# Optimal Bounds for the Neuman-Sándor Mean in terms of the Convex Combination of the First and Second Seiffert Means

Authors: Hao-Chuan Cui; Nan Wang; Bo-Yong Long
Journal: Mathematical Problems in Engineering
 (2015)

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## Abstract

We find the least valueα and the greatest value β such that the double inequality αP(a,b)+(1-α)T(a,b)<M(a,b)<βP(a,b)+(1-β)T(a,b) holds for all a,b>0 with a≠b, where M(a,b),P(a,b), and T(a,b) are the Neuman-Sándor mean and the first and second Seiffert means of two positive numbers 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], the first Seiffert mean P(a,b) [2], and the second Seiffert mean T(a,b) [3] are defined by(1)Ma,b=a-b2sinh-1⁡a-b/a+b,(2)Pa,b=a-b4tan-1⁡a/b-π,(3)Ta,b=a-b2tan-1⁡a-b/a+b,respectively. It can be observed that the first Seiffert mean P(a,b) can be rewritten as (see [1])(4)Pa,b=a-b2sin-1⁡a-b/a+b,where sinh-1⁡(x)=log⁡(x+x2+1), tan-1⁡(x)=arctan⁡(x), and sin-1⁡(x)=arcsin⁡(x) are the inverse hyperbolic sine, inverse tangent, and inverse sine functions, respectively.Recently, the meansM, P, and T and other means have been the subject of intensive research. Many remarkable inequalities for means can be found in the literature [4–14].LetH(a,b)=2ab/(a+b), G(a,b)=ab, L(a,b)=(b-a)/(log⁡b-log⁡a), I(a,b)=1/e(bb/aa)1/(b-a), A(a,b)=(a+b)/2, S(a,b)=(a2+b2)/2, and (5)Mpa,b=ap+bp21/p,p≠0,ab,p=0denote the harmonic, geometric, logarithmic, identric, arithmetic, root-square, and pth power means of two positive numbers a and b with a≠b, respectively. Then it is well known that the inequalities (6)Ha,b<Ga,b<La,b<Pa,b<Ia,b<Aa,b<Ma,b<Ta,b<Sa,bhold for a,b>0 with a≠b.Neuman and Sándor [1] established (7)π2Pa,b>sinh-1⁡1Ma,b>π4Ta,bfor all a,b>0 with a≠b and the following Ky Fan inequalities: (8)Ga,bGa′,b′<La,bLa′,b′<Pa,bPa′,b′<Aa,bAa′,b′<Ma,bMa′,b′<Ta,bTa′,b′for all 0<a,b≤1/2 with a≠b, a′=1-a and b′=1-b.Gao [15] proved the optimal double inequalities (9)eπIa,b<Pa,b<Ia,b,Ia,b<Ta,b<2eπIa,bhold for all a,b>0 with a≠b.The following bounds for the Seiffert meansP(a,b) and T(a,b) in terms of power mean were presented by Jagers in [16]: (10)M1/2<Pa,b<M2/3a,bfor all a,b>0 with a≠b. Hästö [17] improved the results of [16] and found the sharp lower power mean bound for the Seiffert mean P(a,b) as follows: (11)Pa,b>Mlog⁡2/log⁡πa,bfor all a,b>0 with a≠b.In [18], the authors proved the sharp double inequality (12)Mlog⁡2/log⁡π-log⁡2<Ta,b<M5/3a,bholds for all a,b>0 with a≠b.LetL¯p(a,b)=(ap+1+bp+1)/(ap+bp) denote the Lehmer mean of two positive numbers a and b with a≠b. In [19] the authors presented the following best possible Lehmer mean bounds for the Seiffert means P(a,b) and T(a,b): (13)L¯-1/6a,b<Pa,b<L¯0a,b,L¯0a,b<Ta,b<L¯1/3a,bfor all a,b>0 with a≠b.Letu, v, and w be bivariate means such that u(a,b)<w(a,b)<v(a,b) for all a,b>0 with a≠b. The problems of finding 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 u(a,b)αv1-α(a,b)<w(a,b)<u(a,b)βv1-β(a,b) hold for all a,b>0 with a≠b have attracted the interest of many mathematicians.In [20, 21] the authors proved that the double inequalities (14)Sa,bα3A1-α3a,b<Ma,b<Sa,bβ3A1-β3a,b,α4Sa,b+1-α4Ga,b<Ma,b<β4Sa,b+1-β4Ga,bhold for all a,b>0 with a≠b if and only if α3≤1/3, β3≥2(log⁡2+2-log⁡3)/log⁡2, α4≤2/3, and β4≥1/2log⁡1+2.In [22], the authors proved that the double inequality (15)α5Aa,b+1-α5Ga,b<Pa,b<β5Aa,b+1-β5Ga,bholds for all a,b>0 with a≠b if and only if α5≤π/2, β5≥2/3.The main purpose of this paper is to find the least valueα and the greatest value β such that the double inequality (16)αPa,b+1-αTa,b<Ma,b<βPa,b+1-βTa,bholds for all a,b>0 with a≠b.


## 2. Lemmas

To establish our main result, we need several lemmas which we present in this section.Lemmas1 and 2 were proved in [23].Lemma 1.
LetH(x)=x/x2+1sinh-1⁡(x)2-1/sinh-1⁡(x). Then H(x) is strictly decreasing on (0,1). Moreover, the inequality(17)Hx>-x3+x39holds for any x∈(0,0.7) and the inequality(18)Hx<-x3+17x390holds for any x∈(0,1).Lemma 2.
LetS(x)=1/sin-1⁡(x)-x/1-x2(sin-1⁡(x))2. Then it holds that(19)Sx>-x3-x33for any x∈(0,0.7) and(20)Sx<-x3-17x390-x57holds for any x∈(0,1).Forx∈(0,1), the power series expansion of the function tan-1(x) is presented as follows:(21)tan-1⁡x=x-x33+x55+⋯=∑n=0∞-1n2n+1x2n+1.Therefore, we have the following.Lemma 3.
Ifx∈(0,1), then it holds that(22)x-x33<tan-1⁡x<x-x33+x55<x,(23)tan-1⁡x<x-x35.Proof.
The inequality of (22) follows from equality (21) directly. The inequality of (23) follows from (24)tan-1⁡x-x+x35x=0=0,tan-1⁡x-x+x55′=-x211+x2+3x25<0for x∈(0,1).Lemma 4.
LetT(x)=1/tan-1⁡(x)-x/x2+1tan-1⁡(x)2. Then for x∈(0,1) it holds that(25)2x3-2x35+x57<Tx<2x3-2x39+2x57.Proof.
To prove inequality (25), it suffices to show(26)t1x>0,(27)t2x<0for x∈(0,1), where(28)t1x=1+x2tan-1⁡x2Tx-2x3+2x35-x57,t2x=1+x2tan-1⁡x2Tx-2x3+2x39-2x57.
Firstly, we prove inequality (26). From the expression of t1(x), we have(29)t10=0,(30)t1′x=tan-1⁡xt1∗x,where(31)t1∗x=2x3+4x35-2x57-tan-1x23+4x25-9x47+x6.Because of (32)23+4x25-9x47+x6>2x43+4x45-9x47+x6=19x4105+x6>0for x(0,1), it follows from (22) and (31) that(33)t1∗x>2x3+4x35-2x57-x23+4x25-9x47+x6=x5-x7>0for any x∈(0,1). Therefore, inequality (26) follows from (29) and (30) together with (33).
Secondly, we prove inequality (27). From the expression of t2(x), we have(34)t20=0,t2′=tan-1⁡xt2∗x,where(35)t2∗x=2x3+4x39-4x57-tan-1⁡x23+4x23+20x463+2x6.Because of(36)23+4x23+20x463+2x6>0for any x∈(0,1), it follows from (22) and (35) together with (36) that(37)t2∗x<2x3+4x39-4x57-x-x3323+4x23+20x463+2x6=-2x33-4x59-358x7189+2x93<-2x33-4x59-358x7189+2x73<-2x33-4x59<0for any x∈(0,1). Therefore, inequality (27) follows from (34) and (37).Lemma 5.
The functionf(x)=λS(x)+(1-λ)T(x)+H(x) is strictly decreasing on (0.7,1), where λ=1/2π4/π-1/sinh-1⁡1=0.2177… and H(x), S(x), and T(x) are defined as in Lemmas 1, 2, and 4, respectively.Proof.
Direct computations lead to(38)T′x=2x-tan-1⁡x1+x22tan-1⁡x2,(39)T′′x=2ψx1+x23tan-14,where (40)ψx=-3x+31-x2tan-1+4xtan-1⁡x2.Then inequality (22) leads to(41)ψx<-3x+31-x2x-x33+x55+4xx-x33+x552=425x3x4-25x24+52for any x∈(0.7,1). Obviously, it holds that(42)x4-25x24+52<0.74-2540.72+52=-0.3224<0for any x∈(0.7,1). Therefore, for any x∈(0.7,1), T′′(x)<0 follows from (39) and (41) together with (42). In other words, T′(x) is strictly decreasing on (0.7,1).
Letϕ(x)=λS(x)+(1-λ)T(x). It was proved that S′(x) is strictly decreasing on (0.7,1) in [23] Lemma 2.4. Thus, from the monotonicity of S′(x) and T′(x), we have (43)ϕ′λ,x<λS′0.7+1-λT′0.7=-0.0507⋯<0for any x∈(0.7,1). That is to say, ϕ(x) is strictly decreasing on (0.7,1). Considering the monotonicity of H(x) in Lemma 1, the proof is completed.Lemma 6.
Letξ(x)=x((1-λ)/7)x4-(13/45-λ/15)x2+1/3-λ, where λ=1/2π(4/π-1/sinh-1⁡(1))=0.2177…. Then ξ(x)>0 for x∈(0,0.7).Proof.
Letζ(x)=((1-λ)/7)x4-(13/45-λ/15)x2+(1/3-λ). Then it is easy to verify that ζ(x) is decreasing on (0,μ), where (44)μ=1373π+14sinh-1⁡110π-2sinh-1⁡1=1.1079….Considering ζ(0.7)=0.0079⋯>0, we have ζ(x)>0 for x∈(0,0,7). Therefore, ξ(x)>0 for x∈(0,0,7).


## 3. Main Results

Theorem 7.
The double inequality(45)αPa,b+1-αTa,b<Ma,b<βPa,b+1-βTa,bholds for any a,b> 0 with a≠b if and only if α≥1/3 and β≤1/2π4/π-1/sinh-1⁡(1)=0.2177….Proof.
BecauseP(a,b), M(a,b), and T(a,b) are symmetric and homogeneous of degree 1, without loss of generality, we assume that a>b. Let p∈(0,1), x=(a-b)/(a+b)∈(0,1), and λ=1/2π4/π-1/sinh-1⁡(1). Then by (1), (3), and (4), direct computations lead to(46)Pa,bAa,b=xsin-1x,Ma,bAa,b=xsinh-1x,Ta,bAa,b=xtan-1x.
Let(47)Dpx≔pPa,b+1-pTa,b-Ma,bAa,b=pxsin-1⁡x+1-pxtan-1⁡x-xsinh-1⁡x.Then it follows that(48)Dp0+=0,Dp1-=4π-2pπ-1sinh-1⁡1,Dλ1-=0.Differentiating Dp(x), we have(49)Dp′x=p1sin-1⁡x-x1-x2sin-1⁡x2+1-p1tan-1⁡x-xx2+1tan-1⁡x2-xx2+1sinh-1⁡x2-1sinh-1⁡x=pSx+1-pTx+Hx,where H(x), S(x), and T(x) are defined as in Lemmas 1, 2, and 4, respectively.
On one hand, from inequalities in Lemmas1, 2, and 4, we clearly see that(50)D1/3′x=13Sx+23Tx+Hx<13-x3-17x390-x57+232x3-2x39+2x57-x3+17x390=-29x-424x3+127x5<-29x<0for x∈(0,1). It leads to(51)D1/3x<D1/30=0for any x∈(0,1). According to (47) and (51), we conclude that(52)Ma,b>13Pa,b+23Ta,bfor all a,b>0 with a≠b.
On the other hand, from inequalities in Lemmas1, 2, and 4, we have(53)Dλ′x>-λx3+x33+1-λ2x3-2x35+x57-x3+x39=x1-λ7x4-1345-λ15x2+13-λfor (0,0.7). According to Lemma 6, we have(54)Dλ′x>ξx>0for (0,0.7). Lemma 5 shows that Dλ′(x) is strictly decreasing on (0.7,1). This fact and (54) together with Dλ′(1-)=-∞ imply that there exists x0∈(0.7,1) such that Dλ(x) is strictly increasing on (0,x0] and strictly decreasing on [x0,1). Equations (47) and (48) together with the piecewise monotonicity of Dλ(x) lead to the conclusion that(55)Ma,b<λPa,b+1-λTa,bfor all a,b>0 with a≠b.
At last, by easy computations, (1), (3), and (4) lead to(56)Ta,b-Ma,bTa,b-Pa,b=x/tan-1⁡x-x/sinh-1⁡xx/tan-1⁡x-x/sin-1⁡x,(57)limx→0+⁡x/tan-1⁡x-x/sinh-1⁡xx/tan-1⁡x-x/sin-1⁡x=13,(58)limx→1-⁡x/tan-1⁡x-x/sinh-1⁡xx/tan-1⁡x-x/sin-1⁡x=λ.Thus, we have the following claims.
Claim 1. If α<1/3, then (56) and (57) imply that there exists σ1∈(0,1) such that M(a,b)<αP(a,b)+(1-α)T(a,b) for all a,b with (a-b)/(a+b)∈(0,σ1).
Claim 2. If β>λ, then (56) and (58) imply that there exists σ2∈(0,1) such that M(a,b)>βP(a,b)+(1-β)T(a,b) for all a,b with (a-b)/(a+b)∈(1-σ2,1).
Inequalities (52) and (55) in conjunction with the above two claims mean the proof is completed.



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