1. Introduction
For a,b>0 with a≠b, the NeumanSándor mean M(a,b) [1], second Seiffert mean T(a,b) [2] are defined by
(1)M(a,b)=ab2sinh1((ab)/(a+b)) ,T(a,b)=ab2arctan((ab)/(a+b)),
respectively. Herein, sinh1(x)=log(x+1+x2) is the inverse hyperbolic sine function.
Let H(a,b)=2ab/(a+b), G(a,b)=ab, L(a,b)=(ab)/(logalogb), P(a,b)=(ab)/[4arctan(a/b)π], I(a,b)=1/e(bb/aa)1/(ba), A(a,b)=(a+b)/2, Q(a,b)=(a2+b2)/2, and C(a,b)=(a2+b2)/(a+b) be the harmonic, geometric, logarithmic, first Seiffert, identric, arithmetic, quadratic, and contraharmonic means of two distinct positive real numbers a and b, respectively. Then it is well known that the inequalities
(2)H(a,b)<G(a,b)<L(a,b)<P(a,b)<I(a,b)<A(a,b)<M(a,b)<T(a,b)<Q(a,b)<C(a,b)
hold for all a,b>0 with a≠b.
Among means of two variables, the NeumanSándor, contraharmonic, and second Seiffert means have attracted the attention of several researchers. In particular, many remarkable inequalities and applications for these means can be found in the literature [3–15].
Neuman and Sándor [1, 16] proved that the inequalities
(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)<2A(a,b)+Q(a,b)3,P(a,b)M(a,b)<A2(a,b),A(a,b)T(a,b)<M2(a,b)<(A2(a,b)+T2(a,b))2
hold for all a,b>0 with a≠b.
Let 0<a, b<1/2 with a≠b, a′=1a and b′=1b. 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′)
can be found in [1].
Li et al. [17] proved that the double inequality Lp0(a,b)<M(a,b)<L2(a,b) holds for all a,b>0 with a≠b, where Lp(a,b)=[(bp+1ap+1)/((p+1)(ba))]1/p (p≠1,0), L0(a,b)=I(a,b) and L1(a,b)=L(a,b) is the pth generalized logarithmic mean of a and b, and p0=1.843⋯ is the unique solution of the equation (p+1)1/p=2log(1+2).
In [18], Neuman proved that the inequalities
(5)αQ(a,b)+(1α)A(a,b)<M(a,b) <βQ(a,b)+(1β)A(a,b),λC(a,b)+(1λ)A(a,b)<M(a,b) <μC(a,b)+(1μ)A(a,b)
hold for all a,b>0 with a≠b if and only if α≤[1log(1+2)]/[(21)log(1+2)]=0.3249⋯, λ≤[1log(1+2)]/log(1+2)=0.1345⋯, β≥1/3 and μ≥1/6.
Zhao et al. [19] found the least values α1,α2,α3 and the greatest values β1,β2,β3 such that the double inequalities
(6)α1H(a,b)+(1α1)Q(a,b)<M(a,b) <β1H(a,b)+(1β1)Q(a,b),α2G(a,b)+(1α2)Q(a,b)<M(a,b) <β2G(a,b)+(1β2)Q(a,b),α3H(a,b)+(1α3)C(a,b)<M(a,b) <β3H(a,b)+(1β3)C(a,b)
hold for all a,b>0 with a≠b.
In [20, 21], the authors proved that the double inequalities
(7)α1T(a,b)+(1α1)G(a,b)<A(a,b) <β1T(a,b)+(1β1)G(a,b),α2Q(a,b)+(1α2)A(a,b)<T(a,b) <β2Q(a,b)+(1β2)A(a,b),Qα3(a,b)A1α3(a,b)<T(a,b) <Qβ3(a,b)A1β3(a,b)
hold for all a,b>0 with a≠b if and only of α1≤3/5, β1≥4/π, α2≤(4π)/[(21)π], β2≥2/3, α3≤2/3, and β3≥42logπ/log2.
For α,β,λ,μ∈(1/2,1), Chu et al. [22, 23] proved that the inequalities
(8)C(αa+(1α)b,αb+(1α)a)<T(a,b) <C(βa+(1β)b,βb+(1β)a),Q(λa+(1λ)b,λb+(1λ)a)<T(a,b) <Q(μa+(1μ)b,μb+(1μ)a)
hold for all a,b>0 with a≠b if and only if α≤(1+4/π1)/2, β≥(3+3)/6, λ≤(1+16/π21)/2 and μ≥(3+6)/6.
The aim of this paper is to find the greatest values r1, r2 and the least values s1, s2 such that the double inequalities
(9)C(r1a+(1r1)b,r1b+(1r1)a) <αA(a,b)+(1α)T(a,b) <C(s1a+(1s1)b,s1b+(1s1)a),(10)C(r2a+(1r2)b,r2b+(1r2)a) <αA(a,b)+(1α)M(a,b) <C(s2a+(1s2)b,s2b+(1s2)a)
hold for any α∈(0,1) and all a,b>0 with a≠b.
2. Lemmas
In order to prove our main results, we need three lemmas, which we present in this section.
Lemma 1 (see [<xref reftype="bibr" rid="B24">24</xref>, Theorem 1.25]).
For ∞<a<b<+∞, let f,g:[a,b]→ℝ be continuous on [a,b] and differentiable on (a,b), let g′(x)≠0 on (a,b). If f′(x)/g′(x) is increasing (decreasing) on (a,b), then so are
(11)f(x)f(a)g(x)g(a), f(x)f(b)g(x)g(b).
If f′(x)/g′(x) is strictly monotone, then the monotonicity in the conclusion is also strict.
Lemma 2.
Let u,α∈(0,1) and
(12)fu,α(x)=ux2(1α)(xarctanx1).
Then fu,α(x)>0 for all x∈(0,1) if and only if u≥(1α)/3 and fu,α(x)<0 for all x∈(0,1) if and only if u≤(1α)(4/π1).
Proof.
From (12), one has
(13)fu,α(0+)=0,(14)fu,α(1)=u(1α)(4π1),(15)fu,α′(x)=2x[u1α2g(x)],
where
(16)g(x)=(1+x2)arctanxxx(1+x2)(arctanx)2.
Let g1(x)=arctanxx/(1+x2) and g2(x)=x(arctanx)2, then
(17)g(x)=g1(x)g2(x), g1(0)=g2(0)=0,(18)g1′(x)g2′(x) =2x22x(1+x2)arctanx+(1+x2)2(arctanx)2 =1((1+x2)arctanx/x)+(1/2)[(1+x2)arctanx/x]2.
It is not difficult to verify that the function (1+x2)arctanx/x is strictly increasing on (0,1). Then (17) and (18) together with Lemma 1 lead to the conclusion that g(x) is strictly decreasing on (0,1). Moreover, making use of L'Hôpital's rule, we get
(19)g(0+)=23,(20)g(1)=4(π2)π2.
We divide the proof into four cases.
Case
1.
u
≥
(
1

α
)
/
3
. Then from (15) and (19) together with the monotonicity of g(x), we clearly see that fu,α(x) is strictly increasing on (0,1). Therefore, fu,α(x)>0 for all x∈(0,1) follows from (13) and the monotonicity of fu,α(x).
Case
2.
u
≤
2
(
1

α
)
(
π

2
)
/
π
2
. Then from (15) and (20) together with the monotonicity of g(x), we clearly see that fu,α(x) is strictly decreasing on (0,1). Therefore, fu,α(x)<0 for all x∈(0,1) follows from (13) and the monotonicity of fu,α(x).
Case
3.
2
(
1

α
)
(
π

2
)
/
π
2
<
u
≤
(
1

α
)
(
4
/
π

1
)
. Then (14) leads to
(21)fu,α(1)≤0.
From (15), (19), and (20) together with the monotonicity of g(x), we clearly see that there exists unique x0∈(0,1) such that fu,α(x) is strictly decreasing on (0,x0] and strictly increasing on [x0,1). Therefore, fu,α(x)<0 for all x∈(0,1) follows from (13) and (21) together with the piecewise monotonicity of fu,α(x).
Case
4
.
(
1

α
)
(
4
/
π

1
)
<
u
≤
(
1

α
)
/
3
. Then (14) leads to
(22)fu,α(1)>0.
It follows from (15), (19), and (20) together with the monotonicity of g(x), there exists unique x1∈(0,1) such that fu,α(x) is strictly decreasing on (0,x1] and strictly increasing on [x1,1). Equation (13) and inequality (22) together with the piecewise monotonicity of fu,α(x) lead to the conclusion that there exists x2∈(x1,1) such that fu,α(x)<0 for x∈(0,x2) and fu,α(x)>0 for x∈(x2,1).
Lemma 3.
Let λ,α∈(0,1) and
(23)φλ,α(x)=λx2(1α)(xsinh1(x)1).
Then φλ,α(x)>0 for all x∈(0,1) if and only if λ≥(1α)/6 and φλ,α(x)<0 for all x∈(0,1) if and only if λ≤(1α)(1log(1+2))/log(1+2).
Proof.
From (23) we get
(24)φλ,α(0+)=0,(25)φλ,α(1)=λ(1α)[1log(1+2)]log(1+2),(26)φλ,α′(x)=2x[λ1α2ψ(x)],
where
(27)ψ(x)=sinh1(x)x/1+x2x(sinh1(x))2.
Let ψ1(x)=sinh1(x)x/1+x2 and ψ2(x)=x(sinh1(x))2, then
(28)ψ(x)=ψ1(x)ψ2(x), ψ1(0)=ψ2(0)=0,ψ1′(x)ψ2′(x) =x2×((1+x2)3/2(sinh1(x))2 + 2x(1+x2 )sinh1(x)(sinh1(x))2)1 =(((1+x2)3/4sinh1(x)/x)2 + 2(1+x2)1/4((1+x2)3/4sinh1(x)/x))1.
It is not difficult to verify that the function (1+x2)3/4sinh1(x)/x is strictly increasing on (0,1). Then (28) together with Lemma 1 leads to the conclusion that ψ(x) is strictly decreasing on (0,1). Moreover, making use of L'Hôpital's rule, we have
(29)ψ(0+)=13,(30)ψ(1)=2log(1+2)12log2(1+2).
We divide the proof into four cases.
Case
1.
λ
≥
(
1

α
)
/
6
. Then from (26) and (29) together with the monotonicity of ψ(x), we clearly see that φλ,α(x) is strictly increasing on (0,1). Therefore, φλ,α(x)>0 for all x∈(0,1) follows from (24) and the monotonicity of φλ,α(x).
Case
2. λ≤(1α)[2log(1+2)1]/[22log2(1+2)]. Then from (26) and (30) together with the monotonicity of ψ(x), we clearly see that φλ,α(x) is strictly decreasing on (0,1). Therefore, φλ,α(x)<0 for all x∈(0,1) follows from (24) and the monotonicity of φλ,α(x).
Case
3.
(
(
1

α
)
[
2
log
(
1
+
2
)

1
]
/
2
2
log
2
(
1
+
2
)
)
<
λ
≤
(
(
1

α
)
[
1

log
(
1
+
2
)
]
/
log
(
1
+
2
)
)
. Then (25) leads to
(31)φλ,α(1)≤0.
From (26), (29), and (30) together with the monotonicity of ψ(x), we clearly see that there exists x3∈(0,1) such that φλ,α(x) is strictly decreasing on (0,x3] and strictly increasing on [x3,1). Therefore, φλ,α(x)<0 for all x∈(0,1) follows from (24) and (31) together with the piecewise monotonicity of φλ,α(x).
Case
4.
(
(
1

α
)
[
1

log
(
1
+
2
)
]
/
log
(
1
+
2
)
)
<
λ
<
(
(
1

α
)
/
6
)
. Then (25) leads to
(32)φλ,α(1)>0.
It follows from (26), (29), and (30) together with the monotonicity of ψ(x), there exists x4∈(0,1) such that φλ,α(x) is strictly decreasing on (0,x4] and strictly increasing on [x4,1). Equation (24) and inequality (32) together with the piecewise monotonicity of φλ,α(x) lead to the conclusion that there exists x5∈(x4,1) such that φλ,α(x)<0 for x∈(0,x5) and φλ,α(x)>0 for x∈(x5,1).