1. Introduction
For p∈ℝ the pth power mean Mp(a,b), Neuman-Sándor Mean M(a,b) [1], and identric mean I(a,b) of two positive numbers a and b are defined by
(1)Mp(a,b)={(ap+bp2)1/p,p≠0,ab,p=0,(2)M(a,b)={a-b2sinh-1((a-b)/(a+b)),a≠b,a,a=b,(3)I(a,b)={1e(bbaa)1/(b-a),a≠b,a,a=b,
respectively, where sinh-1(x)=log(x+1+x2) is the inverse hyperbolic sine function.

The main properties for Mp(a,b) and I(a,b) are given in [2]. It is well known that Mp(a,b) is continuously and strictly increasing with respect to p∈ℝ for fixed a,b>0 with a≠b. Recently, the power, Neuman-Sándor, and identric means have been a subject of intensive research. In particular, many remarkable inequalities for these means can be found in the literature [3–26].

Let H(a,b)=2ab/(a+b), G(a,b)=ab, L(a,b)=(b-a)/(logb-loga), P(a,b)=(a-b)/[4arctan(a/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, geometric, logarithmic, first Seiffert, arithmetic, second Seiffert, quadratic, and contraharmonic means of two positive numbers a and b with a≠b, respectively. Then, it is well known that the inequalities
(4)H(a,b)=M-1(a,b)<G(a,b)=M0(a,b)<L(a,b)H(a,b)<P(a,b)<I(a,b)<A(a,b)=M1(a,b)<M(a,b)H(a,b)<T(a,b)<Q(a,b)=M2(a,b)<C(a,b),
hold for all a,b>0 with a≠b.

The following sharp bounds for L, I, (IL)1/2, and (I+L)/2 in terms of power means are presented in [27–32]:
(5)M0(a,b)<L(a,b)<M1/3(a,b),M2/3(a,b)<I(a,b)<Mlog2(a,b),M0(a,b)<I1/2(a,b)L1/2(a,b)<M1/2(a,b),12[I(a,b)+L(a,b)]<M1/2(a,b),
for all a,b>0 with a≠b.

Pittenger [31] found the greatest value r1 and the least value r2 such that the double inequality
(6)Mr1(a,b)≤Lp(a,b)≤Mr2(a,b),
holds for all a,b>0, where Lr(a,b) is the rth generalized logarithmic means which is defined by
(7)Lr(a,b)={[br+1-ar+1(r+1)(b-a)]1/r,a≠b, r≠-1, r≠0,1e(bbaa)1/(b-a),a≠b, r=0,b-alogb-loga,a≠b, r=-1,a,a=b.

The following sharp power mean bounds for the first Seiffert mean P(a,b) are given in [10, 33]:
(8)Mlog2/logπ(a,b)<P(a,b)<M2/3(a,b),
for all a,b>0 with a≠b.

In [17], the authors answered the question: for α∈(0,1), what are the greatest value p and the least value q such that the double inequality (9)Mp(a,b)<Pα(a,b)G1-α(a,b)<Mq(a,b)
holds for all a,b>0 with a≠b?

Neuman and Sándor [1] established that
(10)A(a,b)<M(a,b)<A(a,b)log(1+2),π4T(a,b)<M(a,b)<T(a,b),M(a,b)<2A(a,b)+Q(a,b)3,
for all a,b>0 with a≠b.

Let 0<a,b≤1/2 with a≠b, a′=1-a and b′=1-b. Then, the Ky Fan inequalities
(11)G(a,b)G(a′,b′)<L(a,b)L(a′,b′)<P(a,b)P(a′,b′)<A(a,b)A(a′,b′)<M(a,b)M(a′,b′)<T(a,b)T(a′,b′)
were presented in [1].

In [24], Li et al. found the best possible bounds for the Neuman-Sándor mean M(a,b) in terms of the generalized logarithmic mean Lr(a,b). Neuman [25] and Zhao et al. [26] proved that the inequalities
(12)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),α1H(a,b)+(1-α1)Q(a,b)<M(a,b)<β1H(a,b)+(1-β1)Q(a,b),α2G(a,b)+(1-α2)Q(a,b)<M(a,b)<β2G(a,b)+(1-β2)Q(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)], β≥1/3, λ≤[1-log(1+2)]/log(1+2), μ≥1/6, α1≥2/9, β1≤1-1/[2log(1+2)], α2≥1/3, and β2≤1-1/[2log(1+2)].

In [7], Sándor and Trif proved that the inequalities
(13)e((a-b)2/6(a+b)2)<A(a,b)I(a,b)<e((a-b)2/24ab),e((a-b)2/3(a+b)2)<I(a,b)G(a,b)<e((a-b)2/12ab),e((a-b)4/30(a+b)4)<I(a,b)A2/3(a,b)G1/3(a,b)<e((a-b)4/120ab(a+b)4)
hold for all a,b>0 with a≠b.

Neuman and Sándor [15] and Gao [20] proved that α1=1, β1=e/2, α2=1, β2=22/e, α3=1, β3=3/e, α4=e/π, β4=1, α5=1, and β5=2e/π are the best possible constants such that the double inequalities α1<A(a,b)/I(a,b)<β1, α2<I(a,b)/M2/3(a,b)<β2, α3<I(a,b)/He(a,b)<β3, α4<P(a,b)/I(a,b)<β4, and α5<T(a,b)/I(a,b)<β5 hold for all a,b>0 with a≠b, where He(a,b)=(a+ab+b)/3=(2A(a,b)+G(a,b))/3 is the Heronian mean of a and b.

In [34], Sándor established that
(14)He(a,b)<M2/3(a,b),
for all a,b>0 with a≠b.

It is not difficult to verify that the inequality
(15)2A(a,b)+Q(a,b)3<[He(a2,b2)]1/2
holds for all a,b>0 with a≠b.

From inequalities (10), (14), and (15), one has
(16)M(a,b)<[M2/3(a2,b2)]1/2=M4/3(a,b),
for all a,b>0 with a≠b.

It is the aim of this paper to find the best possible lower power mean bound for the Neuman-Sándor mean M(a,b) and to present the sharp constants α and β such that the double inequality
(17)α<M(a,b)I(a,b)<β
holds for all a,b>0 with a≠b.

2. Main Results
Theorem 1.
p
0
=
(
log
2
)
/
log
[
2
log
(
1
+
2
)
]
=
1.224
…
is the greatest value such that the inequality
(18)M(a,b)>Mp0(a,b)
holds for all a,b>0 with a≠b.

Proof.
From (1) and (2), we clearly see that both M(a,b) and Mp(a,b) are symmetric and homogenous of degree one. Without loss of generality, we assume that b=1 and a=x>1.

Let p0=(log2)/log [2log(1+2)], then from (1) and (2) one has
(19)logM(x,1)-logMp0(x,1) =logx-12sinh-1((x-1)/(x+1))-1p0logxp0+12.
Let
(20)f(x)=logx-12sinh-1((x-1)/(x+1))-1p0logxp0+12.
Then, simple computations lead to
(21)limx→1+f(x)=0,(22)limx→+∞f(x)=1p0log2-log [2sinh-1(1)]=0,(23)f′(x)=(1+xp0-1)f1(x)(x-1)(xp0+1)sinh-1((x-1)/(x+1)),
where
(24)f1(x)=-2(x-1)(xp0+1)(x+1)(xp0-1+1)1+x2+sinh-1(x-1x+1),f1(1)=0,(25)limx→+∞f1(x)=-2+sinh-1(1)=-0.5328⋯<0,(26)f1′(x)=2(x-1)f2(x)(x+1)2(xp0-1+1)2(1+x2)3/2,
where
(27)f2(x)=1+x+2x2+(p0-1)xp0-2-xp0-1+xp0+1-(p0-1)xp0+2-2x2p0-2-x2p0-1-x2p0,f2(1)=0,(28)limx→+∞f2(x)=-∞,(29)f2′(x)=1+4x+(p0-1)(p0-2)xp0-3-(p0-1)xp0-2+(p0+1)xp0-(p0-1)(p0+2)xp0+1-4(p0-1)x2p0-3-(2p0-1)x2p0-2-2p0x2p0-1,f2′(1)=4(4-3p0)>0,(30)limx→+∞f2′(x)=-∞,(31)f2′′(x)=4+(p0-1)(p0-2)(p0-3)xp0-4-(p0-1)(p0-2)xp0-3+p0(p0+1)xp0-1-(p0-1)(p0+2)(p0+1)xp0-4(p0-1)(2p0-3)x2p0-4-2(2p0-1)(p0-1)x2p0-3-2p0(2p0-1)x2p0-2,f2′′(1)=4(2p0-1)(4-3p0)>0,(32)limx→+∞f2′′(x)=-∞,(33)f2′′′(x)=(p0-1)xp0-5f3(x),
where
(34)f3(x)=-(2-p0)(3-p0)(4-p0)-(2-p0)(3-p0)x+p0(p0+1)x3-p0(p0+1)(p0+2)x4-8(3-2p0)(2-p0)xp0+2(2p0-1)(3-2p0)xp0+1-4p0(2p0-1)xp0+2<-(2-p0)(3-p0)(4-p0)-(2-p0)(3-p0)x+p0(p0+1)x4-p0(p0+1)(p0+2)x4-8(3-2p0)(2-p0)xp0+2(2p0-1)(3-2p0)xp0+2-4p0(2p0-1)xp0+2=-(2-p0)(3-p0)(4-p0)-(2-p0)(3-p0)x-p0(p0+1)2x4-8(3-2p0)(2-p0)xp0-2(2p0-1)(4p0-3)xp0+2<0,
for x>1.

Equation (33) and inequality (34) imply that f2′′(x) is strictly decreasing on [1,+∞). Then, the inequality (31) and (32) lead to the conclusion that there exists x1>1, such that f2′(x) is strictly increasing on [1,x1] and strictly decreasing on [x1,+∞).

From (29) and (30) together with the piecewise monotonicity of f2′(x), we clearly see that there exists x2>x1>1, such that f2(x) is strictly increasing on [1,x2] and strictly decreasing on [x2,+∞).

It follows from (26)–(28) and the piecewise monotonicity of f2(x) that there exists x3>x2>1, such that f1(x), is strictly increasing on [1,x3] and strictly decreasing on [x3,+∞).

From (23)–(25) and the piecewise monotonicity of f1(x) we see that there exists x4>x3>1, such that f(x) is strictly increasing on (1,x4] and strictly decreasing on [x4,+∞).

Therefore, M(x,1)>Mp0(x,1) for x>1 follows easily from (19)–(22) and the piecewise monotonicity of f(x).

Next, we prove that p0=(log2)/log [2log(1+2)]=1.224… is the greatest value such that M(x,1)>Mp0(x,1) for all x>1.

For any ε>0 and x>1, from (1) and (2), one has
(35)limx→+∞Mp0+ε(x,1)M(x,1)=limx→+∞[(1+xp0+ε2)1/(p0+ε) 2sinh-1((x-1)/(x+1))x-1]=2-1/(p0+ε)×2sinh-1(1)=2ε/p0(p0+ε) >1.

Inequality (35) implies that for any ε>0, there exists X=X(ε)>1, such that M(x,1)<Mp0+ε(x,1) for x∈(X,+∞).

Remark 2.
4
/
3
is the least value such that inequality (16) holds for all a,b>0 with a≠b, namely, M4/3(a,b) is the best possible upper power mean bound for the Neuman-Sándor mean M(a,b).

In fact, for any ε∈(0,4/3) and x>0, one has
(36)M4/3-ε(1+x,1)-M(1+x,1) =[(1+x)4/3-ε+12]1/(4/3-ε)-x2sinh-1(x/(2+x)).

Letting x→0 and making use of Taylor expansion, we get
(37)[(1+x)4/3-ε+12]1/(4/3-ε)-x2sinh-1(x/(2+x)) =[1+4-3ε6x+(4-3ε)(1-3ε)36x2+o(x2)]1/(4/3-ε) -xx-(1/2)x2+(5/24)x3+o(x3) =[1+12x+1-3ε24x2+o(x2)] -[1+12x+124x2+o(x2)]=-ε8x2+o(x2).

Equations (36) and (37) imply that for any ε∈(0,4/3) there exists δ=δ(ε)>0, such that M(1+x,1)>M(4/3)-ε(1+x,1) for x∈(0,δ).

Theorem 3.
For all a,b>0 with a≠b, one has
(38)1<M(a,b)I(a,b)<e2log(1+2),
with the best possible constants 1 and e/[2log(1+2)]=1.5419….

Proof.
From (2) and (3), we clearly see that both M(a,b) and I(a,b) are symmetric and homogenous of degree one. Without loss of generality, we assume that b=1 and a=x>1. Let
(39)f(x)=M(x,1)I(x,1)=e(x-1)2xx/(x-1)sinh-1((x-1)/(x+1)).
Then, simple computations lead to
(40)f′(x)f(x)=logx(x-1)2sinh-1((x-1)/(x+1))f1(x),
where
(41)f1(x)=sinh-1(x-1x+1)-2(x-1)2(x+1)1+x2logx,limx→1+f1(x)=0,(42)f1′(x)=2f2(x)x(x+1)2(1+x2)3/2log2x,
where
(43)f2(x)=x(x+1)(1+x2)log2x-x(3x3-x2+x-3)logx+(x-1)2(x+1)(1+x2),f2(1)=0,(44)f2′(x)=(4x3+3x2+2x+1)log2x+5(-2x3+x2+1)logx+5x4-7x3+x2-x+2,f2′(1)=0,(45)f2′′(x)=2(6x2+3x+1)log2x+2(-11x2+8x+2+x-1)logx+20x3-31x2+7x-1+5x-1,f2′′(1)=0,(46)f2′′′(x)=6(4x+1)log2x+2(-10x+14+2x-1-x-2)logx+60x2-84x+23+4x-1-3x-2,f2′′′(1)=0,(47)f2(4)(x)=24log2x+4(7+3x-1-x-2+x-3)logx+120x-104+28x-1+4x-3>0
for x>1.

From (46) and (47), we clearly see that f2′′(x) is strictly increasing on [1,+∞). Then, (45) leads to the conclusion that f2′(x) is strictly increasing on [1,+∞).

Equations (43) and (44) together with the monotonicity of f2′(x) impliy that f2(x)>0 for x>1. Then, (42) leads to the conclusion that f1(x) is strictly increasing on [1,+∞).

It follows from equations (40) and (41) together with the monotonicity of f1(x) that f(x) is strictly increasing on (1,+∞).

Therefore, Theorem 3 follows from (39) and the monotonicity of f(x) together with the facts that
(48)limx→+∞f(x)=e2log(1+2),limx→1+f(x)=1.