JAM Journal of Applied Mathematics 1687-0042 1110-757X Hindawi Publishing Corporation 807623 10.1155/2013/807623 807623 Research Article Optimal Bounds for Neuman Means in Terms of Harmonic and Contraharmonic Means He Zai-Yin 1 Chu Yu-Ming 2 Wang Miao-Kun 2 Lin Chong 1 Department of Mathematics Huzhou Teachers College Huzhou 313000 China hutc.zj.cn 2 School of Mathematics and Computation Science Hunan City University Yiyang 413000 China hncu.net 2013 28 11 2013 2013 07 08 2013 11 11 2013 2013 Copyright © 2013 Zai-Yin He et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

For a,b>0 with ab, the Schwab-Borchardt mean SB(a,b) is defined as SB(a,b)={b2-a2/cos-1(a/b)  if  a<b,a2-b2/cosh-1(a/b)  if  a>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.

1. Introduction

For a,b>0 with ab, 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 mean SB(a,b) was firstly investigated in . In , 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 .

Let H(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 ab.

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 means SAH, 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 ab.

For fixed a,b>0 with ab, 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.

Let α1,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 ab 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 ab if and only if α2[1-1-2/π]/2 and β2[3-3]/6.

Theorem 2.

Let α3,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 ab 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 ab 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 [<xref ref-type="bibr" rid="B1">16</xref>, Theorem 1.25]).

For -<a<b<, let f,g:[a,b] 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 function φ(x)=(xcosh(x)-sinh(x))/[x(cosh(x)-1)] is strictly increasing from (0,) onto (2/3,1).

(2) The function ϕ(x)=(x-sin(x))/[x(1-cos(x))] is strictly increasing from (0,π/2) onto (1/3,(π-2)/π).

(3) The function ξ(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 function η(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), let φ1(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 xtanh(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 let ϕ1(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 xtan(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 Theorems <xref ref-type="statement" rid="thm1.1">1</xref> and <xref ref-type="statement" rid="thm1.2">2</xref> Proof of Theorem <xref ref-type="statement" rid="thm1.1">1</xref>.

Without loss of generality, we assume that a>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 Theorem <xref ref-type="statement" rid="thm1.2">2</xref>.

Without loss of generality, we assume that a>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).

Acknowledgments

This research was supported by the Natural Science Foundation of China under Grants 61374086 and 11171307, the Natural Science Foundation of Zhejiang Province under Grant LY13A010004, and the Natural Science Foundation of Huzhou Teachers College under Grant KX21063.

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