We present the necessary and sufficient condition for the monotonicity of the ratio of the power and second Seiffert means. As applications, we get the sharp upper and lower bounds for the second Seiffert mean in terms of the power mean.

1. Introduction

Throughout this paper, we assume that a,b>0 with a≠b. The second Seiffert mean T(a,b) and rth power mean Mr(a,b) of a and b are defined by
(1)T(a,b)=a-b2arctan((a-b)/(a+b)),(2)Mr(a,b)=(ar+br2)1/r(r≠0),M0(a,b)=ab,
respectively.

It is well-known that the power mean Mr(a,b) is strictly increasing with respect to r∈R for fixed a,b>0 with a≠b. In the recent past, both mean values have been the subject of intensive research. In particular, many remarkable inequalities for T and Mr can be found in the literature [1–5].

Seiffert [6] proved that the double inequality
(3)M1(a,b)<T(a,b)<M2(a,b)
holds for all a,b>0 with a≠b.

In [7], Hästö proved that the function T(1,x)/Mp(1,x) is strictly increasing on [1,∞) if p≤1 and presented an improvement for the first inequality in (3).

Costin and Toader [8] proved that the inequality
(4)T(a,b)>M3/2(a,b)
holds for all a,b>0 with a≠b.

In [9], Witkowski proved that the double inequality
(5)22πM2(a,b)<T(a,b)<4πM1(a,b)
holds for all a,b>0 with a≠b.

Recently, the following optimal estimations for the second Seiffert mean by power means were obtained independently in [10, 11]:
(6)Mlog2/(logπ-log2)(a,b)<T(a,b)<M5/3(a,b)
for all a,b>0 with a≠b.

The main purpose of this paper is to give the necessary and sufficient condition for the monotonicity of the function T(1,x)/Mp(1,x) on (0,1) and present the best possible parameters α and β such that the double inequality
(7)αM5/3(a,b)<T(a,b)≤βMlog2/(logπ-log2)(a,b)
holds for all a,b>0 with a≠b.

2. Main Results

In order to prove our main results we first establish a lemma.

Lemma 1.

Let f(p,x) be defined on R×(0,1) by
(8)f(p,x)=(1-x)(1+xp)(1+x2)(1+xp-1)-arctan1-x1+x.
Then there exists λ∈(0,1) such that f(p,x) is strictly decreasing with respect to x on (0,λ] and strictly increasing with respect to x on [λ,1) if p∈(1,5/3).

Proof.

Let
(9)f1(p,x)=(1-p)xp+(1+p)xp-1-(1+p)xp-2+(p-1)xp-3-2x2p-3+2.
Then,
(10)f1(p,1)=0,f1(p,0+)=∞,(11)∂f(p,x)∂x=-x(1-x)(1+x2)2(1+xp-1)2f1(p,x),(12)x4-p∂f1(p,x)∂x=-2(2p-3)xp-p(p-1)x3+(p-1)(p+1)x2-(p+1)(p-2)x+(p-1)(p-3)∶=f2(p,x),(13)f2(p,0)=(p-1)(p-3)<0,f2(p,1)=2(5-3p)>0,(14)∂f2(p,x)∂x=-2p(2p-3)xp-1-3p(p-1)x2+2(p-1)(p+1)x-(p+1)(p-2).

We divide two cases to prove that ∂f2(p,x)/∂x>0 for all x∈(0,1) and p∈(1,5/3).

Case 1.

Consider that p∈(1,3/2]. From (14) we clearly see that
(15)∂3f2(p,x)∂x3=-2p(p-1)[3+(2-p)(3-2p)xp-3]<0,(16)∂f2∂x(p,0)=(p+1)(2-p)>0,∂f2∂x(p,1)=2p(5-3p)>0.

Equation (15) implies that ∂f2(p,x)/∂x is strictly concave with respect to x on the interval (0,1). Then (16) and the basic properties of concave function lead to the conclusion that
(17)∂f2∂x(p,x)>(1-x)∂f2∂x(p,0)+x∂f2∂x(p,1)>0.

Case 2.

Consider that p∈(3/2,5/3). Making use of the weighted arithmetic-geometric inequality λa+(1-λ)b≥aλb1-λ(0≤λ≤1) we get
(18)xp-1≤(p-1)x+(2-p).

Equations (14) and (18) lead to
(19)∂f2(p,x)∂x≥-2p(2p-3)[(p-1)x+(2-p)]-3p(p-1)x2+2(p-1)(p+1)x-(p+1)(p-2)=-3p(p-1)x2-2(p-1)(2p2-4p-1)x+(p-2)(4p2-7p-1)∶=f3(p,x).

Note that
(20)∂2f3(p,x)∂x2=-6p(p-1)<0,f3(p,1)=2p(5-3p)>0,f3(p,0)=4(p-2)(p-65+78)(p+65+78)>0.

It follows from (20) and the concavity of the function f3(p,x) with respect to x on the interval (0,1) that
(21)f3(p,x)>(1-x)f3(p,0+)+xf3(p,1)>0.
Therefore, ∂f2(p,x)/∂x>0 follows from (19) and (21).

Next we prove the desired result. From (12) and (13) together with the fact that ∂f2(p,x)/∂x>0 we clearly see that there exists λ1∈(0,1) such that f1(p,x) is strictly decreasing with respect to x on (0,λ1] and strictly increasing with respect to x on [λ1,1). Therefore, Lemma 1 follows easily from (10) and (11) together with the piecewise monotonicity of f1(p,x) with respect to x on the interval (0,1).

Theorem 2.

Let F(p,x) be defined on R×(0,1) by
(22)F(p,x)=logT(1,x)Mp(1,x)=log1-x2arctan((1-x)/(1+x))-1plog1+xp2(p≠0),(23)F(0,x)=limp→0F(p,x)=log1-x2arctan((1-x)/(1+x))-12logx.
Then the following statements are true.

F(p,x) is strictly increasing with respect to x on (0,1) if and only if p≥5/3.

F(p,x) is strictly decreasing with respect to x on (0,1) if and only if p≤1.

If p∈(1,5/3), then there exists μ∈(0,1) such that F(p,x) is strictly increasing with respect to x on (0,μ] and strictly decreasing with respect to x on [μ,1).

Proof.

It follows from (22) and (23) that
(24)∂F(p,x)∂x=1+xp-1x(1-x)(1+xp)arctan((1-x)/(1+x))f(p,x),
where f(p,x) is defined by (8). And
(25)∂f(p,x)∂x=-x(1-x)(1+x2)2(1+xp-1)g(p,x),
where
(26)g(p,x)=(1-p)xp+(1+p)xp-1-2x2p-3-(1+p)xp-2+(p-1)xp-3+2.

(1) If F(p,x) is strictly increasing with respect to x on (0,1), then (24) leads to f(p,x)>0 for all x∈(0,1). Making use of L’Höspital’s rule and (8) we get
(27)limx→1-f(p,x)(1-x)3=124(3p-5)≥0,
which implies that p≥5/3.

If p≥5/3, then from (8) and (26) together with the fact that the function p→(1+xp)/(1+xp-1) is strictly increasing on R we get
(28)f(p,x)≥f(53,x),(29)g(53,x)=23x-4/3(1-x1/3)3(1+x2/3)×(1+3x1/3+5x2/3+3x+x4/3)>0
for all x∈(0,1).

Equations (8) and (25) together with inequality (29) lead to the conclusion that
(30)f(53,x)>f(53,1)=0
for all x∈(0,1).

Therefore, F(p,x) is strictly increasing with respect to x on (0,1) which follows easily from (24), (28), and (30).

(2) If F(p,x) is strictly decreasing with respect to x on (0,1), then (24) implies that f(p,x)<0 for all x∈(0,1). In particular, we have f(p,0+)≤0 and p≤1. Indeed, if p>1, then (8) leads to the conclusion that f(p,0+)=1-π/4>0.

If p≤1, then from (8) and (26) together with the fact that the function p→(1+xp)/(1+xp-1) is strictly increasing on R we get
(31)f(p,x)≤f(1,x),(32)g(1,x)=4(1-1x)<0
for all x∈(0,1).

Equations (8) and (25) together with inequality (32) lead to the conclusion that
(33)f(1,x)<f(1,1)=0
for all x∈(0,1).

Therefore, F(p,x) is strictly decreasing with respect to x on (0,1) which follows easily from (24), (31), and (33).

(3) If p∈(1,5/3), then (8) leads to
(34)f(p,0)=1-π4>0,f(p,1)=0.

It follows from Lemma 1 and (34) that we clearly see that there exists μ∈(0,1) such that f(p,x)>0 for x∈(0,μ) and f(p,x)<0 for x∈(μ,1). Then from (24) we get Theorem 2(3) immediately.

Theorem 3.

For all a,b>0 with a≠b, the double inequality
(35)αM5/3(a,b)<T(a,b)≤βMlog2/(logπ-log2)(a,b)
holds with the best possible constants β=eF(log2/(logπ-log2),μ)=1.0136… and α=28/5/π=0.9649…, where μ is the solution of the equation f(log2/(logπ-log2),x)=0 on (0,1) and f(p,x) and F(p,x) are defined by (8) and (22), respectively.

Proof.

Without loss of generality, we assume that a>b>0. Let x=b/a∈(0,1); then from (1) and (2) we get
(36)logT(a,b)-logMp(a,b)=F(p,x).

If p=5/3, then from (22) and Theorem 2(1) we get
(37)F(53,x)>F(53,0)=log28/5π.

Therefore, the first inequality in (35) with the best possible constant α=28/5/π follows from (36) and (37) together with the monotonicity of F(5/3,x) given in Theorem 2(1).

If p=log2/(logπ-log2)∈(1,5/3), then Lemma 1 and (24) together with (34) imply that there exists μ∈(0,1) such that f(log2/(logπ-log2),x)=0, and F(log2/(logπ-log2),x) is strictly increasing on (0,μ] and strictly decreasing on [μ,1). Therefore, we have
(38)F(log2logπ-log2,x)≤F(log2logπ-log2,μ).

Making use of MATHEMATICA software, numerical computations show that
(39)0.186930110570624<μ<0.186930110570625,eF(log2/(logπ-log2),μ)=1.0136….

Therefore, the second inequality in (35) with the best possible constant β=eF(log2/(logπ-log2),μ)=1.0136… follows from (36) and (38) together with the piecewise monotonicity of F(log2/(logπ-log2),x).

Corollary 4.

The double inequality
(40)Q2(a,b)Lp-1(a,b)<T(a,b)<Q2(a,b)Lq-1(a,b)
holds for all a,b>0 with a≠b if and only if p≥5/3 and q≤1, where Q(a,b)=(a2+b2)/2 and Lp(a,b)=(ap+1+bp+1)/(ap+bp) are, respectively, the quadratic and pth Lehmer means of a and b.

Proof.

Without loss of generality, we assume that a>b>0. Let x=b/a∈(0,1). Then from Theorem 2 and (24) we clearly see that the f(p,x)>0 if and only if p≥5/3 and f(p,x)<0 if and only if p≤1. Then (8) leads to the conclusion that the inequalities
(41)(1-x)(1+xp)(1+x2)(1+xp-1)>arctan1-x1+x,(42)(1-x)(1+xq)(1+x2)(1+xq-1)<arctan1-x1+x
hold for all x∈(0,1) if and only p≥5/3 and q≤1.

Therefore, Corollary 4 follows easily from inequalities (41) and (42) together with (1).

Corollary 5.

Let a1,b1,a2,b2>0 with a1/b1<a2/b2. Then Theorem 2 leads to the following Ky Fan type inequality:
(43)T(a1,b1)T(a2,b2)<(>)Mp(a1,b1)Mp(a2,b2)
if p≥5/3(p≤1).

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This research was supported by the Natural Science Foundation of China under Grants 61374086, 11371125, and 11171307, the Natural Science Foundation of Hunan Province under Grant 12C0577, and the Natural Science Foundation of Zhejiang Province under Grant LY13A010004.

NeumanE.SándorJ.On the Schwab-Borchardt meanNeumanE.SándorJ.On the Schwab-Borchardt mean. IIJiangW.-D.QiF.Some sharp inequalities involving Seiffert and other means and their concise proofsJiangW.-D.Some sharp inequalities involving reciprocals of the Seiffert and other meansWitkowskiA.Optimal weighted harmonic interpolations between Seiffert meansSeiffertH.-J.Aufgabe β16HästöP. A.A monotonicity property of ratios of symmetric homogeneous meansCostinI.ToaderG.A nice separation of some Seiffert-type means by power meansWitkowskiA.Interpolations of Schwab-Borchardt meanCostinI.ToaderG.Optimal evaluations of some Seiffert-type means by power meansYangZ.-H.Sharp bounds for the second Seiffert mean in terms of power meanshttp://arxiv.org/pdf/1206.5494v1.pdf