then simple computation yields
(9)E′(x)=(1+x)log(1+x)-xx2(1+x)E(x).
If we define
(10)F(x)=(1+x)log(1+x)-x,
then
(11)F′(x)=log(1+x).
Equation (11) implies that
(12)F′(x)<0,x∈(-1,0),F′(x)>0,x∈(0,+∞).
Equations (10) and (12) lead to
(13)F(x)>limx→0F(x)=0
for x∈(-1,0)∪(0,+∞).
From (9) and (10) together with (13), we clearly see that
(14)E′(x)>0
for x∈(-1,0)∪(0,+∞).
Therefore, the continuity ofE follows from (6) and (7), and the strict monotonicity of E follows from (8), (14) and the continuity of E.Remark 2.
From Lemma1, we clearly see that for any fixed λ∈(0,1), there exists a unique x∈(-1,∞) such E(x)=λ. In particular, for λ=1/π, making use of Mathematica software, we get
(15)E(-0.241)-1π=0.000166⋯>0,E(-0.242)-1π=-0.000062⋯<0.
Therefore, the unique solution of the equationE(x)=1/π belongs to the interval (-0.242,-0.241).Lemma 3.
Letp∈(-0.242,-0.241) and let g:[1,+∞)→ℝ be defined with g(t)=pt2p+3+ 2(p-1)t2p+2-(3p+2)t2p+1-p(2p+1)tp+3+(2p2+p+4)tp+2+(2p2+p+4)tp+1-p(2p+1)tp- (3p+2)t2+ 2(p-1)t+p. Then, there exists λ∈(1,+∞) such that g(t)<0 for t∈(1,λ) and g(t)>0 for t∈(λ,+∞).Proof.
Simple computations lead to(16)g(1)=0,(17)limt→+∞g(t)=+∞,(18)g′(t)=p(2p+3)t2p+2+4(p-1)(p+1)t2p+1m-(2p+1)(3p+2)t2p-p(p+3)(2p+1)tp+2m+(p+2)(2p2+p+4)tp+1+(p+1)m×(2p2+p+4)tp-p2(2p+1)tp-1m-2(3p+2)t+2(p-1),(19)g′(1)=0,(20)limt→+∞g′(t)=+∞,(21)g′′(t)=2p(p+1)(2p+3)t2p+1m+4(p-1)(p+1)(2p+1)t2p-2pm×(2p+1)(3p+2)t2p-1m-p(p+2)(p+3)(2p+1)tp+1m+(p+1)(p+2)(2p2+p+4)tpm+p(p+1)(2p2+p+4)tp-1m-p2(p-1)(2p+1)tp-2-2(3p+2),(22)g′′(1)=0,(23)limt→+∞g′′(t)=+∞.
Leth(t)=t3-pg′′′(t). Further computations lead to
(24)h(t)=2p(p+1)(2p+1)(2p+3)tp+3N+8p(p-1)(p+1)(2p+1)tp+2m-2p(2p-1)(2p+1)(3p+2)tp+1m-p(p+1)(p+2)(p+3)m×(2p+1)t3+p(p+1)(p+2)m×(2p2+p+4)t2+p(p-1)m×(p+1)(2p2+p+4)tm-p2(p-2)(p-1)(2p+1),(25)h(1)=0,(26)limt→+∞h(t)=+∞,(27)h′(t)=2p(p+1)(p+3)(2p+1)(2p+3)tp+2m+8p(p-1)(p+1)(p+2)(2p+1)tp+1m-2p(p+1)(2p-1)(2p+1)(3p+2)tpm-3p(p+1)(p+2)(p+3)(2p+1)t2m+2p(p+1)(p+2)(2p2+p+4)t+pm×(p-1)(p+1)(2p2+p+4),(28)h′(1)=24p3(p+1)<0,(29)limt→+∞h′(t)=+∞,(30)h′′(t)=2p(p+1)(p+2)(p+3)(2p+1)(2p+3)tp+1n+8p(p-1)(p+1)2(p+2)(2p+1)tpn-2p2(p+1)(2p-1)(2p+1)(3p+2)tp-1n-6p(p+1)(p+2)(p+3)(2p+1)tm+2p(p+1)(p+2)(2p2+p+4),(31)h′′(1)=p3(72p2+156p+84)<0,(32)limt→+∞h′′(t)=+∞,(33)h′′′(t)=2p(p+1)2(p+2)(p+3)(2p+1)m×(2p+3)tp+8p2(p-1)(p+1)2m×(p+2)(2p+1)tp-1-2p2(p-1)m×(p+1)(2p-1)(2p+1)m×(3p+2)tp-2-6p(p+1)m×(p+2)(p+3)(2p+1),(34)h′′′(1)=p2(112p4+312p3+352p2+192p+40)>0,(35)h(4)(t)=2p2(p+1)2(p+2)(p+3)m×(2p+1)(2p+3)tp-1m+8p2(p-1)2(p+1)2(p+2)(2p+1)tp-2m+2p2(1-p)(2-p)(1-2p)M×(p+1)(2p+1)(3p+2)tp-3>0.
Inequalities (34) and (35) imply that
(36)h′′′(t)>0
for t∈(1,∞).
From (31) and (32) together with (36), we clearly see that there exists t0∈(1,∞) such that h′′(t)<0 for t∈(1,t0) and h′′(t)>0 for t∈(t0,∞). Hence, h′ is strictly decreasing on (1,t0) and strictly increasing on (t0,∞).
It follows from (28) and (29) together with the monotonicity of h′ that there exists t1∈(1,∞) such that h is strictly decreasing on (1,t1) and strictly increasing on (t1,∞).
Equations (25)-(26) and the monotonicity of h lead to the conclusion that there exists t2∈(1,∞) such that h(t)<0 for t∈(1,t2) and h(t)>0 for t∈(t2,∞). Therefore, g′′ is strictly decreasing on (1,t2) and strictly increasing on (t2,∞).
From (22) and (23) together with the monotonicity of g′′, we clearly see that there exists t3∈(1,∞) such that g′ is strictly decreasing on (1,t3) and strictly increasing on (t3,∞).
Equations (19)-(20) and the monotonicity of g′ imply that there exists t4∈(1,∞) such that g is strictly decreasing on (1,t4) and strictly increasing on (t4,∞).
Therefore, Lemma3 follows from (16) and (17) together with the monotonicity of g.


## 3. Main Results

Theorem 4.
LetE be as in Lemma 1 and let p be the unique solution of the equation E(x)=1/π. Then, for all a,b>0, a≠b, the inequality
(37)Lp(a,b)<P(a,b)
holds, and Lp(a,b) is the best possible lower generalized logarithmic mean bound for the Seiffert mean P(a,b).Proof.
From Remark2, it follows that
(38)p∈(-0.242,-0.241).
We first prove that the inequality (37) holds. Without loss of generality, we assume that a>b. If t=a/b>1, then, from (1) and (2), we have
(39)logLp(a,b)-logP(a,b)mm=1plogtp+1-1(p+1)(t-1)mmm+log(4arctant-π)-log(t-1),
If
(40)f(t)=1plogtp+1-1(p+1)(t-1)m+log(4arctant-π)-log(t-1),
then simple computations lead to
(41)limt→1f(t)=limt→+∞f(t)=0,(42)f′(t)=(p+1)(tp-1)p(tp+1-1)(t-1)(4arctant-π)f1(t),
where
(43)f1(t)=-4arctant+π+2p(tp+1-1)(t-1)(p+1)t(t+1)(tp-1),limt→1f1(t)=0,limt→+∞f1(t)=+∞,(44)f1′(t)=f2(t)(1+p)(t+1)2(tp-1)2t3/2,
where
(45)f2(t)=pt2p+3+2(p-1)t2p+2-(3p+2)t2p+1m-p(2p+1)tp+3+(2p2+p+4)tp+2m+(2p2+p+4)tp+1-p(2p+1)tpm-(3p+2)t2+2(p-1)t+p.
From (38) and (44)-(45) together with Lemma 3, we clearly see that there exists λ∈(1,+∞) such that f1 is strictly decreasing on (1,λ) and strictly increasing on (λ,∞). Then, (38) and (42)-(43) together with the monotonicity of f1 imply that there exists μ∈(1,+∞) such that f is strictly decreasing on (1,μ) and strictly increasing on (μ,+∞).
Therefore,Lp(a,b)<P(a,b) follows from (39)–(41) and the monotonicity of f.
Next, we prove thatLp(a,b) is the best possible lower generalized logarithmic mean bound for the Seiffert mean P(a,b).
For any0<ε<-p, from (1) and (2), we get
(46)limt→+∞Lp+ε(t,1)P(t,1)=π(p+1+ε)-1/(p+ε).
Lemma 1 and (46) lead to
(47)limt→+∞Lp+ε(t,1)P(t,1)>1.
Inequality (47) implies that, for 0<ε<-p, there exists T=T(ε)>1 such that Lp+ε(t,1)>P(t,1) for t∈(T,∞).Theorem 5.
L
0
(
a
,
b
) is the best possible upper generalized logarithmic mean bound for the Seiffert mean P(a,b).Proof.
For any0<ε<1 and x>0, from (1) and (2), we have
(48)[P(1+x,1)]ε-[L-ε(1+x,1)]εm=f(x)[(1+x)1-ε-1][4arctan1+x-π]ε,
where f(x)=[(1+x)1-ε-1]xε-(1-ε)(4arctan1+x-π)εx.
Whenx→0, then making use of the Taylor expansion, we get
(49)f(x)=[(1-ε)x-12ε(1-ε)x2mm+16ε(1-ε)(1+ε)x3+o(x3)]xεm-(1-ε)[x-12x2+724x3+o(x3)]εx=124ε2(1-ε)x3+ε+o(x3+ε).
Equations (48) and (49) imply that for any 0<ε<1, there exists δ=δ(ε)>0 such that
(50)P(1+x,1)>L-ε(1+x,1)
for x∈(0,δ).
Therefore, Theorem5 follows from inequalities (4) and (50).



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