# Best Possible Bounds for Neuman-Sándor Mean by the Identric, Quadratic and Contraharmonic Means

Authors: Tie-Hong Zhao; Yu-Ming Chu; Yun-Liang Jiang; Yong-Min Li
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We prove that the double inequalitiesIα1(a,b)Q1-α1(a,b)<M(a,b)<Iβ1(a,b)Q1-β1(a,b),Iα2(a,b)C1-α2(a,b)<M(a,b)<Iβ2(a,b)C1-β2(a,b) hold for all a,b>0 with a≠b if and only if α1≥1/2, β1≤log[2log(1+2)]/(1-log2), α2≥5/7, and β2≤log[2log(1+2)], where I(a,b), M(a,b), Q(a,b), and C(a,b) are the identric, Neuman-Sándor, quadratic, and contraharmonic means of a and b, respectively.


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## Body

## 1. Introduction

Forp∈ℝ and a,b>0 with a≠b, the identric mean I(a,b), Neuman-Sándor mean M(a,b) [1], quadratic mean Q(a,b), contraharmonic mean C(a,b), and pth power mean Mp(a,b) are defined by
(1)I(a,b)=1e(bbaa)1/(b-a),M(a,b)=a-b2sinh-1[(a-b)/(a+b)],Q(a,b)=a2+b22,C(a,b)=a2+b2a+b,Mp(a,b)={(ap+bp2)1/p,p≠0,ab,p=0,
respectively, where sinh-1(x)=log(x+1+x2) is the inverse hyperbolic sine function.Recently, the identric, Neuman-Sándor, quadratic, and contraharmonic means have attracted the interest of numerous eminent mathematicians. In particular, many remarkable inequalities for these means can be found in the literature [1–18].LetH(a,b)=2ab/(a+b), G(a,b)=ab, L(a,b)=(b-a)/(logb-loga), P(a,b)=(a-b)/(4arctana/b-π), A(a,b)=(a+b)/2, and T(a,b)=(a-b)/[2arctan((a-b)/(a+b))] be the harmonic, geometric, logarithmic, first Seiffert, arithmetic, and second Seiffert means of two distinct positive numbers a and b, respectively. Then it is well known that the inequalities
(2)H(a,b)=M-1(a,b)<G(a,b)=M0(a,b)<L(a,b)<P(a,b)<I(a,b)<A(a,b)<M1(a,b)<M(a,b)<T(a,b)<Q(a,b)=M2(a,b)<C(a,b)
hold for all a,b>0 with a≠b.Neuman and Sándor [1, 8] established that
(3)A(a,b)<M(a,b)<A(a,b)log(1+2),π4T(a,b)<M(a,b)<T(a,b),M(a,b)<A2(a,b)P(a,b),A(a,b)T(a,b)<M(a,b)<A2(a,b)+T2(a,b)2,M(a,b)<2A(a,b)+Q(a,b)3
for all a,b>0 with a≠b.Let0<a,b≤1/2 with a≠b, a′=1-a, and b′=1-b. Then the Ky Fan inequalities
(4)G(a,b)G(a′,b′)<L(a,b)L(a′,b′)<P(a,b)P(a′,b′)<A(a,b)A(a′,b′)<M(a,b)M(a′,b′)<T(a,b)T(a′,b′)
were presented in [1].Li et al. [19] found the best possible bounds for the Neuman-Sándor mean in terms of the generalized logarithmic mean Lr(a,b). Neuman [20] and Zhao et al. [21] proved that the inequalities
(5)αQ(a,b)+(1-α)A(a,b)<M(a,b)<βQ(a,b)+(1-β)A(a,b),λC(a,b)+(1-λ)A(a,b)<M(a,b)<μC(a,b)+(1-μ)A(a,b),α1H(a,b)+(1-α1)Q(a,b)<M(a,b)<β1H(a,b)+(1-β1)Q(a,b),α2C(a,b)+(1-α2)Q(a,b)<M(a,b)<β2C(a,b)+(1-β2)Q(a,b)
hold for all a,b>0 with a≠b if and only if α≤[1-log(1+2)]/[(2-1)log(1+2)], β≥1/3, λ≤[1-log(1+2)]/log(1+2), μ≥1/6, α1≥2/9, β1≤1-1/[2log(1+2)], α2≥1/3, and β2≤1-1/[2log(1+2)].In [22], Chu and Long gave the best possible constants p,q,α, and β such that the double inequalities Mp(a,b)<M(a,b)<Mq(a,b) and αI(a,b)<M(a,b)<βI(a,b) hold for all a,b>0 with a≠b.The ratio of identric means leads to the weighted geometric mean(6)I(a2,b2)I(a,b)=(aabb)1/(a+b),
which has been investigated in [23–25]. Alzer [26] proved that the inequalities
(7)A(a,b)G(a,b)<I(a,b)L(a,b)<I(a,b)+L(a,b)2<A(a,b)+G(a,b)2
hold for all a,b>0 with a≠b.The following sharp bounds forI, (IL)1/2, and (I+L)/2 in terms of the power mean and the convex combination of arithmetic and geometric means are given in [27] as
(8)M2/3(a,b)<I(a,b)<Mlog2(a,b),M0(a,b)<I(a,b)L(a,b)<M1/2(a,b),Mlog2/(1+log2)(a,b)<I(a,b)+L(a,b)2<M1/2(a,b),23A(a,b)+13G(a,b)<I(a,b)<2eA(a,b)+(1-2e)G(a,b)
for all a,b>0 with a≠b.Chu et al. [28] presented the optimal constants α1,β1,α2, and β2 such that the double inequalities
(9)α1Q(a,b)+(1-α1)A(a,b)<2π∫0π/2‍a2cos2θ+b2sin2θdθ<β1Q(a,b)+(1-β1)A(a,b),Qα2(a,b)A1-α2(a,b)<2π∫0π/2‍a2cos2θ+b2sin2θdθ<Qβ2(a,b)A1-β2(a,b)
hold for all a,b>0 with a≠b.The aim of this paper is to find the best possible constantsα1,β1,α2 and β2 such that the double inequalities
(10)Iα1(a,b)Q1-α1(a,b)<M(a,b)<Iβ1(a,b)Q1-β1(a,b),Iα2(a,b)C1-α2(a,b)<M(a,b)<Iβ2(a,b)C1-β2(a,b)
hold for all a,b>0 with a≠b. All numerical computations are carried out using mathematica software.


## 2. Lemmas

In order to prove our main results, we need several lemmas, which we present in this section.Lemma 1.
The double inequality(11)x+x33-2x515<1+x2sinh-1(x)<x+x33-2x515+8x7105
holds for x∈(0,1).Proof.
To prove Lemma1, it suffices to prove that
(12)η1(x)=1+x2sinh-1(x)-(x+x33-2x515)>0,(13)η2(x)=1+x2sinh-1(x)-(x+x33-2x515+8x7105)<0
for x∈(0,1).
From the expressions ofη1(x) and η2(x), we get
(14)η1(0)=η2(0)=0,η1′(x)=xη1*(x)1+x2,η2′(x)=xη2*(x)1+x2,
where
(15)η1*(x)=sinh-1(x)-(x-2x33)1+x2,η2*(x)=sinh-1(x)-(x-2x33+8x515)1+x2,η1*(0)=η2*(0)=0,(16)η1*′(x)=8x431+x2>0,(17)η2*′(x)=-16x651+x2<0,
for x∈(0,1).
Therefore, inequality (12) follows from (14)–(16), and inequality (13) follows from (14)–(17).Lemma 2.
Let(18)L(x)=log[(1+x)1+x(1-x)1-x]-2x-xlog(1-x2).
Then
(19)L(x)>2x33+2x55+2x77
for x∈(0,1), and
(20)L(x)<2x33+2x55+2x77+x9
for x∈(0,3/4).Proof.
To prove inequalities (19) and (20), it suffices to show that
(21)ι1(x)≔log[(1+x)1+x(1-x)1-x]-2x-xlog(1-x2)-(2x33+2x55+2x77)>0
for x∈(0,1), and
(22)ι2(x)≔log[(1+x)1+x(1-x)1-x]-2x-xlog(1-x2)-(2x33+2x55+2x77+x9)<0
for x∈(0,3/4).
From (21) and (22), one has
(23)ι1(0+)=ι2(0+)=0,(24)ι1′(x)=2x81-x2>0
for x∈(0,1), and
(25)ι2′(x)=-9x81-x2(79-x2)<0
for x∈(0,3/4).
Therefore, inequality (21) follows from (23) and (24), and inequality (22) follows from (23) and (25).Lemma 3.
Let(26)Φ1(x)=11+x2sinh-1(x)-1x(1+x2).
Then the double inequality(27)2x3-34x345+x52<Φ1(x)<2x3-34x345+4x55
holds for x∈(0,0.7).Proof.
To prove inequality (27), it suffices to show that
(28)ϕ1(x)=x1+x2-sinh-1(x)-x(1+x2)sinh-1(x)(2x3-34x345+x52)>0,(29)ϕ2(x)=x1+x2-sinh-1(x)-x(1+x2)sinh-1(x)(2x3-34x345+4x55)<0
for x∈(0,0.7).
First, we prove inequality (28). From the expression of ϕ1(x), we have
(30)ϕ1(0)=0,ϕ1(0.7)=0.0033⋯,(31)ϕ1′(x)=xϕ1*(x)901+x2,
where
(32)ϕ1*(x)=120x+8x3+23x5-45x7-2(60-16x2-69x4+180x6)×1+x2sinh-1(x).
Equation (32) leads to
(33)ϕ1*(0.6)=3.017⋯,ϕ1*(0.7)=-3.551⋯,ϕ1*′(x)=-xϕ1**(x)1+x2,
where
(34)ϕ1**(x)=-56x-309x3+422x5+675x7+2(28-324x2+735x4+1260x6)×1+x2sinh-1(x).
Note that(35)60-16x2-69x4+180x6>0
for x∈(0,0.6], and
(36)28-324x2+735x4+1260x6>0
for x∈[0.6,0.7).
It follows from (32) and (34)–(36) together with Lemma 1 that
(37)ϕ1*(x)>120x+8x3+23x5-45x7-2(60-16x2-69x4+180x6)(x+x33)=x53(515-1077x2-360x4)≥x53[515-1077×925-360×81625]=10078x5375>0
for x∈(0,0.6], and
(38)ϕ1**(x)>-56x-309x3+422x5+675x7+2(28-324x2+735x4+1260x6)×(x+x33-2x515)=x315×(-14075+25028x2+56571x4+9660x6-5040x8)>x315[-14075+25028×(0.6)2+56571×(0.6)4]=1416676x39375>0
for x∈[0.6,0.7).
From (33), (37), and (38), we clearly see that there exists x1∈(0.6,0.7) such that ϕ1*(x)>0 for x∈(0,x1) and ϕ1*(x)<0 for x∈(x1,0.7). Then (31) leads to the conclusion that ϕ1(x) is strictly increasing on (0,x1] and strictly decreasing on [x1,0.7).
Therefore, inequality (28) follows from (30) and the piecewise monotonicity of ϕ1(x).
Next, we prove inequality (29). From the expression of ϕ2(x), we get
(39)ϕ2(0)=0,ϕ2′(x)=-2xϕ2*(x)451+x2,
where
(40)ϕ2*(x)=x(18x6+x4-2x2-30)+2(15-4x2+3x4+72x6)×1+x2sinh-1(x).
It follows from Lemma1 and (40) that
(41)ϕ2*(x)>x(18x6+x4-2x2-30)+2(15-4x2+3x4+72x6)(x+x33-2x515)=x515(5+2476x2+708x4-288x6)>0
for x∈(0,0.7).
Therefore, inequality (29) follows from (39) together with (41).Lemma 4.
Let(42)Φ2(x)=11+x2sinh-1(x)-1-x2x(1+x2).
Then the double inequality
(43)5x3-79x345+11x510<Φ2(x)<5x3-79x345+9x55
holds for x∈(0,3/4).Proof.
To prove Lemma4, it suffices to prove that
(44)φ1(x)≔x1+x2-(1-x2)sinh-1(x)-x(1+x2)sinh-1(x)(5x3-79x345+11x510)>0,(45)φ2(x)≔x1+x2-(1-x2)sinh-1(x)-x(1+x2)sinh-1(x)(5x3-79x345+9x55)<0
for x∈(0,3/4).
We first prove inequality (44). From the expression of φ1(x), we obtain
(46)φ1(0)=0,φ1(34)=0.008457⋯>0,(47)φ1′(x)=xφ1*(x)901+x2,
where
(48)φ1*(x)=120x+8x3+59x5-99x7-2×(60-16x2-177x4+396x6)1+x2sinh-1(x).
Equation (48) leads to
(49)φ1*(0.66)=6.02⋯>0,φ1*(34)=-19.299⋯<0,(50)φ1*′(x)=-xϕ1**(x)1+x2,
where
(51)φ1**(x)=-56x-705x3+836x5+1485x7+14(4-108x2+213x4+396x6)×1+x2sinh-1(x).
Note that(52)60-16x2-177x4+396x6>60-16×(0.66)2-177×(0.66)4=19.4451>0
for x∈(0,0.66), and
(53)4-108x2+213x4+396x6>4-108×(34)2+213×(0.66)4+396×(0.66)6=16.3972>0
for x∈[0.66,3/4).
It follows from Lemma1, (48), and (51)–(53) that
(54)φ1*(x)>120x+8x3+59x5-99x7-2(60-16x2-177x4+396x6)×(x+x33-2x515+8x7105)=x5105[46165-82573x2-32420x4=x5105+7584x6+6336x6(1-x2)]>x5105[46165-82573×(0.66)2-32420×(0.66)4]=x5105×4044.5917⋯>0
for x∈(0,0.66), and
(55)φ1**(x)>-56x-705x3+836x5+1485x7+14(4-108x2+213x4+396x6)(x+x33-2x515)=x315[-32975+49598x2+123369x4+10668x6+11088x6(1-x2)]>x315[-32975+49598×(0.66)2+123369×(0.66)4]=x315×12038.83⋯>0
for x∈[0.66,3/4).
From (50) and (55), we know that φ1*(x) is strictly decreasing on [0.66,3/4), and this in conjunction with (49) and (54) leads to the conclusion that there exists x1∈(0.66,3/4) such that φ1*(x)>0 for x∈(0,x1) and φ1*(x)<0 for x∈(x1,3/4). Then (47) implies that φ1(x) is strictly increasing on (0,x1] and strictly decreasing on [x1,3/4). Therefore, inequality (44) follows from (46) and the piecewise monotonicity of φ1(x).
Next, we prove inequality (45). From the expression of φ2(x) one has
(56)φ2(0)=0,ϕ2′(x)=-xϕ2*(x)451+x2,
where
(57)ϕ2*(x)=-60x-4x3+2x5+81x7+4(15-4x2+3x4+162x6)×1+x2sinh-1(x).
It follows from Lemma1 and (52) that
(58)ϕ2*(x)>-60x-4x3+2x5+81x7+4(15-4x2+3x4+162x6)(x+x33-2x515)=x515(10+11027x2+3216x4-1296x6)>0
for x∈(0,3/4).
Therefore, inequality (45) follows from (56) together with (58).Lemma 5.
LetL(x) be defined as in Lemma 2 and
(59)Υ1(x)=L(x)2x2+x1+x2.
Then the double inequality(60)4x3-4x35+4x55<Υ1(x)<4x3-4x35+8x57
holds for x∈(0,0.7).Proof.
From Lemma2, one has
(61)Υ1(x)-(4x3-4x35+4x55)>12x2(2x33+2x55+2x77)+x1+x2-(4x3-4x35+4x55)=23x535(1+x2)(1223-x2)>0,Υ1(x)-(4x3-4x35+8x57)<12x2(2x33+2x55+2x77+x9)+x1+x2-(4x3-4x35+8x57)=-x7(1-x2)2(1+x2)<0
for x∈(0,0.7).
Therefore, Lemma5 follows easily from (61).Lemma 6.
LetL(x) be defined as in Lemma 2 and
(62)Υ2(x)=L(x)2x2+2x1+x2.
Then the double inequality(63)7x3-9x35+7x55<Υ2(x)<7x3-9x35+15x57
holds for x∈(0,3/4).Proof.
It follows from Lemma2 that
(64)Υ2(x)-(7x3-9x35+7x55)>12x2(2x33+2x55+2x77)+2x1+x2-(7x3-9x35+7x55)=44x535(1+x2)(1322-x2)>0,Υ2(x)-(7x3-9x35+15x57)<12x2(2x33+2x55+2x77+x9)+2x1+x2-(7x3-9x35+15x57)=-x7(3-x2)2(1+x2)<0
for x∈(0,3/4).
Therefore, Lemma6 follows from (64).Lemma 7.
The inequality(65)x31+x2>[sinh-1(x)]3
holds for x∈(0,1).Proof.
Let(66)ζ(x)=x31+x2-[sinh-1(x)]3.
Then
(67)ζ(0)=0,ζ′(x)=ζ1(x)(1+x2)3/2,
where
(68)ζ1(x)=x2(3+2x2)-3[1+x2sinh-1(x)]2.
It follows from Lemma1 and (68) that
(69)ζ1(x)>x2(3+2x2)-3(x+x33-2x515+8x7105)2=x6[37525+(208525+36x2175+64x63675)=x6×(1-x2)+32x6735]>0
for x∈(0,1).
Therefore, Lemma7 follows from (67) together with (69).Lemma 8.
Let(70)μ1(x)=1+3x2(x+x3)2-1(1+x2)[sinh-1(x)]2-x(1+x2)3/2sinh-1(x).
Thenμ1(x)<0.2 for x∈[0.7,1).Proof.
Let(71)ω1(x)=1x2-1[sinh-1(x)]2,ω2(x)=21+x2-xsinh-1(x).
Then
(72)μ1(x)=ω1(x)1+x2+ω2(x)(1+x2)3/2.
Lemma7 and x>sinh-1(x) give ω1(x)<0 and
(73)ω1′(x)=2x3[sinh-1(x)]3[x31+x2-(sinh-1(x))3]>0
for x∈(0,1). This in turn implies that
(74)[ω1(x)1+x2]′=ω1′(x)(1+x2)-2xω1(x)(1+x2)2>0
for x∈(0,1).
On the other hand, from the expression ofω2(x), we get
(75)ω2(1)=0.2796⋯>0,ω2′(x)=-2x(1+x2)3/2+ω2*(x)[sinh-1(x)]2,
where
(76)ω2*(x)=x1+x2-sinh-1(x),ω2*(0)=0,ω2*′(x)=-x2(1+x2)3/2<0
for x∈(0,1).
From (75)–(76), we clearly see that ω2′(x)<0 and ω2(x)>0 for x∈(0,1). This in turn implies that
(77)[ω2(x)(1+x2)3/2]′=ω2′(x)(1+x2)3/2-3x1+x2ω2(x)(1+x2)3<0
for x∈(0,1).
Equation (72) together with inequalities (74) and (77) lead to the conclusion that
(78)μ1(x)≤ω1(1)2+ω2(0.7)[1+(0.7)2]3/2=0.167⋯<0.2
for x∈[0.7,1).Lemma 9.
Let(79)μ2(x)=1+4x2-x4(x+x3)2-1(1+x2)[sinh-1(x)]2-x(1+x2)3/2sinh-1(x).
Then μ2(x)<0.51 for x∈[0.65,1).Proof.
Let(80)τ1(x)=1x2-1[sinh-1(x)]2=μ1(x),τ2(x)=3-x21+x2-xsinh-1(x),
then
(81)μ2(x)=τ1(x)1+x2+τ2(x)(1+x2)3/2.
From (74), we clearly see that
(82)[τ1(x)1+x2]′=[ω1(x)1+x2]′>0
for x∈(0,1).
On the other hand, from the expression ofτ2(x) together with Lemma 1, we get
(83)τ2(1)=0.2796⋯>0,τ2′(x)=-1sinh-1(x)-xτ2*(x)(1+x2)3/2[sinh-1(x)]2,τ2*(x)=(5+x2)[sinh-1(x)]2-(1+x2),τ2*(0.65)=0.6033⋯,τ2*′(x)=2x[sinh-1(x)]2+2[5+x21+x21+x2sinh-1(x)-x]>0
for x∈(0,1).
From (83), we clearly see that τ2′(x)<0 and τ2(x)>0 for x∈[0.65,1). This in turn implies that
(84)[τ2(x)(1+x2)3/2]′=τ2′(x)(1+x2)3/2-3x1+x2τ2(x)(1+x2)3<0
for x∈[0.65,1).
Equation (81) together with inequalities (82) and (84) lead to the conclusion that
(85)μ2(x)≤τ1(1)2+τ2(0.65)[1+(0.65)2]3/2=0.503⋯<0.51
for x∈[0.65,1).Lemma 10.
LetL(x) be defined as in Lemma 2 and
(86)ν1(x)=2(1+x4)(1-x2)(1+x2)2-L(x)x3.
Then ν1(x)>1.2 for x∈[0.7,1).Proof.
Differentiatingν1(x) yields
(87)ν1′(x)=3L(x)x4-2+8x2-20x4-6x8x(1-x2)2(1+x2)3.
It follows from (19) and (87) that (88)ν1′(x)>1x[3(23+2x25+2x47)-2+8x2-20x4-6x8(1-x2)2(1+x2)3]=2x(-84+316x2-97x4+68x6+26x8+36x10+15x12)35(1-x2)2(1+x2)3>2x35(1-x2)2(1+x2)3[-84+316×(0.7)2-3495>2x35(1-x2)2(1+x2)3+68x4(x2-25)]>2x35(1-x2)2(1+x2)3>0for x∈[0.7,1).
Therefore,ν1(x)≥ν1(0.7)=1.214⋯>1.2 for x∈[0.7,1) follows from (88).Lemma 11.
LetL(x) be defined as in Lemma 2 and
(89)ν2(x)=3-2x2+3x4(1-x2)(1+x2)2-L(x)x3.
Then ν2(x)>1.38 for x∈[0.65,1).Proof.
Differentiatingν2(x) yields
(90)ν2′(x)=3L(x)x4-2(1+7x2-17x4+5x6-4x8)x(1-x2)2(1+x2)3.
It follows from (19) and (90) together with the monotonicity of the function 561x2-272x4 on [0.65,1) that (91)ν2′(x)>1x[3(23+2x25+2x47)-2(1+7x2-17x4+5x6-4x8)(1-x2)2(1+x2)3]=2x(-189+561x2-272x4+103x6+26x8+36x10+15x12)35(1-x2)2(1+x2)3>2x[-189+561×(0.65)2-272×(0.65)4+103×(0.65)6]35(1-x2)2(1+x2)3=2x×7.23⋯35(1-x2)2(1+x2)3>0for x∈[0.65,1).
Equation (91) leads to the conclusion that ν2(x)≥ν2(0.65)=1.389⋯>1.38 for x∈[0.65,1).Lemma 12.
LetΦ1(x) and Υ1(x) be defined, respectively, as in Lemmas 3 and 5, and Θ1(x;p)=Φ1(x)-pΥ1(x). Then Θ1(x;p) is strictly decreasing on [0.7,1) if p>1/6.Proof.
DifferentiatingΘ1(x;p) with respect to x and making use of Lemmas 8 and 10, we get
(92)dΘ1(x;p)dx=Φ1′(x)-pΥ1′(x)=μ1(x)-pν1(x)<0.2-16×1.2=0
for x∈[0.7,1) and p>1/6. This in turn implies that Θ1(x;p) is strictly decreasing on [0.7,1) if p>1/6.Lemma 13.
LetΦ2(x) and Υ2(x) be defined, respectively, as in Lemmas 4 and 6, and Θ2(x;q)=Φ2(x)-qΥ2(x). Then Θ2(x;q) is strictly decreasing on [0.65,1) if q>2/5.Proof.
DifferentiatingΘ2(x;q) with respect to x and making use of Lemmas 9 and 11, we have
(93)dΘ1(x;q)dx=Φ2′(x)-qΥ2′(x)=μ2(x)-qν2(x)<0.51-25×1.38=-0.042<0
for x∈[0.65,1) and q>2/5. This in turn implies that Θ2(x;q) is strictly decreasing on [0.65,1) if q>2/5.


## 3. Main Results

Theorem 14.
The double inequality(94)Iα1(a,b)Q1-α1(a,b)<M(a,b)<Iβ1(a,b)Q1-β1(a,b)
holds for all a,b>0 with a≠b if and only if β1≤log[2log(1+2)]/(1-log2)=0.337⋯ and α1≥1/2.Proof.
SinceI(a,b), M(a,b), and Q(a,b) are symmetric and homogeneous of degree one, then without loss of generality, we assume that a>b. Let p∈(0,1), x=(a-b)/(a+b), and λ1=log[2log(1+2)]/(1-log2). Then x∈(0,1), and
(95)I(a,b)A(a,b)=1e[(1+x)1+x(1-x)1-x]1/2x,M(a,b)A(a,b)=xsinh-1(x),Q(a,b)A(a,b)=1+x2,(96)log[Q(a,b)]-log[M(a,b)]log[Q(a,b)]-log[I(a,b)]=log1+x2-logx+log[sinh-1(x)]log1+x2-log[(1+x)1+x/(1-x)1-x]/(2x)+1,(97)limx→0+log1+x2-logx+log[sinh-1(x)]log1+x2-log[(1+x)1+x/(1-x)1-x]/(2x)+1=12,(98)limx→1-log1+x2-logx+log[sinh-1(x)]log1+x2-log[(1+x)1+x/(1-x)1-x]/(2x)+1=λ1.
The difference between the convex combination oflog[I(a,b)],log[Q(a,b)] and log[M(a,b)] is as follows:
(99)plog[I(a,b)]+(1-p)log[Q(a,b)]-log[M(a,b)]=p2xlog[(1+x)1+x(1-x)1-x]-p+(1-p)log1+x2-log[xsinh-1(x)]:=Dp(x).
Equation (99) leads to
(100)Dp(0+)=0,Dp(1-)=log[2log(1+2)]-p(1-log2),Dλ1(1-)=0,(101)Dp′(x)=-1+px2x+x3+11+x2sinh-1(x)-L(x)2x2=Φ1(x)-pΥ1(x)=Θ1(x;p),
where L(x),Φ1(x),Υ1(x), and Θ1(x;p) are defined as in Lemmas 2, 3, 5, and 12, respectively.
It follows from (101) together with Lemmas 3 and 5 that
(102)D1/2′(x)<2x3-34x345+4x55-12(4x3-4x35+4x55)=-2x25(89-x2)<0
for x∈(0,0.7). Moreover, we see clearly, from Lemma 12, that D1/2′(x) is strictly decreasing on [0.7,1) and so D1/2′(x)<D1/2′(0.7)=-0.109⋯<0 for x∈[0.7,1). This in conjunction with (100) and (102) implies that
(103)D1/2(x)<0
for x∈(0,1).
On the other hand, (101) and Lemmas 3 and 5 together with the monotonicity of the function -2(17-18λ1)x2/45+(7-16λ1)x4/14 on (0,0.7) lead to
(104)Dλ1′(x)>2x3-34x345+x52-λ1(4x3-4x35+8x57)=x[2(1-2λ1)3-2(17-18λ1)45x2+7-16λ114x4]>x[2(1-2λ1)3-2(17-18λ1)45×(0.7)2>x+7-16λ114×(0.7)4]=(74969-218832λ1)x180000>0
for x∈(0,0.7).
It follows from Lemma12 that Dλ1′(x) is strictly decreasing on [0.7,1). Note that
(105)Dλ1′(0.7)=0.0229⋯>0,Dλ1′(1-)=-∞.
From (104) and (105) together with the monotonicity of Dλ1′(x) on [0.7,1), we clearly see that there exists c1∈(0.7,1) such that Dλ1(x) is strictly increasing on (0,c1] and strictly decreasing on [c1,1). This in conjunction with (100) implies that
(106)Dλ1(x)>0
for x∈(0,1).
Equation (99) together with inequalities (103) and (106) gives rise to
(107)M(a,b)>I1/2(a,b)Q1/2(a,b),M(a,b)<Iλ1(a,b)Q1-λ1(a,b).
Therefore, Theorem14 follows from (107) together with the following statements.(i)
Ifα1<1/2, then (96) and (97) imply that there exists δ1∈(0,1) such that M(a,b)<Iα1(a,b)Q1-α1(a,b) for all a,b>0 with (a-b)/(a+b)∈(0,δ1).(ii)
Ifβ1>λ1, then (96) and (98) imply that there exists δ2∈(0,1) such that M(a,b)>Iβ1(a,b)Q1-β1(a,b) for all a,b>0 with (a-b)/(a+b)∈(1-δ2,1).Theorem 15.
The double inequality(108)Iα2(a,b)C1-α2(a,b)<M(a,b)<Iβ2(a,b)C1-β2(a,b)
holds for all a,b>0 with a≠b if and only if α2≥5/7 and β2≤log[2log(1+2)]=0.566⋯.Proof.
We will follow the same idea in the proof of Theorem14. Since I(a,b), M(a,b), and C(a,b) are symmetric and homogeneous of degree one. Without loss of generality, we assume that a>b. Let q∈(0,1), λ2=log[2log(1+2)], and x=(a-b)/(a+b). Then x∈(0,1).
Making use of (95) together with C(a,b)/A(a,b)=1+x2 gives
(109)log[C(a,b)]-log[M(a,b)]log[C(a,b)]-log[I(a,b)]=log(1+x2)-logx+log[sinh-1(x)]log(1+x2)-log[(1+x)1+x/(1-x)1-x]/(2x)+1,(110)limx→0+log(1+x2)-logx+log[sinh-1(x)]log(1+x2)-log[(1+x)1+x/(1-x)1-x]/(2x)+1=57,(111)limx→1-log(1+x2)-logx+log[sinh-1(x)]log(1+x2)-log[(1+x)1+x/(1-x)1-x]/(2x)+1=λ2.
The difference between the convex combination oflog[I(a,b)],log[C(a,b)] and log[M(a,b)] is as follows:
(112)qlog[I(a,b)]+(1-q)log[C(a,b)]-log[M(a,b)]=q2xlog[(1+x)1+x(1-x)1-x]-q+(1-q)log(1+x2)-log[xsinh-1(x)]:=Eq(x).
Equation (112) leads to
(113)Eq(0+)=0,Eq(1-)=log[2log(1+2)]-q,Eλ2(1-)=0,(114)Eq′(x)=-1-x2+2qx2x+x3+11+x2sinh-1(x)-L(x)2x2=Φ2(x)-qΥ2(x)=Θ2(x;q),
where L(x),Φ2(x),Υ2(x), and Θ2(x;q) are defined as in Lemmas 2, 4, 6, and 13, respectively.
It follows from Lemmas4, 6, and 13 together with (114) that
(115)E5/7′(x)<(5x3-79x345+9x55)-57(7x3-9x35+7x55)=-4x25(3763-x2)<0
for x∈(0,0.65) and E5/7′(x) is strictly decreasing on [0.65,1). Thus, we have E5/7′(x)<E5/7′(0.65)=-0.117⋯ for x∈[0.65,1). This in conjunction with (113) and (115) implies that
(116)E5/7(x)<0
for x∈(0,1).
On the other hand, Lemmas4, 6, and 13 together with (114) lead to
(117)Eλ2′(x)>(5x3-79x345+11x510)-λ2(7x3-9x35+15x57)=x[5-7λ23-79-81λ245x2-150λ2-7770x4]>x[5-7λ23-79-81λ245×(0.65)2>x-150λ2-7770×(0.65)4]=113027173-197098950λ2100800000x>0
for x∈(0,0.65) and Eλ2′(x) is strictly decreasing on [0.65,1). Note that
(118)Eλ2′(0.65)=0.0609⋯,Eλ2′(1-)=-∞.
From (117) and (118) together with the monotonicity of Eλ2′(x) on [0.65,1), we clearly see that there exists c2∈(0.65,1) such that Eλ2(x) is strictly increasing on (0,c2] and strictly decreasing on [c2,1). This in conjunction with (113) implies that
(119)Eλ2(x)>0
for x∈(0,1).
Equation (112) together with inequalities (116) and (119) lead to the conclusion that
(120)M(a,b)>I5/7(a,b)C2/7(a,b),M(a,b)<Iλ2(a,b)C1-λ2(a,b).
Therefore, Theorem15 follows from (120) together with the following statements.(i)
Ifα2<5/7, then (109) and (110) imply that there exists δ3∈(0,1) such that M(a,b)<Iα2(a,b)C1-α2(a,b) for all a,b>0 with (a-b)/(a+b)∈(0,δ3).(ii)
Ifβ2>λ2, then (109) and (111) imply that there exists δ4∈(0,1) such that M(a,b)>Iβ2(a,b)C1-β2(a,b) for all a,b>0 with (a-b)/(a+b)∈(1-δ4,1).



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*Source: 348326-2013-03-25.xml*