We prove that the double inequalities 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 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.
1. Introduction
For p∈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].
Let H(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.
Let 0<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 for I, (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 constants α1,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 Lemma 1, 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 of η1(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 Lemma 1 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 Lemma 4, 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 Lemma 1, (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 Lemma 1 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.
Let L(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 Lemma 2, 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, Lemma 5 follows easily from (61).
Lemma 6.
Let L(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 Lemma 2 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, Lemma 6 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 Lemma 1 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, Lemma 7 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).
Then μ1(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.
Lemma 7 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 of ω2(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 of τ2(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.
Let L(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).
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.
Let L(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).
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.
Let Φ1(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.
Differentiating Θ1(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.
Let Φ2(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.
Differentiating Θ2(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 ResultsTheorem 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.
Since I(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 of log[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 Lemma 12 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, Theorem 14 follows from (107) together with the following statements.
If α1<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).
If β1>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 Theorem 14. 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 of log[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 Lemmas 4, 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, Lemmas 4, 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, Theorem 15 follows from (120) together with the following statements.
If α2<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).
If β2>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).
Acknowledgments
This research was supported by the Natural Science Foundation of China under Grants 11171307, 61174076, and 61173123 and the Natural Science Foundation of Zhejiang Province under Grants Z1110551 and LY12F02012.
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