We present the sharp bounds for the Neuman means SHA, SAH, SCA and SAC in terms of the arithmetic, harmonic, and contraharmonic means. Our results are the refinements or improvements of the results given by Neuman.
1. Introduction
For a,b>0 with a≠b, the Schwab-Borchardt mean SB(a,b) of a and b is given by
(1)SB(a,b)={b2-a2cos-1(a/b),a<b,a2-b2cosh-1(a/b),a>b,
where cos-1(x) and cosh-1(x)=log(x+x2-1) are the inverse cosine and inverse hyperbolic cosine functions, respectively.
It is well-known that the mean SB(a,b) is strictly increasing in both a and b, nonsymmetric and homogeneous of degree 1 with respect to a and b. Many symmetric bivariate means are special cases of the Schwab-Borchardt mean; for example,
(2)P(a,b)=a-b2sin-1[(a-b)/(a+b)]=SB(G,A)isthefirstSeiffertmean,T(a,b)=a-b2tan-1[(a-b)/(a+b)]=SB(A,Q)isthesecondSeiffertmean,M(a,b)=a-b2sinh-1[(a-b)/(a+b)]=SB(Q,A)istheNeuman-Sándormean,L(a,b)=a-b2tanh-1[(a-b)/(a+b)]=SB(A,G)isthelogarithmicmean,
where G(a,b)=ab, A(a,b)=(a+b)/2, and Q(a,b)=(a2+b2)/2 denote the classical geometric mean, arithmetic mean, and quadratic mean of a and b, respectively. The Schwab-Borchardt mean SB(a,b) was investigated in [1, 2].
Let H(a,b)=2ab/(a+b) and C(a,b)=(a2+b2)/(a+b) be the harmonic and contraharmonic means of two positive numbers a and b, respectively. Then, it is well-known that
(3)H(a,b)<G(a,b)<L(a,b)<P(a,b)<A(a,b)<M(a,b)H(a,b)<T(a,b)<Q(a,b)<C(a,b),
for a,b>0 with a≠b.
Recently, the Schwab-Borchardt mean and its special cases have been the subject of intensive research. Neuman and Sándor [3, 4] proved that the inequalities
(4)P(a,b)>2πA(a,b),A(a,b)log(1+2)>M(a,b)>π4log(1+2)T(a,b),T(A(a,b),G(a,b))<P(a,b),T(a,b)>T(A(a,b),Q(a,b)),L(a,b)<L(A(a,b),G(a,b)),M(a,b)<L(A(a,b),Q(a,b)),L(a,b)>H(P(a,b),G(a,b)),P(a,b)>H(L(a,b),A(a,b)),M(a,b)>H(T(a,b),A(a,b)),T(a,b)>H(M(a,b),Q(a,b)),G(a,b)P(a,b)<L2(a,b)<G2(a,b)+P2(a,b)2,L(a,b)A(a,b)<P2(a,b)<L2(a,b)+A2(a,b)2,A(a,b)T(a,b)<M2(a,b)<A2(a,b)+T2(a,b)2,M(a,b)Q(a,b)<T2(a,b)<M2(a,b)+Q2(a,b)2,Q1/3(a,b)A2/3(a,b)<M(a,b)<13Q(a,b)+23A(a,b)
hold for all a, b>0 with a≠b. In [5], the author proved that the double 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)
hold for all a, b>0 with a≠b if and only if α≤[1-log(1+2)]/[(2-1)log(1+2)]=0.3249⋯, β≥1/3, λ≤[1-log(1+2)]/log(1+2)=0.1345⋯, and μ≥1/6. Chu and Long [6] found that the double inequality
(6)Mp(a,b)<M(a,b)<qI(a,b)
holds for all a,b>0 with a≠b if and only if p≤log2/log[2log(1+2)]=1.224⋯ and q≥e/[2log(1+2)], where Mp(a,b)=[(ap+bp)/2]1/p(p≠0) and M0(a,b)=ab is the pth power mean of a and b. Zhao et al. [7] presented the least values α1, α2, and α3 and the greatest values β1, β2, and β3 such that the double inequalities
(7)α1H(a,b)+(1-α1)Q(a,b)<M(a,b)<β1H(a,b)+(1-β1)Q(a,b),α2G(a,b)+(1-α2)Q(a,b)<M(a,b)<β2G(a,b)+(1-β2)Q(a,b),α3H(a,b)+(1-α3)C(a,b)<M(a,b)<β3H(a,b)+(1-β3)C(a,b)
hold for all a, b>0 with a≠b.
Very recently, the bivariate means SAH, SHA, SCA, and SAC derived from the Schwab-Borchardt mean are defined by Neuman [8, 9] as follows:
(8)SAH=SB(A,H),SHA=SB(H,A),SCA=SB(C,A),SAC=SB(A,C).
We call the means SAH, SHA, SCA, and SAC given in (8) the Neuman means. Moreover, let v=(a-b)/(a+b)∈(-1,1); then the following explicit formulas for SAH, SHA, SAC, and SCA are found by Neuman [8]:
(9)SAH=Atanh(p)p,SHA=Asin(q)q,(10)SCA=Asinh(r)r,SAC=Atan(s)s,
where p, q, r, and s are defined implicitly as sech(p)=1-v2, cos(q)=1-v2, cosh(r)=1+v2, and sec(s)=1+v2, respectively. Clearly, p∈(0,∞), q∈(0,π/2), r∈(0,log(2+3)), and s∈(0,π/3).
In [8, 9], Neuman proved that the inequalities
(11)H(a,b)<SAH(a,b)<L(a,b)<SHA(a,b)<P(a,b),T(a,b)<SCA(a,b)<Q(a,b)<SAC(a,b)<C(a,b),(12)H1/3(a,b)A2/3(a,b)<SHA(a,b)<13H(a,b)+23A(a,b),C1/3(a,b)A2/3(a,b)<SCA(a,b)<13C(a,b)+23A(a,b),A1/3(a,b)H2/3(a,b)<SAH(a,b)<13A(a,b)+23H(a,b),A1/3(a,b)C2/3(a,b)<SAC(a,b)<13A(a,b)+23C(a,b)
hold for a, b>0 with a≠b.
He et al. [10] found the greatest values α1, α2∈[0,1/2], α3, α4∈[1/2,1], and the least values β1, β2∈[0,1/2], β3, β4∈[1/2,1] such that the double inequalities
(13)H(α1a+(1-α1)b,α1b+(1-α1)a)<SAH(a,b)<H(β1a+(1-β1)b,β1b+(1-β1)a),H(α2a+(1-α2)b,α2b+(1-α2)a)<SHA(a,b)<H(β2a+(1-β2)b,β2b+(1-β2)a),C(α3a+(1-α3)b,α3b+(1-α3)a)<SCA(a,b)<C(β3a+(1-β3)b,β3b+(1-β3)a),C(α4a+(1-α4)b,α4b+(1-α4)a)<SAC(a,b)<C(β4a+(1-β4)b,β4b+(1-β4)a)
hold for all a,b>0 with a≠b.
Motivated by inequalities (12), it is natural to ask what the greatest values α1, α2, α3, and α4 and the least values β1, β2, β3, and β4 are such that the double inequalities
(14)α1[H(a,b)3+2A(a,b)3]+(1-α1)H1/3(a,b)A2/3(a,b)<SHA(a,b)<β1[H(a,b)3+2A(a,b)3]+(1-β1)H1/3(a,b)A2/3(a,b),α2[C(a,b)3+2A(a,b)3]+(1-α2)C1/3(a,b)A2/3(a,b)<SCA(a,b)<β2[C(a,b)3+2A(a,b)3]+(1-β2)C1/3(a,b)A2/3(a,b),α3[A(a,b)3+2H(a,b)3]+(1-α3)A1/3(a,b)H2/3(a,b)<SAH(a,b)<β3[A(a,b)3+2H(a,b)3]+(1-β3)A1/3(a,b)H2/3(a,b),α4[A(a,b)3+2C(a,b)3]+(1-α4)A1/3(a,b)C2/3(a,b)<SAC(a,b)<β4[A(a,b)3+2C(a,b)3]+(1-β4)A1/3(a,b)C2/3(a,b)
hold for all a, b>0 with a≠b.
The purpose of this paper is to answer these questions. All numerical computations are carried out using MATHEMATICA software. Our main results are the following Theorems 1–4.
Theorem 1.
The double inequality
(15)α1[H(a,b)3+2A(a,b)3]+(1-α1)H1/3(a,b)A2/3(a,b)<SHA(a,b)<β1[H(a,b)3+2A(a,b)3]+(1-β1)H1/3(a,b)A2/3(a,b)
holds for all a,b>0 with a≠b if and only if α1≤4/5 and β1≥3/π.
Theorem 2.
The two-sided inequality
(16)α2[C(a,b)3+2A(a,b)3]+(1-α2)C1/3(a,b)A2/3(a,b)<SCA(a,b)<β2[C(a,b)3+2A(a,b)3]+(1-β2)C1/3(a,b)A2/3(a,b)
holds true for all a,b>0 with a≠b if and only if α2≤3[23log(2+3)-3]/[(323-4)log(2+3)]=0.7528⋯ and β2≥4/5.
Theorem 3.
The double inequality
(17)α3[A(a,b)3+2H(a,b)3]+(1-α3)A1/3(a,b)H2/3(a,b)<SAH(a,b)<β3[A(a,b)3+2H(a,b)3]+(1-β3)A1/3(a,b)H2/3(a,b)
holds for all a,b>0 with a≠b if and only if α3≤0 and β3≥4/5.
Theorem 4.
The two-sided inequality
(18)α4[A(a,b)3+2C(a,b)3]+(1-α4)A1/3(a,b)C2/3(a,b)<SAC(a,b)<β4[A(a,b)3+2C(a,b)3]+(1-β4)A1/3(a,b)C2/3(a,b)
holds true for all a,b>0 with a≠b if and only if α4≤4/5 and β2≥3(33-43π)/[(5-343)π]=0.8400⋯.
2. Two Lemmas
In order to prove our main results, we need two lemmas, which we present in this section.
Lemma 5.
Let p∈ℝ and
(19)f(x)=p2x6+2p2x5+3(-p2+4p-2)x4f(x)=+2(-2p2+9p-6)x3+(4p2+6p-9)x2f(x)=+6(p-1)x+3(p-1).
Then, the following statements are true.
If p=4/5, then f(x)<0 for all x∈(0,1) and f(x)>0 for all x∈(1,23).
If p=3/π, then there exists λ1∈(0,1) such that f(x)<0 for x∈(0,λ1) and f(x)>0 for x∈(λ1,1).
If p=3[23log(2+3)-3]/[(323-4)log(2+3)]=0.7528⋯, then there exists λ2∈(1,23) such that f(x)<0 for x∈(1,λ2) and f(x)>0 for x∈(λ2,23).
Proof.
For part (1), if p=4/5, then (19) becomes
(20)f(x)=125(x-1)(16x5+48x4+90x3+86x2wf(x)=125(x-10)+45x+1516x5+48x4).
Therefore, part (1) follows easily from (20).
For part (2), if p=3/π, then simple computations lead to
(21)-p2+4p-2=-2π2+12π-9π2>0,(22)-2p2+9p-6=-6π2+27π-18π2>0,(23)4p2+6p-9=-9π2+18π+36π2>0,(24)f(0)=-3(π-3)π<0,(25)f(1)=9(15-4π)π>0,(26)f′(x)=6p2x5+10p2x4+12(-p2+4p-2)x3+6(-2p2+9p-6)x2+2(4p2+6p-9)x+6(p-1),(27)f′(0)=6(3-π)π<0,(28)f′(1)=12(30-7π)π>0,(29)f′′(x)=30p2x4+40p2x3+36(-p2+4p-2)x2+12(-2p2+9p-6)x+2(4p2+6p-9).
It follows from (21)–(23) and (29) that f′(x) is strictly increasing on (0,1). Then, (27) and (28) lead to the conclusion that there exists x0∈(0,1) such that f(x) is strictly decreasing in (0,x0] and strictly increasing in [x0,1).
Therefore, part (2) follows from (24) and (25) together with the piecewise monotonicity of f(x).
For part (3), if p=3[23log(2+3)-3]/[(323-4)log(2+3)]=0.7528⋯, then numerical computations lead to
(30)-p2+4p-2=0.444⋯>0,(31)4p2+6p-9=-2.215⋯<0,(32)6(p-1)=-1.483⋯<0,(33)f(1)=9(5p-4)=-2.120⋯<0,(34)f(23)=1.669⋯>0.
It follows from (26) and (30)–(32) that
(35)f′(x)>6p2x2+10p2x2+12(-p2+4p-2)x2+6(-2p2+9p-6)x2+2(4p2+6p-9)x2+6(p-1)x2=12(10p-7)x2>0
for x∈(1,23).
Therefore, part (3) follows easily from (33)–(35).
Lemma 6.
Let p∈ℝ and
(36)g(x)=3(1-p)x6+6(1-p)x5+(-4p2-6p+9)x4+2(2p2-9p+6)x3+3(p2-4p+2)x2-2p2x-p2.
Then, the following statements are true.
If p=4/5, then g(x)<0 for all x∈(0,1) and g(x)>0 for all x∈(1,23).
If p=3(33-43π)/[(5-343)π]=0.8400⋯, then there exists λ3∈(1,23) such that g(x)<0 for x∈(1,λ3) and g(x)>0 for x∈(λ3,23).
Proof.
For part (1), if p=4/5, then (36) becomes
(37)g(x)=125(x-1)(15x5+45x4+86x3+90x2w+48x+1615x5+45x4).
Therefore, part (1) follows from (37).
For part (2), if p=3(33-43π)/[(5-343)π]=0.8400⋯, then numerical computations lead to
(38)-4p2-6p+9=1.137⋯>0,(39)p2-4p+2=-0.654⋯<0,(40)g(1)=9(4-5p)=-1.801⋯<0,(41)g(23)=1.635⋯>0,(42)g′(x)=18(1-p)x5+30(1-p)x4+4(-4p2-6p+9)x3+6(2p2-9p+6)x2+6(p2-4p+2)x-2p2.
From (38) and (39) together with (42), we clearly see that
(43)g′(x)>18(1-p)x2+30(1-p)x2+4(-4p2-6p+9)x2+6(2p2-9p+6)x2+6(p2-4p+2)x2-2p2x2=6(22-25p)x2>0
for x∈(1,23).
Therefore, part (2) follows from (40) and (41) together with (43).
3. Proofs of Theorems 1–4Proof of Theorem 1.
Without loss of generality, we assume that a>b. Let v=(a-b)/(a+b), λ=v2-v2, x=1-λ26, and p∈{4/5,3/π}. Then, v, λ, x∈(0,1),
(44)SHA(a,b)-H1/3(a,b)A2/3(a,b)H(a,b)/3+2A(a,b)/3-H1/3(a,b)A2/3(a,b)=λ/sin-1(λ)-(1-λ2)1/62/3+(1-λ2)1/2/3-(1-λ2)1/6,(45)SHA(a,b)-[p(13H(a,b)+2A(a,b)3)SHA(a,b)-e+(1-p)H1/3(a,b)A2/3(a,b)(13H(a,b)+2A(a,b)3)]=A(a,b)[λsin-1(λ)-p((1-λ2)1/23+23)SHA(a,)--(1-p)(1-λ2)1/6((1-λ2)1/23+23)]=(A(a,b)[p((1-λ2)1/2+2)+3(1-p)(1-λ2)1/6])×(3sin-1(λ))-1F(x),
where
(46)F(x)=31-x6px3+3(1-p)x+2p-sin-1(1-x6),(47)F(0)=32p-π2,(48)F(1)=0,(49)F′(x)=3(x-1)21-x6[px3+3(1-p)x+2p]2f(x),
where f(x) is defined as in Lemma 5.
We divide the proof into two cases.
Case 1(p=4/5). Then, from Lemma 5(1) and (49), we clearly see that F(x) is strictly decreasing in (0,1). Therefore,
(50)SHA(a,b)>45[13H(a,b)+23A(a,b)]+15H1/3(a,b)A2/3(a,b)
for all a,b>0 with a≠b follows from (45) and (48) together with the monotonicity of F(x).
Case 2 (p=3/π). Then, from (47) and (49) and Lemma 5(2), we know that
(51)F(0)=0
and there exists λ1∈(0,1) such that F(x) is strictly decreasing in (0,λ1] and strictly increasing in [λ1,1). Therefore,
(52)SHA(a,b)<3π[13H(a,b)+23A(a,b)]+(1-3π)H1/3(a,b)A2/3(a,b)
for all a,b>0 with a≠b follows from (45) and (48) together with (51) and the piecewise monotonicity of F(x).
Note that
(53)limλ→0+λ/sin-1(λ)-(1-λ2)1/62/3+(1-λ2)1/2/3-(1-λ2)1/6=45,(54)limλ→1-λ/sin-1(λ)-(1-λ2)1/62/3+(1-λ2)1/2/3-(1-λ2)1/6=3π.
Therefore, Theorem 1 follows from (50) and (52)–(54) together with the following statements.
If α>4/5, then (44) and (53) imply that there exists small enough δ>0 such that SHA(a,b)<α(H(a,b)/3+2A(a,b)/3)+(1-α)H1/3(a,b)A2/3(a,b) for all a>b>0 with b/a∈(0,δ).
If β<3/π, then (44) and (54) imply that there exists large enough M>1 such that SHA(a,b)>β(H(a,b)/3+2A(a,b)/3)+(1-β)H1/3(a,b)A2/3(a,b) for all a>b>0 with a/b∈(M,+∞).
Proof of Theorem 2.
Without loss of generality, we assume that a>b. Let v=(a-b)/(a+b), μ=v2+v2, x=1+μ26, and p∈{3[23log(2+3)-3]/[(323-4)log(2+3)],4/5}. Then, v∈(0,1), μ∈(0,3), x∈(1,23),
(55)SCA(a,b)-C1/3(a,b)A2/3(a,b)C(a,b)/3+2A(a,b)/3-C1/3(a,b)A2/3(a,b)=μ/sinh-1(μ)-(1+μ2)1/62/3+(1+μ2)1/2/3-(1+μ2)1/6,(56)SCA(a,b)-[p(13C(a,b)+2A(a,b)3)wSCA(a,b)-+(1-p)C1/3(a,b)A2/3(a,b)(13C(a,b)+2A(a,b)3)]=A(a,b)[μsinh-1(μ)-p((1+μ2)1/23+23)SCA(a,b)-ww-(1-p)(1+μ2)1/6((1+μ2)1/23+23)]=(A(a,b)[p((1+μ2)1/2+2)+3(1-p)(1+μ2)1/6])×(3sinh-1(μ))-1G(x),
where
(57)G(x)=3x6-1px3+3(1-p)x+2p-sinh-1(x6-1),(58)G(1)=0,(59)G(23)=33(4-323)p+323-log(1+3),(60)G′(x)=-3(x-1)2x6-1[px3+3(1-p)x+2p]2f(x),
where f(x) is defined as in Lemma 5.
We divide the proof into two cases.
Case 1 (p=3[23log(2+3)-3]/[(323-4)log(2+3)]=0.7528⋯). Then, from (59) and (60) together with Lemma 5(3), we clearly see that there exists λ2∈(1,23) such that G(x) is strictly increasing in (1,λ2] and strictly decreasing in [λ2,32), and
(61)G(23)=0.
Therefore,
(62)SCA(a,b)>3(23log(2+3)-3)(323-4)log(2+3)[13C(a,b)+23A(a,b)]+(1-3(23log(2+3)-3)(323-4)log(2+3))×C1/3(a,b)A2/3(a,b)
for all a,b>0 with a≠b follows easily from (56) and (58) together with (61) and the piecewise monotonicity of G(x).
Case 2 (p=4/5). Then, Lemma 5(1) and (60) lead to the conclusion that G(x) is strictly decreasing in (1,23). Therefore,
(63)SCA(a,b)<45[13C(a,b)+23A(a,b)]+15C1/3(a,b)A2/3(a,b)
for all a,b>0 with a≠b follows from (56) and (58) together with the monotonicity of G(x).
Note that
(64)limμ→0+μ/sinh-1(μ)-(1+μ2)1/62/3+(1+μ2)1/2/3-(1+μ2)1/6=45,(65)limμ→3-μ/sinh-1(μ)-(1+μ2)1/62/3+(1+μ2)1/2/3-(1+μ2)1/6=3(23log(2+3)-3)(323-4)log(2+3).
Therefore, Theorem 2 follows from (55) and (62)–(65).
Proof of Theorem 3.
Without loss of generality, we assume that a>b. Let v=(a-b)/(a+b), λ=v2-v2, x=1-λ26, and p∈{4/5,0}. Then, v,λ,x∈(0,1) and (9) leads to
(66)SAH(a,b)=A(a,b)λtanh-1(λ).
It follows from (66) that
(67)SAH(a,b)-A1/3(a,b)H2/3(a,b)A(a,b)/3+2H(a,b)/3-A1/3(a,b)H2/3(a,b)=λ/tanh-1(λ)-(1-λ2)1/31/3+2(1-λ2)1/2/3-(1-λ2)1/3,(68)SAH(a,b)-[p(13A(a,b)+2H(a,b)3)SAH(a,b)-w+(1-p)A1/3(a,b)H2/3(a,b)(13A(a,b)+2H(a,b)3)]=A(a,b)[λtanh-1(λ)-p(2(1-λ2)1/23+13)SAH(a,b)-w-(1-p)(1-λ2)1/3(2(1-λ2)1/23+13)]=A(a,b)[p(2(1-λ2)1/2+1)+3(1-p)(1-λ2)1/3]3tanh-1(λ)×H(x),
where
(69)H(x)=31-x62px3+3(1-p)x2+p-tanh-1(1-x6)(70)H(1)=0,(71)H′(x)=-3(1-x)2x1-x6[2px3+3(1-p)x2+p]2g(x),
where g(x) is defined as in Lemma 6.
If p=4/5, then Lemma 6(1) and (71) lead to the conclusion that H(x) is strictly increasing in (0,1). Therefore,
(72)SAH(a,b)<45(13A(a,b)+2H(a,b)3)SAH(a,b)<+15A1/3(a,b)H2/3(a,b)
for all a, b>0 with a≠b follows from (68) and (70) together with the monotonicity of H(x).
Note that
(73)limλ→0+λ/tanh-1(λ)-(1-λ2)1/31/3+2(1-λ2)1/2/3-(1-λ2)1/3=45,(74)limλ→1-λ/tanh-1(λ)-(1-λ2)1/31/3+2(1-λ2)1/2/3-(1-λ2)1/3=0.
Therefore, Theorem 3 follows from (12) and (67) together with (72)–(74).
Proof of Theorem 4.
Without loss of generality, we assume that a>b. Let v=(a-b)/(a+b), μ=v2+v2, x=1+μ26, and p∈{3(33-43π)/[(5-343)π],4/5}. Then, v∈(0,1), μ∈(0,3), and x∈(1,23) and (10) leads to
(75)SAC(a,b)=A(a,b)μtan-1(μ).
It follows from (75) that
(76)SAC(a,b)-A1/3(a,b)C2/3(a,b)A(a,b)/3+2C(a,b)/3-A1/3(a,b)C2/3(a,b)=μ/tan-1(μ)-(1+μ2)1/31/3+2(1+μ2)1/2/3-(1+μ2)1/3,(77)SAC(a,b)-[p(13A(a,b)+2C(a,b)3)SAC(a,b)-e+(1-p)A1/3(a,b)C2/3(a,b)(13A(a,b)+2C(a,b)3)]=A(a,b)[μtan-1(μ)-p(2(1+μ2)1/23+13)SAC(a,b)-w-(1-p)(1+μ2)1/3(2(1+μ2)1/23+13)]=A(a,b)[p(2(1+μ2)1/2+1)+3(1-p)(1+μ2)1/3]3tan-1(μ)×J(x),
where
(78)J(x)=3x6-12px3+3(1-p)x2+p-tan-1(x6-1),(79)J(1)=0,(80)J(23)=33(5-343)p+343-π3,(81)J′(x)=3(x-1)2x6-1[2px3+3(1-p)x2+p]2g(x),
where g(x) is defined as in Lemma 6.
We divide the proof into two cases.
Case 1 (p=4/5). Then, (81) and Lemma 6(1) lead to the conclusion that J(x) is strictly increasing in (1,23). Therefore,
(82)SAC(a,b)>45(13A(a,b)+2C(a,b)3)SAC(a,b)>+15A1/3(a,b)C2/3(a,b)
for all a,b>0 with a≠b follows easily from (77) and (79) together with the monotonicity of J(x).
Case 2 (p=3(33-43π)/(5-343)π). Then, (80) and (81) together with Lemma 6(2) lead to the conclusion that there exists λ3∈(1,23) such that J(x) is strictly decreasing in (1,λ3] and strictly increasing in [λ3,23), and
(83)J(23)=0.
Therefore,
(84)SAC(a,b)<3(33-43π)(5-343)π(13A(a,b)+2C(a,b)3)SAC(a,b)<+(1-3(33-43π)(5-343)π)A1/3(a,b)C2/3(a,b)
for all a,b>0 with a≠b follows easily from (77) and (79) together with (83) and the piecewise monotonicity of J(x).
Note that
(85)limμ→0+μ/tan-1(μ)-(1+μ2)1/31/3+2(1+μ2)1/2/3-(1+μ2)1/3=45,(86)limμ→3-μ/tan-1(μ)-(1+μ2)1/31/3+2(1+μ2)1/2/3-(1+μ2)1/3=3(33-43π)(5-343)π.
Therefore, Theorem 4 follows from (76) and (82) together with (84)–(86).
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgments
This research was supported by the Natural Science Foundation of China under Grants 61374086 and 11171307, the Natural Science Foundation of the Open University of China under Grant Q1601E-Y, and the Natural Science Foundation of Zhejiang Broadcast and TV University under Grant XKT-13Z04.
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