# Optimal Bounds for Neuman Means in Terms of Harmonic and Contraharmonic Means

Authors: Zai-Yin He; Yu-Ming Chu; Miao-Kun Wang
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

Fora,b>0 with a≠b, the Schwab-Borchardt mean SB(a,b) is defined as SB(a,b)={b2-a2/cos-1(a/b)  ifa<b,a2-b2/cosh-1(a/b)  ifa>b. In this paper, we find the greatest values of
α1 and α2 and the least values of
β1 and β2 in [0,1/2] such that H(α1a+(1-α1)b,α1b+(1-α1)a)<SAH(a,b)<H(β1a+(1-β1)b,β1b+(1-β1)a) and H(α2a+(1-α2)b,α2b+(1-α2)a)<SHA(a,b)<H(β2a+(1-β2)b,β2b+(1-β2)a). Similarly, we also find the greatest values of
α3 and α4 and the least values of
β3 and β4 in [1/2,1] such that C(α3a+(1-α3)b,α3b+(1-α3)a)<SCA(a,b)<C(β3a+(1-β3)b,β3b+(1-β3)a) and C(α4a+(1-α4)b,α4b+(1-α4)a)<SAC(a,b)<C(β4a+(1-β4)b,β4b+(1-β4)a). Here, H(a,b)=2ab/(a+b), A(a,b)=(a+b)/2, and C(a,b)=(a2+b2)/(a+b) are the harmonic, arithmetic, and contraharmonic means, respectively, and SHA(a,b)=SB(H,A), SAH(a,b)=SB(A,H), SCA(a,b)=SB(C,A), and SAC(a,b)=SB(A,C) are four Neuman means derived from the Schwab-Borchardt mean.


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## Body

## 1. Introduction

Fora,b>0 with a≠b, the Schwab-Borchardt mean SB(a,b) is defined as
(1)SB(a,b)={b2-a2cos-1(a/b)ifa<b,a2-b2cosh-1(a/b)ifa>b.
It is well known that the mean SB(a,b) is strictly increasing in both a and b, nonsymmetric, and homogeneous of degree 1 in its variables. Several symmetric bivariate means are special cases of the Schwab-Borchardt mean; for example,
(2)P(a,b)=a-b2sin-1[(a-b)/(a+b)]=SB(G,A)isthefirstSeiffertmean,T(a,b)=a-b2tan-1[(a-b)/(a+b)]=SB(A,Q)isthesecondSeiffertmean,M(a,b)=a-b2sinh-1[(a-b)/(a+b)]=SB(Q,A)istheNeuman-Sa´ndormean,L(a,b)=a-b2tanh-1[(a-b)/(a+b)]=SB(A,G)isthelogarithmicmean,
where G(a,b)=ab, A(a,b)=(a+b)/2 and Q(a,b)=(a2+b2)/2 denote the classical geometric mean, arithmetic mean, and quadratic mean, respectively.The Schwab-Borchardt meanSB(a,b) was firstly investigated in [1–4]. In [3], the authors pointed out that the logarithmic mean, two Seiffert means, and the Neuman-Sándor mean are particular cases of the Schwab-Borchardt mean. Later, SB and its special cases have been the subject of intensive research. In particular, many inequalities for them can be found in the literature [3–13].LetH(a,b)=2ab/(a+b), C(a,b)=(a2+b2)/(a+b) be the harmonic and contraharmonic means of two positive numbers a and b, respectively. Then, it is well known that
(3)H<G<L<P<A<M<T<Q<C.
for a,b>0 with a≠b.Recently, the second author of this paper reviewed two elegant papers [14, 15] by Neuman and found that the bivariate means SAH, SHA, SCA, and SAC, derived from the Schwab-Borchardt mean are very interesting. They are defined as follows:
(4)SAH=SB(A,H),SHA=SB(H,A),SCA=SB(C,A),SAC=SB(A,C).We call the meansSAH, SHA, SCA, and SAC, defined in (4) the Neuman means. Moreover, if we let v=(a-b)/(a+b)∈(-1,1), then explicit formulas for SAH, SHA, SAC, and SCA are in the following:
(5)SAH=Atanh(p)p,SHA=Asin(q)q,(6)SCA=Asinh(r)r,SAC=Atan(s)s,
where p, q, r, and s are defined implicitly as sech(p)=1-v2, cos(q)=1-v2, cosh(r)=1+v2 and sec(s)=1+v2, respectively. Clearly, p∈(0,∞), q∈(0,π/2), r∈(0,log(2+3)), and s∈(0,π/3).Neuman [14, 15] presented several optimal bounds for SHA, SAH, SCA, and SAC. The bounding quantities are arithmetic convex, geometric convex, and harmonic convex combinations of their generating means. Besides, he also proved that
(7)H<SAH<L<SHA<P,T<SCA<Q<SAC<C,
for a,b>0 with a≠b.For fixeda,b>0 with a≠b, x∈[0,1/2] and y∈[1/2,1]. Let
(8)f(x)=H(xa+(1-x)b,xb+(1-x)a),g(y)=C(ya+(1-y)b,yb+(1-y)a).
Then, it is not difficult to verify that f(x) and g(y) are continuous and strictly increasing on [0,1/2] and [1/2,1], respectively. Note that f(0)=H<SAH<SHA<A=f(1/2), g(1/2)=A<SCA<SAC<C=g(1). Therefore, it is natural to ask what are the greatest values of α1 and α2 and the least values of β1 and β2 in [0,1/2] such that H(α1a+(1-α1)b,α1b+(1-α1)a)<SAH(a,b)<H(β1a+(1-β1)b,β1b+(1-β1)a) and H(α2a+(1-α2)b,α2b+(1-α2)a)<SHA(a,b)<H(β2a+(1-β2)b,β2b+(1-β2)a)? And what are the greatest values of α3 and α4 and the least values of β3 and β4 in [1/2,1] such that C(α3a+(1-α3)b,α3b+(1-α3)a)<SCA(a,b)<C(β3a+(1-β3)b,β3b+(1-β3)a) and C(α4a+(1-α4)b,α4b+(1-α4)a)<SAC(a,b)<C(β4a+(1-β4)b,β4b+(1-β4)a)? The main purpose of this paper is to answer these questions. Our main results are in Theorems 1 and 2.Theorem 1.
Letα1,α2,β1,β2∈[0,1/2]. Then, the double inequality
(9)H(α1a+(1-α1)b,α1b+(1-α1)a)<SAH<H(β1a+(1-β1)b,β1b+(1-β1)a)
holds for all a,b>0 with a≠b if and only if α1=0 and β1≥[3-6]/6. Also the double inequality
(10)H(α2a+(1-α2)b,α2b+(1-α2)a)<SHA<H(β2a+(1-β2)b,β2b+(1-β2)a)
holds for all a,b>0 with a≠b if and only if α2≤[1-1-2/π]/2 and β2≥[3-3]/6.Theorem 2.
Letα3,α4,β3,β4∈[1/2,1]. Then, the double inequality
(11)C(α3a+(1-α3)b,α3b+(1-α3)a)<SCA<C(β3a+(1-β3)b,β3b+(1-β3)a)
holds for all a,b>0 with a≠b if and only if α3≤[1+3/log(2+3)-1]/2 and β3≥(3+3)/6. Also the double inequality
(12)C(α4a+(1-α4)b,α4b+(1-α4)a)<SAC<C(β4a+(1-β4)b,β4b+(1-β4)a)
holds for all a,b>0 with a≠b if and only if α4≤[1+33/π-1]/2 and β4≥(3+6)/6.


## 2. Two Lemmas

In order to prove the desired theorems, we first give two lemmas.Lemma 1 (see [16, Theorem 1.25]).
For-∞<a<b<∞, let f,g:[a,b]→ℝ be continuous on [a,b], and be 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
(13)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.
(1) The functionφ(x)=(xcosh(x)-sinh(x))/[x(cosh(x)-1)] is strictly increasing from (0,∞) onto (2/3,1).
(2) The functionϕ(x)=(x-sin(x))/[x(1-cos(x))] is strictly increasing from (0,π/2) onto (1/3,(π-2)/π).
(3) The functionξ(x)=(sinh(x)-x)/[x(cosh(x)-1)] is strictly decreasing from (0,log(2+3)) onto ([3-log(2+3)]/log(2+3),1/3).
(4) The functionη(x)=(sin(x)-xcos(x))/[x(1-cos(x))] is strictly decreasing from (0,π/3) onto ((33-π)/π,2/3).Proof.
From part (1), letφ1(x)=xcosh(x)-sinh(x) and φ2(x)=x(cosh(x)-1). Then, φ(x)=φ1(x)/φ2(x), φ1(0)=φ2(0)=0, and
(14)φ1′(x)φ2′(x)=xsinh(x)cosh(x)-1+xsinh(x)=11+(cosh(x)-1)/(xsinh(x))=11+(1/2)tanh(x/2)/(x/2).
It is well known that x→tanh(x)/x is strictly decreasing on (0,∞). Then, Lemma 1 and (14) lead to the conclusion that φ(x) is strictly increasing on (0,∞). Moreover, by l’Hôptial’s rule we have φ(0+)=2/3 and limx→+∞φ(x)=1.
From part (2), similarly letϕ1(x)=x-sin(x) and ϕ2(x)=x(1-cos(x)). Then ϕ(x)=ϕ1(x)/ϕ2(x), ϕ1(0)=ϕ2(0)=0 and
(15)ϕ1′(x)ϕ2′(x)=1-cos(x)1-cos(x)+xsin(x)=11+xsin(x)/(1-cos(x))=11+2(x/2)/tan(x/2).
It is well known that x→tan(x)/x is strictly increasing on (0,π/2). Then, by Lemma 1 and (15) we know that ϕ(x) is strictly increasing on (0,π/2). Clearly, ϕ(π/2)=(π-2)/π, while by l’Hôptial’s rule we have ϕ(0+)=1/3.
Parts (3) and (4) have been proven in [14, Theorem 3].


## 3. Proofs of Theorems1 and 2

Proof of Theorem1.
Without loss of generality, we assume thata>b. Let v=(a-b)/(a+b)∈(0,1) and λ∈[0,1/2]; then,
(16)H(λa+(1-λ)b,λb+(1-λ)a)-SAH=A[1-(1-2λ)2v2]-Atanh(p)p=A[1-(1-2λ)2(1-sech(p))-tanh(p)p]=A(1-sech(p))[pcosh(p)-sinh(p)p(cosh(p)-1)-(1-2λ)2]
provided that sech(p)=1-v2(p>0). Thus, inequality (9) follows from (16) and Lemma 2(1). Similarly,
(17)H(λa+(1-λ)b,λb+(1-λ)a)-SHA=A[1-(1-2λ)2v2]-Asin(q)q=A[1-(1-2λ)2(1-cos(q))-sin(q)q]=A(1-cos(q))[q-sin(q)q(1-cos(q))-(1-2λ)2]
provided that cos(q)=1-v2(q∈(0,π/2)). Thus, inequality (10) follows from (17) and Lemma 2(2).Proof of Theorem2.
Without loss of generality, we assume thata>b. Let v=(a-b)/(a+b)∈(0,1) and μ∈[1/2,1], then
(18)C(μa+(1-μ)b,μb+(1-μ)a)-SCA=A[1+(1-2μ)2v2]-Asinh(r)r=A[1+(1-2μ)2(cosh(r)-1)-sinh(r)r]=A(cosh(r)-1)[(1-2μ)2-sinh(r)-rr(cosh(r)-1)]
provided that cosh(r)=1+v2(r∈(0,cosh-1(2))). Thus, inequality (11) follows from (18) and Lemma 2(3). Similarly,
(19)C(μa+(1-μ)b,μb+(1-μ)a)-SAC=A[1+(1-2μ)2v2]-Atan(s)s=A[1+(1-2μ)2(sec(s)-1)-tan(s)s]=A(sec(s)-1)[(1-2μ)2-sin(s)-scos(s)s(1-cos(s))]
provided that sec(s)=1+v2(s∈(0,π/3)). Thus, inequality (12) follows from (19) and Lemma 2(4).



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