# Bounds for the Combinations of Neuman-Sándor, Arithmetic, and Second Seiffert Means in terms of Contraharmonic Mean

Authors: Zai-Yin He; Wei-Mao Qian; Yun-Liang Jiang; Ying-Qing Song; Yu-Ming Chu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We give the greatest valuesr1, r2 and the least values s1, s2 in (1/2, 1) such that the double inequalities C(r1a+(1-r1)b,r1b+(1-r1)a)<αA(a,b)+(1-α)T(a,b)<C(s1a+(1-s1)b,s1b+(1-s1)a) and C(r2a+(1-r2)b,r2b+(1-r2)a)<αA(a,b)+(1-α)M(a,b)<C(s2a+(1-s2)b,s2b+(1-s2)a) hold for any α∈(0,1) and all a,b>0 with a≠b, where A(a,b), M(a,b), C(a,b), and T(a,b) are the arithmetic, Neuman-Sándor, contraharmonic, and second Seiffert means of a and b, respectively.


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## Body

## 1. Introduction

Fora,b>0 with a≠b, the Neuman-Sándor mean M(a,b) [1], second Seiffert mean T(a,b) [2] are defined by
(1)M(a,b)=a-b2sinh-1((a-b)/(a+b)),T(a,b)=a-b2arctan((a-b)/(a+b)),
respectively. Herein, sinh-1(x)=log(x+1+x2) is the inverse hyperbolic sine function.LetH(a,b)=2ab/(a+b), G(a,b)=ab, L(a,b)=(a-b)/(loga-logb), P(a,b)=(a-b)/[4arctan(a/b)-π], I(a,b)=1/e(bb/aa)1/(b-a), 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 Neuman-Sándor, 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.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′)
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+1-ap+1)/((p+1)(b-a))]1/p(p≠-1,0), L0(a,b)=I(a,b) and L-1(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)+(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⋯, λ≤[1-log(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)+(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.In [20, 21], the authors proved that the double inequalities
(7)α1T(a,b)+(1-α1)G(a,b)<A(a,b)<β1T(a,b)+(1-β1)G(a,b),α2Q(a,b)+(1-α2)A(a,b)<T(a,b)<β2Q(a,b)+(1-β2)A(a,b),Qα3(a,b)A1-α3(a,b)<T(a,b)<Qβ3(a,b)A1-β3(a,b)
hold for all a,b>0 with a≠b if and only of α1≤3/5, β1≥4/π,   α2≤(4-π)/[(2-1)π], β2≥2/3, α3≤2/3, and β3≥4-2logπ/log2.Forα,β,λ,μ∈(1/2,1), Chu et al. [22, 23] proved that the inequalities
(8)C(αa+(1-α)b,αb+(1-α)a)<T(a,b)<C(βa+(1-β)b,βb+(1-β)a),Q(λa+(1-λ)b,λb+(1-λ)a)<T(a,b)<Q(μa+(1-μ)b,μb+(1-μ)a)
hold for all a,b>0 with a≠b if and only if α≤(1+4/π-1)/2, β≥(3+3)/6, λ≤(1+16/π2-1)/2 and μ≥(3+6)/6.The aim of this paper is to find the greatest valuesr1, r2 and the least values s1, s2 such that the double inequalities
(9)C(r1a+(1-r1)b,r1b+(1-r1)a)<αA(a,b)+(1-α)T(a,b)<C(s1a+(1-s1)b,s1b+(1-s1)a),(10)C(r2a+(1-r2)b,r2b+(1-r2)a)<αA(a,b)+(1-α)M(a,b)<C(s2a+(1-s2)b,s2b+(1-s2)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 [24, 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.
Letu,α∈(0,1) and
(12)fu,α(x)=ux2-(1-α)(xarctanx-1).
Then fu,α(x)>0 for all x∈(0,1) if and only if u≥(1-α)/3 and fu,α(x)<0  for all x∈(0,1) if and only if u≤(1-α)(4/π-1).Proof.
From (12), one has
(13)fu,α(0+)=0,(14)fu,α(1-)=u-(1-α)(4π-1),(15)fu,α′(x)=2x[u-1-α2g(x)],
where
(16)g(x)=(1+x2)arctanx-xx(1+x2)(arctanx)2.
Letg1(x)=arctanx-x/(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(π-2)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.
Letλ,α∈(0,1) and
(23)φλ,α(x)=λx2-(1-α)(xsinh-1(x)-1).
Then φλ,α(x)>0 for all x∈(0,1) if and only if λ≥(1-α)/6 and φλ,α(x)<0 for all x∈(0,1) if and only if λ≤(1-α)(1-log(1+2))/log(1+2).Proof.
From (23) we get
(24)φλ,α(0+)=0,(25)φλ,α(1-)=λ-(1-α)[1-log(1+2)]log(1+2),(26)φλ,α′(x)=2x[λ-1-α2ψ(x)],
where
(27)ψ(x)=sinh-1(x)-x/1+x2x(sinh-1(x))2.
Letψ1(x)=sinh-1(x)-x/1+x2 and ψ2(x)=x(sinh-1(x))2, then
(28)ψ(x)=ψ1(x)ψ2(x),ψ1(0)=ψ2(0)=0,ψ1′(x)ψ2′(x)=x2×((1+x2)3/2(sinh-1(x))2+2x(1+x2)sinh-1(x)(sinh-1(x))2)-1=(((1+x2)3/4sinh-1(x)/x)2+2(1+x2)1/4((1+x2)3/4sinh-1(x)/x))-1.
It is not difficult to verify that the function(1+x2)3/4sinh-1(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≤(1-α)[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).


## 3. Main Results

Theorem 4.
Ifα∈(0,1) and r1,s1∈(1/2,1), then the double inequality
(33)C(r1a+(1-r1)b,r1b+(1-r1)a)<αA(a,b)+(1-α)T(a,b)<C(s1a+(1-s1)b,s1b+(1-s1)a)
holds for all a,b>0 with a≠b if and only if r1≤[1+(1-α)(4-π)/π]/2 and s1≥[1+(1-α)/3]/2.Proof.
SinceA(a,b), T(a,b), and C(a,b) are symmetric and homogeneous of degree one, without loss of generality, we assume that a>b. Let p∈(1/2,1) and x=(a-b)/(a+b), then x∈(0,1) and
(34)C(pa+(1-p)b,pb+(1-p)a)-[αA(a,b)+(1-α)T(a,b)]=A(a,b)[(2p-1)2x2-(1-α)(xarctanx-1)].
Therefore, Theorem4 follows easily from Lemma 2 and (34).Theorem 5.
Ifα∈(0,1) and r2,s2∈(1/2,1), then the double inequality
(35)C(r2a+(1-r2)b,r2b+(1-r2)a)<αA(a,b)+(1-α)M(a,b)<C(s2a+(1-s2)b,s2b+(1-s2)a)
holds for all a,b>0 with a≠b if and only if s2≥[1+(1-α)/6]/2 and r2≤[1+(1-α)(1-log(1+2))/log(1+2)]/2.Proof.
SinceA(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∈(1/2,1) and x=(a-b)/(a+b), then x∈(0,1) and
(36)C(qa+(1-q)b,qb+(1-q)a)-[αA(a,b)+(1-α)M(a,b)]=A(a,b)[(2q-1)2x2-(1-α)(xsinh-1(x)-1)].
Therefore, Theorem5 follows easily from Lemma 3 and (36).Remark 6.
Ifα=0, then Theorem 4 reduces to the first double inequality in (8).Corollary 7.
Ifλ,μ∈(1/2,1), then the double inequality
(37)C(λa+(1-λ)b,λb+(1-λ)a)<M(a,b)<C(μa+(1-μ)b,μb+(1-μ)a)
holds for all a,b>0 with a≠b if and only if λ≤[1+1/log(1+2)-1]/2 and μ≥(6+6)/12.Proof.
Corollary7 follows easily from Theorem 5 with α=0.



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