# Bounds for the Arithmetic Mean in Terms of the Neuman-Sándor and Other Bivariate Means

Authors: Fan Zhang; Yu-Ming Chu; Wei-Mao Qian
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

We present the largest valuesα1, α2, and α3 and the smallest values β1, β2, and β3 such that the double inequalities α1M(a,b)+(1-α1)H(a,b)<A(a,b)<β1M(a,b)+ (1-β1)H(a,b), α2M(a,b)+(1-α2) H-(a,b) < A(a,b)<β2M(a,b)+(1-β2)H-(a,b), and α3M(a,b)+(1-α3)He(a,b)< A(a,b)<β3M (a,b)+(1-β3)He(a,b) hold for all a,b>0 with a≠b, where M(a,b), A(a,b), He(a,b), H(a,b) and H-(a,b) denote the Neuman-Sándor, arithmetic, Heronian, harmonic, and harmonic root-square 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] is defined by
(1)M(a,b)=a-b2sinh-1((a-b)/(a+b)),
where sinh-1(x)=log(x+x2+1) is the inverse hyperbolic sine function.Recently, the Neuman-Sándor mean has been the object intensive research. In particular, many remarkable inequalities for the Neuman-Sándor meanM(a,b) can be found in the literature [1–10].LetH¯(a,b)=2ab/a2+b2, H(a,b)=2ab/(a+b), G(a,b)=ab, He(a,b)=(a+ab+b)/3, L(a,b)=(b-a)/(logb-loga), P(a,b)=(a-b)/[4arctana/b-π], A(a,b)=(a+b)/2, T(a,b)=(a-b)/ [2arctan((a-b)/(a+b))], Q(a,b)=(a2+b2)/2, and C(a,b)=(a2+b2)/(a+b) be the harmonic root-square, harmonic, geometric, Heronian, logarithmic, first Seiffert, arithmetic, second Seiffert, quadratic, and contraharmonic means of a and b, respectively. Then it is known that the inequalities
(2)H¯(a,b)<H(a,b)<G(a,b)<L(a,b)<He(a,b)<P(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.Neuman and Sándor [1, 2] proved that the inequalities
(3)π4log(1+2)T(a,b)<M(a,b)<A(a,b)log(1+2),2T2(a,b)-Q2(a,b)<M(a,b)<T2(a,b)Q(a,b),H(T(a,b),A(a,b))<M(a,b)<L(A(a,b),Q(a,b)),H(M(a,b),Q(a,b))<T(a,b),M(a,b)<A2(a,b)P(a,b),A2/3(a,b)Q1/3(a,b)<M(a,b)<2A(a,b)+Q(a,b)3,A(a,b)T(a,b)<M(a,b)<A2(a,b)+T2(a,b),G(x,y)G(1-x,1-y)<L(x,y)L(1-x,1-y)<P(x,y)P(1-x,1-y)<A(x,y)A(1-x,1-y)<M(x,y)M(1-x,1-y)<T(x,y)T(1-x,1-y),1A(1-x,1-y)-1A(x,y)<1M(1-x,1-y)-1M(x,y)<1T(1-x,1-y)-1T(x,y),A(x,y)A(1-x,1-y)<M(x,y)M(1-x,1-y)<T(x,y)T(1-x,1-y)
hold for all a,b>0 and x,y∈(0,1/2] with a≠b and x≠y. All the results stated above are in fact particular cases of more general and stronger results for the Schwab-Borchardt means [1, 2]. Some of them are based on the sequential method of Sándor [11]. In particular, Neuman and Sándor [1] also found that the inequality
(4)M(a,b)<L(an,bn)
holds for all n≥0 and a,b>0 with a≠b, where a0=Q(a,b), b0=A(a,b), an+1=(an+bn)/2, and bn+1=anbn.Li et al. [3] 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)=1/e(bb/aa)1/(b-a), and L-1(a,b)=(b-a)/(logb-loga) is the pth generalized logarithmic mean of a and b; p0=1.843⋯ is the unique solution of the equation (p+1)1/p=2log(1+2).In [4], Neuman 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.The main purpose of this paper is to find the largest valuesα1, α2, and α3 and the smallest values β1, β2, and β3 such that the double inequalities
(6)α1M(a,b)+(1-α1)H(a,b)<A(a,b)<β1M(a,b)+(1-β1)H(a,b),α2M(a,b)+(1-α2)H¯(a,b)<A(a,b)<β2M(a,b)+(1-β2)H¯(a,b),α3M(a,b)+(1-α3)He(a,b)<A(a,b)<β3M(a,b)+(1-β3)He(a,b)
hold for all a,b>0 with a≠b. All numerical computations are carried out using Mathematical software.


## 2. Lemmas

In order to establish our main results we need several lemmas, which we present in this section.Lemma 1 (see [12, 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 an,bn>0 for all n∈{0,1,2,…}. Let h(x)=f(x)/g(x); then the following hold.(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 2.
The function(7)g(t)=t[cosh(2t)+2cosh(t)-3]sinh(2t)+tcosh(2t)-3t
is strictly decreasing on (0,log(1+2)), where sinh(t)=(et-e-t)/2 and cosh(t)=(et+e-t)/2 denote the hyperbolic sine and hyperbolic cosine functions, respectively.Proof.
Making use of power seriessinh(t)=∑n=0∞t2n+1/(2n+1)! and cosh(t)=∑n=0∞t2n/(2n)!, the function g(t) can be written as
(8)g(t)=∑n=0∞[(22n+2+2)/(2n+2)!]t2n∑n=0∞[(2n+5)22n+2/(2n+3)!]t2n.
Let(9)an=22n+2+2(2n+2)!,bn=(2n+5)22n+2(2n+3)!.
Then simple computation leads to(10)an+1bn+1-anbn=cn(2n+5)(2n+7)22n+3,
where
(11)cn=22n+5-(12n2+60n+59).
It follows from (11) that
(12)c0=-27,c1=-3,(13)cn≥128n2-(12n2+60n+59)=116n2-60n-59>0
for all n≥2.
Equations (10) and (12) together with inequality (13) lead to the conclusion that the sequence {an/bn}n=0∞ is strictly decreasing for 0≤n≤1 and strictly increasing for n≥2. Then from Lemma 1(2) and (8) together with (9) we clearly see that there exists t0∈(0,∞) such that g(t) is strictly decreasing on (0,t0) and strictly increasing on (t0,∞).
Lett*=log(1+2). Then simple computations lead to
(14)sinh(t*)=1,cosh(t*)=2,sinh(2t*)=22,cosh(2t*)=3.
It is not difficult to verify that(15)g′(t*)=-2t*2+(2-2)t*+1=-0.03734⋯<0.
From the piecewise monotonicity ofg(t) and inequality (15) we clearly see that t*=log(1+2)<t0, which implies that g(t) is strictly decreasing on (0,log(1+2)).Lemma 3.
The inequality(16)1-t2>1-t22-t48-t616-5t8128-35t10128
holds for all t∈(0,1).Proof.
Simple computations lead to(17)(1-t22-t48-t616-5t8128-35t10128)2=1-t2-t1016384×(8064-4704t2-1200t4-585t6-350t8-1225t10)<1-t2-t1016384(8064-4704-1200-585-350-1225)<1-t2
for all t∈(0,1).Lemma 4.
The functionf(t)=[t-sinh-1(t)]/[(1-1-t2)sinh-1(t)] is strictly decreasing in (0,1).Proof.
Differentiatingf(t) gives
(18)f′(t)=(t(1-t2-1-t2)+1+t2[tsinh-1(t)+1-t2-1]sinh-1(t))×((1-1-t2)21-t4[sinh-1(t)]2)-1.
Making use of the power series
(19)1+t2=∑n=0∞‍(-1)n-1(2n)!(2n-1)4n(n!)2t2n=1+t22-t48+t616-5t8128+7t10256-⋯,(20)1-t2=∑n=0∞‍(2n)!(1-2n)4n(n!)2t2n=1-t22-t48-t616-5t8128-7t10256-⋯,(21)sinh-1(t)=∑n=0∞‍(-1)n(2n)!(2n+1)4n(n!)2t2n+1=t-t36+3t540-5t7112+⋯,
we get
(22)1+t2sinh-1(t)=t+t33-2t515+8t7105-1091t913440+⋯hhhhhhhhhhhhhl<t+t33-2t515+8t7105,tsinh-1(t)+1-t2-1hhhh=t22-7t424+t680-75t8896+⋯<t22-7t434+t680,
for t∈(0,1).
Let(23)g(t)=t(1-t2-1-t2)+1+t2[tsinh-1(t)+1-t2-1]sinh-1(t)=-(1-1-t2)[t1-t2+1+t2sinh-1(t)]+t1+t2[sinh-1(t)]2.
We divide the proof into two cases.
Case 1 (
t
∈
(
0
,
10
/
5
)
). Then from Lemma 3, (22), and (23) we have
(24)g(t)<t[1-t2-(1-t22-t48-t616-5t8128-35t10128)]+(t+t33-2t515+8t7105)×(t22-7t424+t680)=t2201600(24235t2+50309t4+192t6-17920)<t2201600[24235(105)2+50309(105)4+192(105)6-17920]<0.
Case 2 (
t
∈
[
10
/
5,1
)
). Then from Lemma 3, (19), (20), and (23) together with sinh-1(10/5)=0.5964⋯ we get
(25)g(t)<-[1-(1-t22-t48-t616)]×[t(1-t22-t48-t616-5t8128-35t10128)+0.596(1+t22-t48)]+(0.596)2(1+t22-t48+t616)≔g1(t)=0.356409t-0.298t2-0.3217955t3-0.2235t4+0.08448875t5-0.03725t6+0.0847755625t7-0.0093125t8+0.078125t9+0.00465625t10+0.03515625t11+0.1455078125t13+0.03662109375t15+0.01708984375t17.
Numerical computations show that(26)g1(105)=-0.0000435,g1(1)=-0.0510683⋯,(27)g1′(105)=-0.52459⋯,g1′(1)=2.4665⋯,(28)g1′′(105)=-1.86198⋯,g1′′(1)=43.2711⋯,(29)g1′′′(105)=4.63578⋯,(30)g1(4)(t)>(-5.364+9.653865t)+t2(-13.41+71.2114725t-15.645t2)>[-5.364+9.653865×(105)]+t2[(-13.41+71.2114725×(105)-15.645)]=0.7416⋯+t2×15.9830⋯>0.
It follows from (29) and (30) that g1′′(t) is strictly increasing in [10/5,1). Then (28) leads to the conclusion that there exists t0∈(10/5,1) such that g1′(t) is strictly decreasing in [10/5,t0] and strictly increasing in [t0,1).
From (27) and the piecewise monotonicity of g1′(t) we clearly see that there exists t1∈(t0,1) such that g1(t) is strictly decreasing in [10/5,t1] and strictly increasing in [t1,1). Therefore,
(31)g(t)<g1(t)<0
for t∈[10/5,1) follows from (25) and (26) together with the piecewise monotonicity of g1(t).Lemma 5.
Letp∈(0,1), λ0=log(1+2)/[3-2log(1+2)]=0.7123⋯, and
(32)φp(t)=3p1-t2(p-1)t+2p-1-sinh-1(1-t2).
Then φ1/2(t)<0 and φλ0(t)>0 for all t∈(0,1).Proof.
We first prove thatφ1/2(t)<0 for t∈(0,1). From (32) one has
(33)φ1/2(1)=0,(34)φ1/2′(t)=4t-2t2+(1/4)t3-(3t-3/4)2-t21-t22-t2(2-(1/2)t)2.
Let
(35)f(t)=4t-2t2+14t3-(3t-34)2-t2.
Then
(36)f(1)=0,f′(t)=24t2-3t-24+(16-16t+3t2)2-t242-t2.
We divide the proof into two cases.
Case 1 (
t
∈
(
0,0.7
)
). Then we clearly see that
(37)24t2-3t-24+(16-16t+3t2)2-t22-t2<24t2-3t+242+(16-16t+3t2)=27t2-(322+16)t+16-122=27(t-32+32+53762-5870108)×(t-32+32-53762-5870108)=27(t-0.72105⋯)(t+0.04985⋯)<0.
Case 2 (
t
∈
[
0.7,1
)
). Then we get
(38)[24t2-3t-24+(16-16t+3t2)2-t2]2-t2=(24t2-3t-24)2-t2+32-32t-10t2+16t3-3t4<24t2-3t-24+32-32t-10t2+16t3-3t4=-(1-t)(27t+13t2-3t3-8)<0.
It follows from (32) and (33) together with Cases 1 and 2 that f(t)<0 for t∈(0,1). Then from (34) and (35) we know that φ1/2(t) is strictly decreasing in (0,1).
Therefore,φ1/2(t)<0 for t∈(0,1) follows from (33) and the monotonicity of φ1/2(t).
Next, we prove thatφλ0(t)>0 for t∈(0,1). From (32) we clearly see that we only have to prove that(39)φλ0(1-t2)=3λ0t(λ0-1)1-t2+2λ0+1-sinh-1(t)>0
for all t∈(0,1).
Inequality (39) can be rewritten as
(40)λ0>(1-1-t2)sinh-1(t)3t-(2+1-t2)sinh-1(t).
Let
(41)h(t)=(1-1-t2)sinh-1(t)3t-(2+1-t2)sinh-1(t)=11+3[t-sinh-1(t)]/[(1-1-t2)sinh-1(t)].
Then inequality (40) follows from Lemma 4 and (41) together with h(1)=λ0.


## 3. Main Results

Theorem 6.
The double inequality(42)α1M(a,b)+(1-α1)H(a,b)<A(a,b)<β1M(a,b)+(1-β1)H(a,b)
holds for all a,b>0 with a≠b if and only if α1≤6/7=0.8571⋯ and β1≥log(1+2)=0.8813⋯.Proof.
SinceH(a,b), M(a,b), and A(a,b) are symmetric and homogeneous of degree one, without loss of generality, we assume that a>b. Let x=(a-b)/(a+b) and t=sinh-1(x). Then x∈(0,1), t∈(0,log(1+2)), and
(43)A(a,b)-H(a,b)M(a,b)-H(a,b)=x2x/sinh-1(x)-(1-x2)=tcosh(2t)-t2sinh(t)+tcosh(2t)-3t.
Let
(44)f(t)=tcosh(2t)-t2sinh(t)+tcosh(2t)-3t.
Then f(t) can be rewritten as
(45)f(t)=t∑n=0∞(22n/(2n)!)t2n-t2∑n=0∞(t2n+1/(2n+1)!)+∑n=0∞(22n/(2n)!)t2n+1-3t=∑n=1∞(22n/(2n)!)t2n∑n=1∞[(2+(2n+1)22n)/(2n+1)!]t2n+1=∑n=0∞(22n+2/(2n+2)!)t2n∑n=0∞[(2+(2n+3)22n+2)/(2n+3)!]t2n≔∑n=0∞ant2n∑n=0∞bnt2n,an+1bn+1-anbn=(12n+34)22n+2[(2n+3)22n+2+2][(2n+5)22n+4+2]>0
for all n≥0.
It follows from (45) together with Lemma 1(1) that f(t) is strictly increasing in (0,log(1+2)). Note that
(46)limt→0f(t)=a0b0=67,f(log(1+2))=log(1+2).
Therefore, Theorem6 follows from (43) and (44) together with (46) and the monotonicity of f(t).Theorem 7.
The double inequality(47)α2M(a,b)+(1-α2)H¯(a,b)<A(a,b)<β2M(a,b)+(1-β2)H¯(a,b)
holds for all a,b>0 with a≠b if and only if α2≤log(1+2)=0.8813⋯ and β2≥9/10.Proof.
SinceH¯(a,b), M(a,b), and A(a,b) are symmetric and homogeneous of degree one, without loss of generality, we assume that a>b. Let x=(a-b)/(a+b) and t=sinh-1(x). Then x∈(0,1), t∈(0,log(1+2)), and
(48)A(a,b)-H¯(a,b)M(a,b)-H¯(a,b)=(1+x2+x2-1)sinh-1(x)x1+x2-(1-x2)sinh-1(x)=t[cosh(2t)+2cosh(t)-3]sinh(2t)+tcosh(2t)-3t.
Simple computations lead to(49)limt→0t[cosh(2t)+2cosh(t)-3]sinh(2t)+tcosh(2t)-3t=910,limt→log(1+2)t[cosh(2t)+2cosh(t)-3]sinh(2t)+tcosh(2t)-3t=log(1+2).
Therefore, Theorem7 follows from Lemma 2, (48), and (49).Theorem 8.
The double inequality(50)α3M(a,b)+(1-α3)He(a,b)<A(a,b)<β3M(a,b)+(1-β3)He(a,b)
holds for all a,b>0 with a≠b if and only if α3≤1/2 and β3≥log(1+2)/ [3-2log(1+2)]=0.7123⋯.Proof.
SinceH(a,b), M(a,b), and A(a,b) are symmetric and homogeneous of degree one, without loss of generality, we assume that a>b. Let x=(a-b)/(a+b)∈(0,1), 0<p<1, and λ0=log(1+2)/[3-2log(1+2)]; then
(51)A(a,b)-He(a,b)M(a,b)-He(a,b)=(1-1-x2)sinh-1(x)3x-(2+1-x2)sinh-1(x),(52)pM(a,b)+(1-p)He(a,b)-A(a,b)=A(a,b)3[3pxsinh-1(x)-(p-1)1-x2-(1+2p)]=A(a,b)[(1+2p)-(1-p)1-x2]3sinh-1(x)φp(1-x2),
where φp(t) is defined as in Lemma 5.
Note that(53)limx→0(1-1-x2)sinh-1(x)3x-(2+1-x2)sinh-1(x)=12,limx→1(1-1-x2)sinh-1(x)3x-(2+1-x2)sinh-1(x)=λ0.
Therefore, Theorem8 follows from (51)–(53) together with Lemma 5.



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