# On Certain Inequalities for Neuman-Sándor Mean

Authors: Wei-Mao Qian; Yu-Ming Chu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We present several new sharp bounds for Neuman-Sándor mean in terms of arithmetic, centroidal, quadratic, harmonic root square, and contraharmonic means.


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## Body

## 1. Introduction

A binary mapm:ℝ+×ℝ+→ℝ (where ℝ+:=(0,+∞) is the set of positive numbers) is said to be a bivariate mean if the following statements are satisfied for all a,b,λ>0: (i)m(a,a)=a (reflexivity property); (ii)m(a,b)=m(b,a) (symmetry property); (iii)m(λa,λb)=λm(a,b) (homogeneous of order one); (iv)m(a,b) is continuous and strictly increasing with respect to a and b.Letu,v, and w be the bivariate means such that u(a,b)<w(a,b)<v(a,b) for all a,b>0 with a≠b. The problems to find the best possible parameters α and β such that the inequalities αu(a,b)+(1-α)v(a,b)<w(a,b)<βu(a,b)+(1-β)v(a,b) and uα(a,b)v1-α(a,b)<w(a,b)<uβ(a,b)v1-β(a,b) hold for all a,b>0 with a≠b have attracted the interest of many mathematicians.Fora,b>0 with a≠b, the Neuman-Sándor mean M(a,b) [1] is defined by
(1)M(a,b)=a-b2arcsinh[(a-b)/(a+b)],
where arcsinh(x)=log(x+1+x2) is the inverse hyperbolic sine function.Recently, the bounds for the Neuman-Sándor mean in terms of other bivariate means have been the subject of intensive research.LetH¯(a,b)=2ab/a2+b2, H(a,b)=2ab/(a+b), G(a,b)=ab, L(a,b)=(a-b)/(loga-logb), P(a,b)=(a-b)/(4arctana/b-π), I(a,b)=1/e(bb/aa)1/(b-a), A(a,b)=(a+b)/2, T(a,b)=(a-b)/[2arctan((a-b)/(a+b))], E(a,b)=2(a2+ab+b2)/[3(a+b)], Q(a,b)=(a2+b2)/2, and C(a,b)=(a2+b2)/(a+b) be the harmonic root square, harmonic, geometric, logarithmic, first Seiffert, identric, arithmetic, second Seiffert, centroidal, 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)<H(a,b)<G(a,b)<L(a,b)<P(a,b)<I(a,b)<A(a,b)<M(a,b)<T(a,b)<E(a,b)<Q(a,b)<C(a,b)
hold for all a,b>0 with a≠b.Neuman and Sándor [1, 2] 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,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
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. [3] proved that Lp0(a,b)<M(a,b)<L2(a,b) 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 [4], 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.In [5, 6], the authors proved that α=4, β=log2/[2log(1+2)], λ=1, and μ=e/[2log(1+2)] are the best possible constants such that the inequalities M(a,b)>L¯α(a,b), M(a,b)>Mβ(a,b), and λI(a,b)<M(a,b)<μI(a,b) hold for all a,b>0 with a≠b. In here, L¯r(a,b)=[(br-ar)/(r(logb-loga))]1/r and Mp(a,b)=[(ap+bp)/2]1/p are the rth generalized logarithmic and pth power means of a and b, respectively.Zhao et al. [7, 8] found the least values α1, α2, α3, α4, α5 and the greatest values β1, β2, β3, β4, β5 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),Iα4(a,b)Q1-α4(a,b)<M(a,b)<Iβ4(a,b)Q1-β4(a,b),Iα5(a,b)C1-α5(a,b)<M(a,b)<Iβ5(a,b)C1-β5(a,b)
hold for all a,b>0 with a≠b.In [9], the authors proved that if λ,μ∈(1/2,1), then the double inequality 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.The aim of this paper is to present the sharp bounds for Neuman-Sándor mean in terms of the combinations of either arithmetic and centroidal means, or quadratic and harmonic root square means, contraharmonic and harmonic root square means. Our main results are shown in Theorems1–4.Theorem 1.
The double inequality(7)α1E(a,b)+(1-α1)A(a,b)<M(a,b)<β1E(a,b)+(1-β1)A(a,b)
holds for all a,b>0 with a≠b if and only if α1≤(3-3log(1+2))/log(1+2)=0.4037 and β1≥1/2.Theorem 2.
The double inequality(8)Eλ1(a,b)A1-λ1(a,b)<M(a,b)<Eμ1(a,b)A1-μ1(a,b)
holds for all a,b>0 with a≠b if and only if λ1≤log[log(1+2)]/(log3-2log2)=0.4389⋯ and μ1≥1/2.Theorem 3.
The double inequality(9)α2Q(a,b)+(1-α2)H¯(a,b)<M(a,b)<β2Q(a,b)+(1-β2)H¯(a,b)
holds for all a,b>0 with a≠b if and only if α2≤2/[2log(1+2)]=0.8022⋯ and β2≥5/6.Theorem 4.
The double inequality(10)λ2C(a,b)+(1-λ2)H¯(a,b)<M(a,b)<μ2C(a,b)+(1-μ2)H¯(a,b)
holds for all a,b>0 with a≠b if and only if λ2≤1/[2log(1+2)]=0.5672⋯ and μ2≥2/3.


## 2. Lemmas

In order to establish our main results we need several lemmas, which we present in this section.Lemma 5 (see [10, Theorem 1.25]).
For-∞<a<b<∞, let f,g:[a,b]→ℝ be continuous on [a,b] and differentiable on (a,b), and 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 6 (see [11, Lemma 1.1]).
Suppose that the power seriesf(x)=∑n=0∞‍anxn and g(x)=∑n=0∞‍bnxn have the radius of convergence r>0 and bn>0 for all n∈{0,1,2,…}. Let h(x)=f(x)/g(x). Then,(1)
if the sequence{an/bn}n=0∞ is (strictly) increasing (decreasing), then h(x) is also (strictly) increasing (decreasing) on (0,r);(2)
if the sequence{an/bn} is (strictly) increasing (decreasing) for 0<n≤n0 and (strictly) decreasing (increasing) for n>n0, then there exists x0∈(0,r) such that h(x) is (strictly) increasing (decreasing) on (0,x0) and (strictly) decreasing (increasing) on (x0,r).Lemma 7.
The function(12)h(t)=tcosh(3t)+11tcosh(t)-sinh(3t)-9sinh(t)2t[cosh(3t)-cosh(t)]
is strictly decreasing on (0,log(1+2)), where sinh(t)=(et-e-t)/2 and cosh(t)=(et+e-t)/2 are the hyperbolic sine and cosine functions, respectively.Proof.
Let(13)h1(t)=tcosh(3t)+11tcosh(t)-sinh(3t)-9sinh(t),h2(t)=2t[cosh(3t)-cosh(t)].
Then, making use of power series formulas, we have
(14)h1(t)=t∑n=0∞‍(3t)2n(2n)!+11t∑n=0∞‍(t)2n(2n)!-∑n=0∞‍(3t)2n+1(2n+1)!-9∑n=0∞‍(t)2n+1(2n+1)!=∑n=0∞‍(2n+3)(32n+2+11)-(32n+3+9)(2n+3)!t2n+3,h2(t)=∑n=0∞‍2(32n+2-1)(2n+2)!t2n+3.
It follows from (12)–(14) that
(15)h(t)=∑n=0∞‍ant2n∑n=0∞‍bnt2n,
where
(16)an=(2n+3)(32n+2+11)-(32n+3+9)(2n+3)!,bn=2(32n+2-1)(2n+2)!.
Equation (16) leads to
(17)an+1bn+1-anbn=18cn2(2n+3)(2n+5)(32n+2-1)(32n+4-1),
where
(18)cn=3·9n(81·9n-64n2-224n-166)-1.
It is not difficult to verify thatx→81·9x-64x2-224x-166 is positive and strictly increasing in [1,∞). Then, from (18), we get that
(19)cn>0
forn≥1. Note that
(20)c0=-256,c1=7424.
Equations (17) and (20) together with inequality (19) lead to the conclusion that the sequence {an/bn} is strictly decreasing for 0≤n≤1 and strictly increasing for n≥2. Then, from Lemma 6(2) and (15), we clearly see that there exists t0∈(0,∞) such that h(t) is strictly decreasing on (0,t0) and strictly increasing on (t0,∞).
Lett*=log(1+2). Then, simple computations lead to
(21)sinh(t*)=1,cosh(t*)=2,sinh(3t*)=7,cosh(3t*)=52.
Differentiating (12) yields
(22)h′(t)=3tsinh(3t)+11tsinh(t)-2cosh(3t)+2cosh(t)h2(t)-2t[3sinh(3t)-sinh(t)]+2[cosh(3t)-cosh(t)]h2(t)2h1(t).
From (13)  together with (21) and (22), we get
(23)h′(t*)=-32t*2+2t*+2t*2=-0.1529⋯<0.
From the piecewise monotonicity ofh(t) and inequality (23) we clearly see that t0>t*=log(1+2), and the proof of Lemma 7 is completed.Lemma 8.
The function(24)g(t)=tcosh(3t)+3tcosh(t)-2sinh(2t)t[cosh(3t)+2cosh(2t)+3cosh(t)-6]
is strictly increasing from (0,log(1+2)) onto (1/3,1-1/[2log(1+2)]).Proof.
Let(25)g1(t)=tcosh(3t)+3tcosh(t)-2sinh(2t),g2(t)=t[cosh(3t)+2cosh(2t)+3cosh(t)-6].
Then, making use of power series formulas, we have
(26)g1(t)=t∑n=0∞‍(3t)2n(2n)!+3t∑n=0∞‍(t)2n(2n)!-2∑n=0∞‍(2t)2n+1(2n+1)!=∑n=0∞‍(2n+3)(32n+2+3)-22n+4(2n+3)!t2n+3,g2(t)=∑n=0∞‍32n+2+22n+3+3(2n+2)!t2n+3.
It follows from (24)–(26) that
(27)h(t)=∑n=0∞‍dnt2n∑n=0∞‍ent2n,
where
(28)dn=(2n+3)(32n+2+3)-22n+4(2n+3)!,en=32n+2+22n+3+3(2n+2)!.
Note that(29)dnen=(2n+3)(32n+2+3)-22n+4(2n+3)[32n+2+22n+3+3]=1-(16·4n/(2n+3)(9n+1+3))1+(8·4n/(9n+1+3)).
It is not difficult to verify that the functionx→4x/(9x+1+3) is strictly decreasing in (0,∞). Then from (29), we know that the sequence {dn/en} is strictly increasing for n=0,1,2,…. Hence, from Lemma 6(1), (24), and (27) the monotonicity of g(t) follows. Moreover, g(log(1+2))=1-1/[2log(1+2)]=0.4327⋯ and
(30)limt→0g(t)=d0e0=13.


## 3. Proofs of Theorems1–4

Proof of Theorem1.
SinceM(a,b), E(a,b), and A(a,b) are symmetric and homogeneous of degree 1, without loss of generality, we assume that a>b. Let x=(a-b)/(a+b) and t=arcsinh(x). Then, x∈(0,1), t∈(0,log(1+2)), and
(31)M(a,b)-A(a,b)E(a,b)-A(a,b)=x/arcsinh(x)-1(3+x2)/3-1=3[sinh(t)-t]tsinh2(t)=6[sinh(t)-t]t[cosh(2t)-1].
Let(32)F(t)=6[sinh(t)-t]t[cosh(2t)-1].
Then, simple computations lead to
(33)F(t)=∑n=0∞‍an*t2n∑n=0∞‍bn*t2n,
where an*=6/(2n+3)! and bn*=22n+2/(2n+2)!. Note that an*/bn*=6/[(2n+3)22n+2] is strictly decreasing for n=0,1,2,…. Hence, from Lemma 6(1) and (33), we know that F(t) is strictly decreasing in (0,log(1+2)). Moreover,
(34)limt→0F(t)=a0*b0*=12,limt→log(1+2)F(t)=3-3log(1+2)log(1+2)=0.4037⋯.
Therefore, Theorem1 follows from (31), (32), and (34) together with the monotonicity of F(t).Proof of Theorem2.
Without loss of generality, we assume thata>b. Let x=(a-b)/(a+b) and t=arcsinh(x). Then, x∈(0,1), t∈(0,log(1+2)), and
(35)log[M(a,b)]-log[A(a,b)]log[E(a,b)]-log[A(a,b)]=log[x/arcsinh(x)]log(1+(x2/3))=log[sinh(t)/t]log[1+sinh2(t)/3].
LetG1(t)=log[sinh(t)/t], G2(t)=log[1+sinh2(t)/3], and
(36)G(t)=log[sinh(t)/t]log[1+sinh2(t)/3].
Then, G1(0+)=G2(0)=0, G(t)=G1(t)/G2(t), and
(37)G1′(t)G2′(t)=[tcosh(t)-sinh(t)][sinh2(t)+3]2tsinh2(t)cosh(t)=[tcosh(t)-sinh(t)][cosh(2t)+5]2tsinh(2t)sinh(t)=h(t),
where h(t) is defined as in Lemma 7.
It follows from Lemmas5 and 7, (36), and (37) that G(t) is strictly decreasing in (0,log(1+2)). Moreover,
(38)limt→0G(t)=limt→0G1′(t)G2′(t)=12,limt→log(1+2)G(t)=log[log(1+2)]log3-2log2=0.4389⋯.
Therefore, Theorem2 follows easily from (35), (36), and (38) together with the monotonicity of G(t).Proof of Theorem3.
SinceM(a,b), Q(a,b), and H¯(a,b) are symmetric and homogeneous of degree 1, without loss of generality, we assume that a>b. Let x=(a-b)/(a+b) and t=arcsinh(x). Then, x∈(0,1), t∈(0,log(1+2)), and
(39)M(a,b)-Q(a,b)H¯(a,b)-Q(a,b)=x/arcsinh(x)-1+x2(1-x2)/1+x2-1+x2=sinh(t)/t-cosh(t)[1-sinh2(t)]/cosh(t)-cosh(t)=tcosh(2t)+t-sinh(2t)2t[cosh(2t)-1].
Let(40)φ(t)=tcosh(2t)+t-sinh(2t)2t[cosh(2t)-1].
Then, simple computations lead to(41)φ(t)=∑n=0∞‍dn*t2n∑n=0∞‍en*t2n,
where dn*=(2n+1)·22n+2/(2n+3)!, and en*=22n+3/(2n+2)!. Note that dn*/en*=[1-2/(2n+3)]/2 is strictly increasing for n=0,1,2,…. Hence, from Lemma 6(1) and (41), we know that φ(t) is strictly increasing in (0,log(1+2)). Moreover,
(42)limt→0φ(t)=d0*e0*=16,limt→log(1+2)φ(t)=1-22log(1+2)=0.1977⋯.
Therefore, Theorem3 follows from (39), (40), and (42) together with the monotonicity of φ(t).Proof of Theorem4.
SinceM(a,b), C(a,b), and H¯(a,b) are symmetric and homogeneous of degree 1, without loss of generality, we assume that a>b. Let x=(a-b)/(a+b) and t=arcsinh(x). Then, x∈(0,1), t∈(0,log(1+2)), and
(43)M(a,b)-C(a,b)H¯(a,b)-C(a,b)=x/arcsinh(x)-(1+x2)(1-x2)/1+x2-(1+x2)=sinh(t)/t-cosh2(t)[1-sinh2(t)]/cosh(t)-cosh2(t)=tcosh3(t)-sinh(t)cosh(t)t[cosh3(t)+sinh2(t)-1]=g(t),
where g(t) is defined as in Lemma 8.
Therefore, Theorem4 follows from (43) and Lemma 8.



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