# Sharp Geometric Mean Bounds for Neuman Means

Authors: Yan Zhang; Yu-Ming Chu; Yun-Liang Jiang
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We find the best possible constantsα1,α2,β1,β2∈[0,1/2] and α3,α4,β3,β4∈[1/2,1] such that the double inequalities G(α1a+(1-α1)b,α1b + (1-α1)a)<NAG(a,b)<G(β1a + (1-β1)b,β1b+(1-β1)a),G(α2a+(1-α2)b,α2b + (1-α2)a)<NGA(a,b)<G(β2a + (1-β2)b,β2b+(1-β2)a),Q(α3a+(1-α3)b,α3b + (1-α3)a)<NQA(a,b)<Q(β3a + (1-β3)b,β3b+(1-β3)a),Q(α4a+(1-α4)b,α4b + (1-α4)a)<NAQ(a,b)<Q(β4a + (1-β4)b,β4b+(1-β4)a) hold for all a,b>0 with a≠b, where G, A, and Q are, respectively, the geometric, arithmetic, and quadratic means and NAG, NGA, NQA, and NAQ are the Neuman means.


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## Body

## 1. Introduction

Fora,b>0 with a≠b, the Schwab-Borchardt mean SB(a,b) [1, 2] 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. Recently, the Schwab-Borchardt mean has been the subject of intensive research. In particular, many remarkable inequalities for Schwab-Borchardt mean and its generated means can be found in the literature [1–6].Very recently, Neuman [7] found a new bivariate mean N(a,b) derived from the Schwab-Borchardt mean as follows:
(2)N(a,b)=12(a+b2SB(a,b)).LetNAG(a,b)=N(A(a,b),G(a,b)), NGA(a,b)=N(G(a,b),A(a,b)), NQA(a,b)=N(Q(a,b),A(a,b)), and NAQ(a,b)=N(A(a,b),Q(a,b)) be the Neuman means, where G(a,b)=ab, A(a,b)=(a+b)/2, and Q(a,b)=(a2+b2)/2 are the classical geometric, arithmetic, and quadratic means of a and b, respectively. Then the inequalities
(3)G(a,b)<NAG(a,b)<NGA(a,b)<A(a,b)<NQA(a,b)<NAQ(a,b)<Q(a,b)
for all a,b>0 with a≠b, were established by Neuman [7].Leta>b>0 and v=(a-b)/(a+b)∈(0,1). Then the following explicit formulas for NAG(a,b), NGA(a,b), NQA(a,b), and NAQ(a,b) are presented in [7]
(4)NAG(a,b)=12A(a,b)[1+(1-v2)tanh-1vv],(5)NGA(a,b)=12A(a,b)[1-v2+sin-1vv],(6)NQA(a,b)=12A(a,b)[1+v2+sinh⁡-1vv],(7)NAQ(a,b)=12A(a,b)[1+(1+v2)tan-1vv],
where tanh-1(x)=log⁡[(1+x)/(1-x)]/2, sin-1(x), sinh⁡-1(x)=log⁡(x+1+x2), and tan-1(x) are the inverse hyperbolic tangent, inverse sine, inverse hyperbolic sine, and inverse tangent functions, respectively.In [7], Neuman also proved that the double inequalities
(8)α1A(a,b)+(1-α1)G(a,b)<NGA(a,b)<β1A(a,b)+(1-β1)G(a,b),α2Q(a,b)+(1-α2)A(a,b)<NAQ(a,b)<β2Q(a,b)+(1-β2)A(a,b),α3A(a,b)+(1-α3)G(a,b)<NAG(a,b)<β3A(a,b)+(1-β3)G(a,b),α4Q(a,b)+(1-α4)A(a,b)<NQA(a,b)<β4Q(a,b)+(1-β4)A(a,b)
hold for all a,b>0 with a≠b if and only if α1≤2/3, β1≥π/4, α2≤2/3, β2≥(π-2)/[4(2-1)]=0.689…, α3≤1/3, β3≥1/2, α4≤1/3, and β4≥[log⁡(1+2)+2-2]/[2(2-1)]=0.356….Leta,b>0 with a≠b, x∈[0,1/2], y∈[1/2,1],
(9)f(λ)=G[λa+(1-λ)b,λb+(1-λ)a],g(μ)=Q[μa+(1-μ)b,μb+(1-μ)a].
Then, it is not difficult to verify that f(λ) and g(μ) are continuous and strictly increasing on [0,1/2] and [1/2,1], respectively. Note that
(10)f(0)=G(a,b)<NAG(a,b)<NGA(a,b)<A(a,b)=f(12),g(12)=A(a,b)<NQA(a,b)<NAQ(a,b)<Q(a,b)=g(1).Therefore, it is natural to ask what the best possible constantsα1,α2,β1,β2∈[0,1/2] and α3,α4,β3,β4∈[1/2,1] are such that the double inequalities
(11)G(α1a+(1-α1)b,α1b+(1-α1)a)<NAG(a,b)<G(β1a+(1-β1)b,β1b+(1-β1)a),G(α2a+(1-α2)b,α2b+(1-α2)a)<NGA(a,b)<G(β2a+(1-β2)b,β2b+(1-β2)a),Q(α3a+(1-α3)b,α3b+(1-α3)a)<NQA(a,b)<Q(β3a+(1-β3)b,β3b+(1-β3)a),Q(α4a+(1-α4)b,α4b+(1-α4)a)<NAQ(a,b)<Q(β4a+(1-β4)b,β4b+(1-β4)a)
hold for all a,b>0 with a≠b. The main purpose of this paper is to answer this question.


## 2. Lemmas

In order to prove our main results, we need several lemmas, which we present in this section.Lemma 1 (see [8, Theorem 1.25]).
Let-∞<a<b<∞, f,g:[a,b]→(-∞,∞) be continuous on [a,b] and differentiable on (a,b), and g′(x)≠0 on (a,b). If f′(x)/g′(x) is increasing (decreasing) on (a,b), then so are
(12)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 (see [9, 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. If the sequence {an/bn} is (strictly) increasing (decreasing), for all n≥0, then the function f(x)/g(x) is also (strictly) increasing (decreasing) on (0,r).Lemma 3.
The function(13)f1(x)=3sinh⁡2(x)-2xsinh⁡(x)-x2[cosh⁡(x)-1]2
is strictly increasing from (0,∞) onto (8/3,3).Proof.
Making use of the power series expansion, we have(14)3sinh⁡2(x)-2xsinh⁡(x)-x2=32cosh⁡(2x)-2xsinh⁡(x)-x2-32=32∑n=0∞‍22n(2n)!x2n-2x∑n=0∞‍1(2n+1)!x2n+1-x2-32=32∑n=2∞‍22n(2n)!x2n-2∑n=1∞‍1(2n+1)!x2n+2=x4∑n=0∞‍3·22n+3-4(n+2)(2n+4)!x2n,[cosh⁡(x)-1]2=12cosh⁡(2x)-2cosh⁡(x)+32=∑n=1∞‍22n-1-2(2n)!x2n=x4∑n=0∞‍22n+3-2(2n+4)!x2n.
Let
(15)an=3·22n+3-4(n+2)(2n+4)!,bn=22n+3-2(2n+4)!.
Then
(16)bn>0,an+1bn+1-anbn=(6n+1)22n+2+2(2n+4-1)(22n+2-1)>0
for all n≥0. Note that
(17)f1(0+)=a0b0=83,lim⁡x→∞f1(x)=lim⁡n→∞anbn=3.
Therefore, Lemma3 follows easily from Lemma 2 and (14)–(17).Lemma 4.
The function(18)f2(x)=1-(1/4)(cos⁡(x)+x/sin(x))2sin2(x)
is strictly increasing from (0,π/2) onto (1/3,1-π2/16).Proof.
It is not difficult to verify that(19)f2(0+)=13,f2(π2)=1-π216.
Let g(x)=1-[cos⁡(x)+x/sin(x)]2/4 and h(x)=sin2(x). Then,
(20)f2(x)=g(x)h(x),g(0+)=h(0)=0,g′(x)h′(x)=x2-sin2(x)cos⁡2(x)4sin4(x)=(2x)2-sin2(2x)4[1-cos⁡(2x)]2.
From Lemma 1 and (19)-(20), we know that we just need to prove that the function
(21)F(y)=y2-sin2(y)[1-cos⁡(y)]2
is strictly increasing on (0,π).
LetF1(y)=y2-sin2(y), F2(y)=[1-cos⁡(y)]2, F3(y)=y-sin(y)cos⁡(y), and F4(y)=sin(y)(1-cos⁡(y)). Then
(22)F(y)=F1(y)F2(y),F1(0)=F2(0)=F3(0)=F4(0)=0,(23)F1′(y)F2′(y)=F3(y)F4(y).
Note that
(24)F3′(y)F4′(y)=2(1+cos⁡(y))1+2cos⁡(y)=1+11+2cos⁡(y)
is strictly increasing on (0,π). Therefore, the monotonicity of F(y) follows easily from (22) and (23) together with the monotonicity of F3′(y)/F4′(y).Lemma 5.
The function(25)f3(x)=1-(1/4)(cosh⁡(x)+x/sinh⁡(x))2sinh⁡2(x)
is strictly increasing from (0,log⁡(1+2)) onto (-1/3,-(22log⁡(1+2)+log⁡2(1+2)-2)/4).Proof.
Simple computations lead to(26)f3(0+)=-13,f3(log⁡(1+2))=-22log⁡(1+2)+log⁡2(1+2)-24.
Let g1(x)=1-[cosh⁡(x)+x/sinh⁡(x)]2/4 and g2(x)=sinh⁡2(x). Then
(27)f3(x)=g1(x)g2(x),g1(0+)=g2(0)=0,(28)g1′(x)g2′(x)=4x2-sinh⁡2(2x)4[cosh⁡(2x)-1]2=-∑n=0∞(42n+4/(2n+4)!)x2n∑n=0∞((42n+5-4n+4)/(2n+4)!)x2n.
Let
(29)an=42n+4(2n+4)!,bn=42n+5-4n+4(2n+4)!.
Then
(30)bn>0,(31)an+1bn+1-anbn=-3·4n(4n+1-1)(4n+2-1)<0
for all n≥0.
It, from Lemma2 and (28)–(31), that g1′(x)/g2′(x) is strictly increasing on (0,∞). Therefore, Lemma 5 follows easily from (26) and (27) together with Lemma 1 and the monotonicity of g1′(x)/g2′(x).Lemma 6.
The function(32)f4(x)=3sin2(x)-2xsin(x)-x2(1-cos⁡2(x))
is strictly increasing from (0,π/2) on (-8/3,-(π2+4π-12)/4).Proof.
Differentiatingf4(x) gives
(33)f4′(x)=2sin(x)(1-cos⁡(x))3[x2+xsin(x)+4cos⁡(x)-4].
Let
(34)H(x)=x2+xsin(x)+4cos⁡(x)-4.
Then, simple computations lead to
(35)H(0)=0,H′(x)=2x+xcos⁡(x)-3sin(x),(36)H′(0)=0,H′′(x)=2-xsin(x)-2cos⁡(x),(37)H′′(0)=0,H′′′(x)=sin(x)-xcos⁡(x),(38)H′′′(0)=0,(39)H(4)(x)=xsin(x)>0
for x∈(0,π/2).
Therefore, Lemma6 follows easily from (33)–(39) together with the fact that f4(0+)=-8/3 and f4(π/2)=-(π2+4π-12)/4.


## 3. Main Results

Theorem 7.
Letα1,β1∈[0,1/2]. Then the double inequality
(40)G(α1a+(1-α1)b,α1b+(1-α1)a)<NAG(a,b)<G(β1a+(1-β1)b,β1b+(1-β1)a)
holds for all a,b>0 with a≠b if and only if α1≤1/2-3/4=0.06698… and β1≥1/2-6/6=0.09175….Proof.
Since both the geometric meanG(a,b) and arithmetic mean A(a,b) are symmetric and homogeneous of degree 1, without loss of generality, we assume that a>b. Let v=(a-b)/(a+b)∈(0,1) and λ∈[0,1/2]. Then, from (4), one has
(41)G(λa+(1-λ)b,λb+(1-λ)a)-NAG(a,b)=A(a,b)1-v2(1-2λ)2-12A(a,b)[1+1-v2vtanh-1(v)]=(A(a,b)[1v2-14(1v+(1v2-1)hhhhhhhhhhhhhhhhhhh×tanh-1(v)1v)2-(1-2λ)21v]v)×(1v2-(1-2λ)2hhhhhhh+12[1v+(1v2-1)tanh-1(v)]1/v2-(1-2λ)21v2-(1-2λ)2)-1.
Let t=tanh-1(v). Then t∈(0,∞), and
(42)1v2-14(1v+(1v2-1)tanh-1(v))2-(1-2λ)2=3sinh⁡2(t)cosh⁡2(t)-2tsinh⁡(t)cosh⁡(t)-t24sinh⁡4(t)-(1-2λ)2=3sinh⁡2(2t)-4tsinh⁡(2t)-4t24(cosh⁡(2t)-1)2-(1-2λ)2=14f1(2t)-(1-2λ)2,
where f1(t) is defined as in Lemma 3.
Therefore, Theorem7 follows easily from (41) and (42) together with Lemma 3.Theorem 8.
Letα2,β2∈[0,1/2]. Then the double inequality
(43)G(α2a+(1-α2)b,α2b+(1-α2)a)<NGA(a,b)<G(β2a+(1-β2)b,β2b+(1-β2)a)
holds for all a,b>0 with a≠b if and only if α2≤1/2-16-π2/8=0.1905… and β2≥1/2-3/6=0.2113….Proof.
We follow the same idea in the proof of Theorem7. Without loss of generality, we assume that a>b. Let v=(a-b)/(a+b)∈(0,1) and μ∈[0,1/2]. Then, from (5), we get
(44)G(μa+(1-μ)b,μb+(1-μ)a)-NGA(a,b)=A(a,b)1-v2(1-2μ)2-12A(a,b)[1-v2+sin-1(v)v]=(A(a,b)[1v2-14(1v2-1+sin-1(v)v2)2hhhhhhhhhhhhh-(1-2μ)21v2-14(1v2-1+sin-1(v)v2)2]v)×(1v2-(1-2μ)2+12[1v2-1+sin-1(v)v2])-1.
Let t=sin-1(v). Then, t∈(0,π/2) and simple computation leads to
(45)1v2-14(1v2-1+sin-1(v)v2)2-(1-2μ)2=1-(1/4)(cos⁡(t)+t/sin(t))2sin2(t)-(1-2μ)2=f2(t)-(1-2μ)2,
where f2(t) is defined as in Lemma 4.
Therefore, Theorem8 follows easily from (44) and (45) together with Lemma 4.Theorem 9.
Letα3,β3∈[1/2,1]. Then the double inequality
(46)Q(α3a+(1-α3)b,α3b+(1-α3)a)<NQA(a,b)<Q(β3a+(1-β3)b,β3b+(1-β3)a)
holds for all a,b>0 with a≠b if and only if β3≥1/2+3/6=0.7886… and α3≤1/2+22log⁡(1+2)+log⁡2(1+2)-2/4=0.7817….Proof.
Since both the quadratic meanQ(a,b) and arithmetic mean A(a,b) are symmetric and homogeneous of degree 1, without loss of generality, we assume that a>b. Let v=(a-b)/(a+b)∈(0,1) and p∈[1/2,1]. Then, (6) gives
(47)Q(pa+(1-p)b,pb+(1-p)a)-NQA(a,b)=A(a,b)1+v2(2p-1)2-12A(a,b)[1+v2+sinh⁡-1(v)v]=(A(a,b)[1v2-14(1v2+1hhhhhhhhhhhhhhhhhhhh+sinh⁡-1(v)v2)2+(2p-1)214(1v2+1]v14(1v2+1)×(1v2+(2p-1)2+12[1v2+1+sinh⁡-1(v)v2])-1.
Let t=sinh⁡-1(v). Then, t∈(0,log⁡(1+2)) and elementary computations lead to
(48)1v2-14(1v2+1+sinh⁡-1(v)v2)2+(2p-1)2=1-(1/4)[cosh⁡(t)+t/sinh⁡(t)]2sinh⁡2(t)+(2p-1)2=f3(t)+(2p-1)2,
where f3(t) is defined as in Lemma 5.
Therefore, Theorem9 follows easily from (47) and (48) together with Lemma 5.Theorem 10.
Letα4,β4∈[1/2,1]. Then the double inequality
(49)Q(α4a+(1-α4)b,α4b+(1-α4)a)<NAQ(a,b)<Q(β4a+(1-β4)b,β4b+(1-β4)a)
holds for all a,b>0 with a≠b if and only if α4≤1/2+π2+4π-12/8=0.9038… and β4≥1/2+6/6=0.9082….Proof.
We follow the same idea in the proof of Theorem9. Let a>b, v=(a-b)/(a+b)∈(0,1) and q∈[1/2,1]. Then, from (7), we have
(50)Q(qa+(1-q)b,qb+(1-q)a)-NAQ(a,b)=A(a,b)1+v2(2q-1)2-12A(a,b)[1+1+v2vtan-1(v)]=(A(a,b)[1v2-14(1v+(1+1v2)tan-1(v))2hhhhhhhhhhhhh+(2q-1)21v2-14(1v+(1+1v2)tan-1(v))2]v)×(1v2+(2q-1)2+12[1v+(1+1v2)tan-1(v)])-1.
Let t=tan-1(v). Then t∈(0,π/4) and
(51)1v2-14(1v+(1+1v2)tan-1(v))2+(2q-1)2=3sin2(t)cos⁡2(t)-2tsin(t)cos⁡(t)-t24sin4(t)+(2q-1)2=3sin2(2t)-4tsin(2t)-4t24[1-cos⁡(2t)]2+(2q-1)2=14f4(2t)+(2q-1)2,
where f4(t) is defined as in Lemma 6.
Therefore, Theorem10 follows easily from (50) and (51) together with Lemma 6.



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