Bounds for the Combinations of Neuman-Sándor, Arithmetic, and Second Seiffert Means in terms of Contraharmonic Mean

Abstract

We give the greatest values 𝑟₁, 𝑟₂ and the least values 𝑠₁, 𝑠₂ in (1/2, 1) such that the double inequalities 𝐶(𝑟₁𝑎+(1−𝑟₁)𝑏,𝑟₁𝑏+(1−𝑟₁)𝑎)<𝛼𝐴(𝑎,𝑏)+(1−𝛼)𝑇(𝑎,𝑏)<𝐶(𝑠₁𝑎+(1−𝑠₁)𝑏,𝑠₁𝑏+(1−𝑠₁)𝑎) and 𝐶(𝑟₂𝑎+(1−𝑟₂)𝑏,𝑟₂𝑏+(1−𝑟₂)𝑎)<𝛼𝐴(𝑎,𝑏)+(1−𝛼)𝑀(𝑎,𝑏)<𝐶(𝑠₂𝑎+(1−𝑠₂)𝑏,𝑠₂𝑏+(1−𝑠₂)𝑎) hold for any 𝛼∈(0,1) and all 𝑎,𝑏>0 with 𝑎≠𝑏, where 𝐴(𝑎,𝑏), 𝑀(𝑎,𝑏), 𝐶(𝑎,𝑏), and 𝑇(𝑎,𝑏) are the arithmetic, Neuman-Sándor, contraharmonic, and second Seiffert means of 𝑎 and 𝑏, respectively.

1. Introduction

For 𝑎,𝑏>0 with 𝑎≠𝑏, the Neuman-Sándor mean 𝑀(𝑎,𝑏) [1], second Seiffert mean 𝑇(𝑎,𝑏) [2] are defined by 𝑀(𝑎,𝑏)=𝑎−𝑏2sinh⁻¹)/(,𝑇()=((𝑎−𝑏𝑎+𝑏))𝑎,𝑏𝑎−𝑏)/(,2arctan((𝑎−𝑏𝑎+𝑏)) (1) respectively. Herein, sinh⁻¹(𝑥)=log(𝑥+√1+𝑥²) is the inverse hyperbolic sine function.

Let 𝐻(𝑎,𝑏)=2𝑎𝑏/(𝑎+𝑏), 𝐺(𝑎,𝑏)=√𝑎𝑏, 𝐿(𝑎,𝑏)=(𝑎−𝑏)/(log𝑎−log𝑏), 𝑃(𝑎,𝑏)=(𝑎−𝑏)/[4arctan(𝑎/𝑏)−𝜋], 𝐼(𝑎,𝑏)=1/𝑒(𝑏^𝑏/𝑎^𝑎)^{1/(𝑏−𝑎)}, 𝐴(𝑎,𝑏)=(𝑎+𝑏)/2, 𝑄(𝑎,𝑏)=(𝑎²+𝑏²)/2, and 𝐶(𝑎,𝑏)=(𝑎²+𝑏²)/(𝑎+𝑏) be the harmonic, geometric, logarithmic, first Seiffert, identric, arithmetic, quadratic, and contraharmonic means of two distinct positive real numbers 𝑎 and 𝑏, respectively. Then it is well known that the inequalities 𝐻()()()()()()()()()()𝑎,𝑏<𝐺𝑎,𝑏<𝐿𝑎,𝑏<𝑃𝑎,𝑏<𝐼𝑎,𝑏<𝐴𝑎,𝑏<𝑀𝑎,𝑏<𝑇𝑎,𝑏<𝑄𝑎,𝑏<𝐶𝑎,𝑏 (2) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏.

Among means of two variables, the Neuman-Sándor, contraharmonic, and second Seiffert means have attracted the attention of several researchers. In particular, many remarkable inequalities and applications for these means can be found in the literature [3–15].

Neuman and Sándor [1, 16] proved that the inequalities 𝐴()()<𝐴()𝑎,𝑏<𝑀𝑎,𝑏𝑎,𝑏log1+√2,𝜋4𝑇()()(),𝑀()<()()𝑎,𝑏<𝑀𝑎,𝑏<𝑇𝑎,𝑏𝑎,𝑏2𝐴𝑎,𝑏+𝑄𝑎,𝑏3,𝑃(𝑎,𝑏)𝑀(𝑎,𝑏)<𝐴²(),𝐴()𝑇()𝑎,𝑏𝑎,𝑏𝑎,𝑏<𝑀²()<𝐴𝑎,𝑏₂(𝑎,𝑏)+𝑇²(𝑎,𝑏)2 (3) hold 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 𝐺(𝑎,𝑏)𝐺𝑎^′,b^′<𝐿(𝑎,𝑏)𝐿𝑎^′,𝑏^′<𝑃(𝑎,𝑏)𝑃𝑎^′,𝑏^′<𝐴(𝑎,𝑏)𝐴𝑎^′,𝑏^′<𝑀(𝑎,𝑏)𝑀𝑎^′,𝑏^′<𝑇(𝑎,𝑏)𝑇𝑎^′,𝑏^′ (4) can be found in [1].

Li et al. [17] proved that the double inequality 𝐿_𝑝0(𝑎,𝑏)<𝑀(𝑎,𝑏)<𝐿₂(𝑎,𝑏) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏, where 𝐿_𝑝(𝑎,𝑏)=[(𝑏^𝑝+1−𝑎^𝑝+1)/((𝑝+1)(𝑏−𝑎))]^1/𝑝(𝑝≠−1,0), 𝐿₀(𝑎,𝑏)=𝐼(𝑎,𝑏) and 𝐿₋₁(𝑎,𝑏)=𝐿(𝑎,𝑏) is the 𝑝th generalized logarithmic mean of 𝑎 and 𝑏, and 𝑝₀=1.843⋯ is the unique solution of the equation (𝑝+1)^1/𝑝=2log(1+√2).

In [18], Neuman proved that the inequalities ()+()𝐴()()()+𝐴(),()+()𝐴()()()+𝐴()𝛼𝑄𝑎,𝑏1−𝛼𝑎,𝑏<𝑀𝑎,𝑏<𝛽𝑄𝑎,𝑏1−𝛽𝑎,𝑏𝜆𝐶𝑎,𝑏1−𝜆𝑎,𝑏<𝑀𝑎,𝑏<𝜇𝐶𝑎,𝑏1−𝜇𝑎,𝑏 (5) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼≤[1−log(1+√2)]/[(√2−1)log(1+√2)]=0.3249⋯, 𝜆≤[1−log(1+√2)]/log(1+√2)=0.1345⋯, 𝛽≥1/3 and 𝜇≥1/6.

Zhao et al. [19] found the least values 𝛼₁,𝛼₂,𝛼₃ and the greatest values 𝛽₁,𝛽₂,𝛽₃ such that the double inequalities 𝛼₁𝐻(𝑎,𝑏)+1−𝛼₁𝑄(𝑎,𝑏)<𝑀(𝑎,𝑏)<𝛽₁𝐻(𝑎,𝑏)+1−𝛽₁𝑄(),𝛼𝑎,𝑏₂𝐺(𝑎,𝑏)+1−𝛼₂𝑄(𝑎,𝑏)<𝑀(𝑎,𝑏)<𝛽₂𝐺(𝑎,𝑏)+1−𝛽₂𝑄(),𝛼𝑎,𝑏₃𝐻(𝑎,𝑏)+1−𝛼₃𝐶(𝑎,𝑏)<𝑀(𝑎,𝑏)<𝛽₃𝐻(𝑎,𝑏)+1−𝛽₃𝐶(𝑎,𝑏) (6) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏.

In [20, 21], the authors proved that the double inequalities 𝛼₁𝑇(𝑎,𝑏)+1−𝛼₁𝐺(𝑎,𝑏)<𝐴(𝑎,𝑏)<𝛽₁𝑇(𝑎,𝑏)+1−𝛽₁𝐺(),𝛼𝑎,𝑏₂𝑄(𝑎,𝑏)+1−𝛼₂𝐴(𝑎,𝑏)<𝑇(𝑎,𝑏)<𝛽₂𝑄(𝑎,𝑏)+1−𝛽₂𝐴(),𝑄𝑎,𝑏_𝛼3(𝑎,𝑏)𝐴^1−𝛼3(𝑎,𝑏)<𝑇(𝑎,𝑏)<𝑄^𝛽3(𝑎,𝑏)𝐴^1−𝛽3(𝑎,𝑏) (7) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only of 𝛼₁≤3/5, 𝛽₁≥4/𝜋,   𝛼₂≤(4−𝜋)/[(√2−1)𝜋], 𝛽₂≥2/3, 𝛼₃≤2/3, and 𝛽₃≥4−2log𝜋/log2.

For 𝛼,𝛽,𝜆,𝜇∈(1/2,1), Chu et al. [22, 23] proved that the inequalities 𝐶(()()𝑎)()𝑎,𝑄(()()𝑎)()𝑎𝛼𝑎+1−𝛼𝑏,𝛼𝑏+1−𝛼<𝑇𝑎,𝑏<𝐶𝛽𝑎+1−𝛽𝑏,𝛽𝑏+1−𝛽𝜆𝑎+1−𝜆𝑏,𝜆𝑏+1−𝜆<𝑇𝑎,𝑏<𝑄𝜇𝑎+1−𝜇𝑏,𝜇𝑏+1−𝜇 (8) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼≤(1+4/𝜋−1)/2, 𝛽≥(3+√3)/6, 𝜆≤(1+16/𝜋²−1)/2 and 𝜇≥(3+√6)/6.

The aim of this paper is to find the greatest values 𝑟₁, 𝑟₂ and the least values 𝑠₁, 𝑠₂ such that the double inequalities 𝐶𝑟₁𝑎+1−𝑟₁𝑏,𝑟₁𝑏+1−𝑟₁𝑎()+()𝑇()𝑠<𝛼𝐴𝑎,𝑏1−𝛼𝑎,𝑏<𝐶₁𝑎+1−𝑠₁𝑏,𝑠₁𝑏+1−𝑠₁𝑎,𝐶𝑟₂𝑎+1−𝑟₂𝑏,𝑟₂𝑏+1−𝑟₂𝑎()+()𝑀()𝑠<𝛼𝐴𝑎,𝑏1−𝛼𝑎,𝑏<𝐶₂𝑎+1−𝑠₂𝑏,𝑠₂𝑏+1−𝑠₂𝑎 (9)(10) hold for any 𝛼∈(0,1) and all 𝑎,𝑏>0 with 𝑎≠𝑏.

2. Lemmas

In order to prove our main results, we need three lemmas, which we present in this section.

Lemma 1 (see [24, Theorem 1.25]). For −∞<𝑎<𝑏<+∞, let 𝑓,𝑔∶[𝑎,𝑏]→ℝ be continuous on [𝑎,𝑏] and 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 𝑓(𝑥)−𝑓(𝑎)𝑔(𝑥)(𝑎),𝑓(𝑥)(𝑏)−𝑔−𝑓𝑔(𝑥)(𝑏).−𝑔 (11) If 𝑓^′(𝑥)/𝑔^′(𝑥) is strictly monotone, then the monotonicity in the conclusion is also strict.

Lemma 2. Let 𝑢,𝛼∈(0,1) and 𝑓_𝑢,𝛼(𝑥)=𝑢𝑥²−()𝑥1−𝛼.arctan𝑥−1 (12) Then 𝑓_𝑢,𝛼(𝑥)>0 for all 𝑥∈(0,1) if and only if 𝑢≥(1−𝛼)/3 and 𝑓_𝑢,𝛼(𝑥)<0  for all 𝑥∈(0,1) if and only if 𝑢≤(1−𝛼)(4/𝜋−1).

Proof. From (12), one has 𝑓_𝑢,𝛼0⁺𝑓=0,_𝑢,𝛼(1⁻)()4=𝑢−1−𝛼𝜋,𝑓−1_′𝑢,𝛼(𝑥)=2𝑥𝑢−1−𝛼2𝑔(𝑥), (13)(14)(15) where 𝑔(𝑥)=1+𝑥²arctan𝑥−𝑥𝑥1+𝑥²(arctan𝑥)². (16)
Let 𝑔₁(𝑥)=arctan𝑥−𝑥/(1+𝑥²) and 𝑔₂(𝑥)=𝑥(arctan𝑥)², then 𝑔(𝑥)=𝑔₁(𝑥)𝑔₂(𝑥),𝑔₁(0)=𝑔₂(0)𝑔=0,_1′(𝑥)𝑔₂^′(𝑥)=2𝑥²2𝑥1+𝑥²arctan𝑥+1+𝑥²²(arctan𝑥)²=11+𝑥²arctan𝑥/𝑥+(1/2)1+𝑥²arctan𝑥/𝑥². (17)(18)
It is not difficult to verify that the function (1+𝑥²)arctan𝑥/𝑥 is strictly increasing on (0,1). Then (17) and (18) together with Lemma 1 lead to the conclusion that 𝑔(𝑥) is strictly decreasing on (0,1). Moreover, making use of L'Hôpital's rule, we get 𝑔0⁺=23,𝑔(1⁻)=4(𝜋−2)𝜋². (19)(20)
We divide the proof into four cases.
Case 1. 𝑢≥(1−𝛼)/3. Then from (15) and (19) together with the monotonicity of 𝑔(𝑥), we clearly see that 𝑓_𝑢,𝛼(𝑥) is strictly increasing on (0,1). Therefore, 𝑓_𝑢,𝛼(𝑥)>0 for all 𝑥∈(0,1) follows from (13) and the monotonicity of 𝑓_𝑢,𝛼(𝑥).
Case 2. 𝑢≤2(1−𝛼)(𝜋−2)/𝜋². Then from (15) and (20) together with the monotonicity of 𝑔(𝑥), we clearly see that 𝑓_𝑢,𝛼(𝑥) is strictly decreasing on (0,1). Therefore, 𝑓_𝑢,𝛼(𝑥)<0 for all 𝑥∈(0,1) follows from (13) and the monotonicity of 𝑓_𝑢,𝛼(𝑥).
Case 3. 2(1−𝛼)(𝜋−2)/𝜋²<𝑢≤(1−𝛼)(4/𝜋−1). Then (14) leads to 𝑓_𝑢,𝛼(1⁻)≤0. (21)
From (15), (19), and (20) together with the monotonicity of 𝑔(𝑥), we clearly see that there exists unique 𝑥₀∈(0,1) such that 𝑓_𝑢,𝛼(𝑥) is strictly decreasing on (0,𝑥₀] and strictly increasing on [𝑥₀,1). Therefore, 𝑓_𝑢,𝛼(𝑥)<0 for all 𝑥∈(0,1) follows from (13) and (21) together with the piecewise monotonicity of 𝑓_𝑢,𝛼(𝑥).
Case 4. (1−𝛼)(4/𝜋−1)<𝑢≤(1−𝛼)/3. Then (14) leads to 𝑓_𝑢,𝛼(1⁻)>0. (22)
It follows from (15), (19), and (20) together with the monotonicity of 𝑔(𝑥), there exists unique 𝑥₁∈(0,1) such that 𝑓_𝑢,𝛼(𝑥) is strictly decreasing on (0,𝑥₁] and strictly increasing on [𝑥₁,1). Equation (13) and inequality (22) together with the piecewise monotonicity of 𝑓_𝑢,𝛼(𝑥) lead to the conclusion that there exists 𝑥₂∈(𝑥₁,1) such that 𝑓_𝑢,𝛼(𝑥)<0 for 𝑥∈(0,𝑥₂) and 𝑓_𝑢,𝛼(𝑥)>0 for 𝑥∈(𝑥₂,1).

Lemma 3. Let 𝜆,𝛼∈(0,1) and 𝜑_𝜆,𝛼(𝑥)=𝜆𝑥²−()𝑥1−𝛼sinh⁻¹(𝑥)−1. (23) Then 𝜑_𝜆,𝛼(𝑥)>0 for all 𝑥∈(0,1) if and only if 𝜆≥(1−𝛼)/6 and 𝜑_𝜆,𝛼(𝑥)<0 for all 𝑥∈(0,1) if and only if 𝜆≤(1−𝛼)(1−log(1+√2))/log(1+√2).

Proof. From (23) we get 𝜑_𝜆,𝛼0⁺𝜑=0,_𝜆,𝛼(1⁻)(√=𝜆−1−𝛼)1−log1+2log1+√2𝜑_′𝜆,𝛼,(𝑥)=2𝑥𝜆−1−𝛼2𝜓(𝑥), (24)(25)(26) where 𝜓(𝑥)=sinh⁻¹(𝑥)−𝑥/√1+𝑥²𝑥sinh⁻¹(𝑥)². (27)
Let 𝜓₁(𝑥)=sinh⁻¹(𝑥)−𝑥/√1+𝑥² and 𝜓₂(𝑥)=𝑥(sinh⁻¹(𝑥))², then 𝜓(𝑥)=𝜓₁(𝑥)𝜓₂(𝑥),𝜓₁(0)=𝜓₂(0)𝜓=0,_1′(𝑥)𝜓₂^′(𝑥)=𝑥²×1+𝑥^23/2sinh⁻¹(𝑥)²+2𝑥1+𝑥²sinh⁻¹(𝑥)⁻¹=1+𝑥^23/4sinh⁻¹(𝑥)/𝑥²+21+𝑥^21/41+𝑥^23/4sinh⁻¹(𝑥)/𝑥⁻¹. (28)
It is not difficult to verify that the function (1+𝑥²)^3/4sinh⁻¹(𝑥)/𝑥 is strictly increasing on (0,1). Then (28) together with Lemma 1 leads to the conclusion that 𝜓(𝑥) is strictly decreasing on (0,1). Moreover, making use of L'Hôpital's rule, we have 𝜓0⁺=13,𝜓(1⁻)=√2log1+√2−1√2log²1+√2. (29)(30)
We divide the proof into four cases.
Case 1. 𝜆≥(1−𝛼)/6. Then from (26) and (29) together with the monotonicity of 𝜓(𝑥), we clearly see that 𝜑_𝜆,𝛼(𝑥) is strictly increasing on (0,1). Therefore, 𝜑_𝜆,𝛼(𝑥)>0 for all 𝑥∈(0,1) follows from (24) and the monotonicity of 𝜑_𝜆,𝛼(𝑥).
Case 2. 𝜆≤(1−𝛼)[√2log(1+√2)−1]/[2√2log²(1+√2)]. Then from (26) and (30) together with the monotonicity of 𝜓(𝑥), we clearly see that 𝜑_𝜆,𝛼(𝑥) is strictly decreasing on (0,1). Therefore, 𝜑_𝜆,𝛼(𝑥)<0 for all 𝑥∈(0,1) follows from (24) and the monotonicity of 𝜑_𝜆,𝛼(𝑥).
Case 3. ((1−𝛼)[√2log(1+√2)−1]/2√2log²(1+√2))<𝜆≤((1−𝛼)[1−log(1+√2)]/log(1+√2)). Then (25) leads to 𝜑_𝜆,𝛼(1⁻)≤0. (31)
From (26), (29), and (30) together with the monotonicity of 𝜓(𝑥), we clearly see that there exists 𝑥₃∈(0,1) such that 𝜑_𝜆,𝛼(𝑥) is strictly decreasing on (0,𝑥₃] and strictly increasing on [𝑥₃,1). Therefore, 𝜑_𝜆,𝛼(𝑥)<0 for all 𝑥∈(0,1) follows from (24) and (31) together with the piecewise monotonicity of 𝜑_𝜆,𝛼(𝑥).
Case 4. ((1−𝛼)[1−log(1+√2)]/log(1+√2))<𝜆<((1−𝛼)/6). Then (25) leads to 𝜑_𝜆,𝛼(1⁻)>0. (32)
It follows from (26), (29), and (30) together with the monotonicity of 𝜓(𝑥), there exists 𝑥₄∈(0,1) such that 𝜑_𝜆,𝛼(𝑥) is strictly decreasing on (0,𝑥₄] and strictly increasing on [𝑥₄,1). Equation (24) and inequality (32) together with the piecewise monotonicity of 𝜑_𝜆,𝛼(𝑥) lead to the conclusion that there exists 𝑥₅∈(𝑥₄,1) such that 𝜑_𝜆,𝛼(𝑥)<0 for 𝑥∈(0,𝑥₅) and 𝜑_𝜆,𝛼(𝑥)>0 for 𝑥∈(𝑥₅,1).

3. Main Results

Theorem 4. If 𝛼∈(0,1) and 𝑟₁,𝑠₁∈(1/2,1), then the double inequality 𝐶𝑟₁𝑎+1−𝑟₁𝑏,𝑟₁𝑏+1−𝑟₁𝑎()+()𝑇()𝑠<𝛼𝐴𝑎,𝑏1−𝛼𝑎,𝑏<𝐶₁𝑎+1−𝑠₁𝑏,𝑠₁𝑏+1−𝑠₁𝑎 (33) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝑟₁≤[1+(1−𝛼)(4−𝜋)/𝜋]/2 and 𝑠₁≥[1+(1−𝛼)/3]/2.

Proof. Since 𝐴(𝑎,𝑏), 𝑇(𝑎,𝑏), and 𝐶(𝑎,𝑏) are symmetric and homogeneous of degree one, without loss of generality, we assume that 𝑎>𝑏. Let 𝑝∈(1/2,1) and 𝑥=(𝑎−𝑏)/(𝑎+𝑏), then 𝑥∈(0,1) and 𝐶𝑎−[()+()𝑇((𝑝𝑎+1−𝑝𝑏,𝑝𝑏+1−𝑝𝛼𝐴𝑎,𝑏1−𝛼𝑎,𝑏)]=𝐴𝑎,𝑏)2𝑝−1²𝑥²−()𝑥1−𝛼.arctan𝑥−1 (34)
Therefore, Theorem 4 follows easily from Lemma 2 and (34).

Theorem 5. If 𝛼∈(0,1) and 𝑟₂,𝑠₂∈(1/2,1), then the double inequality 𝐶𝑟₂𝑎+1−𝑟₂𝑏,𝑟₂𝑏+1−𝑟₂𝑎()+()𝑀()𝑠<𝛼𝐴𝑎,𝑏1−𝛼𝑎,𝑏<𝐶₂𝑎+1−𝑠₂𝑏,𝑠₂𝑏+1−𝑠₂𝑎 (35) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝑠₂≥[1+(1−𝛼)/6]/2 and 𝑟₂≤[1+(1−𝛼)(1−log(1+√2))/log(1+√2)]/2.

Proof. Since 𝐴(𝑎,𝑏), 𝑀(𝑎,𝑏), and 𝐶(𝑎,𝑏) are symmetric and homogeneous of degree one, without loss of generality, we assume that 𝑎>𝑏. Let 𝑞∈(1/2,1) and 𝑥=(𝑎−𝑏)/(𝑎+𝑏), then 𝑥∈(0,1) and 𝐶𝑎−[()+()𝑀((𝑞𝑎+1−𝑞𝑏,𝑞𝑏+1−𝑞𝛼𝐴𝑎,𝑏1−𝛼𝑎,𝑏)]=𝐴𝑎,𝑏)2𝑞−1²𝑥²−()𝑥1−𝛼sinh⁻¹(𝑥)−1. (36)
Therefore, Theorem 5 follows easily from Lemma 3 and (36).

Remark 6. If 𝛼=0, then Theorem 4 reduces to the first double inequality in (8).

Corollary 7. If 𝜆,𝜇∈(1/2,1), then the double inequality 𝐶(()()𝑎)()𝑎𝜆𝑎+1−𝜆𝑏,𝜆𝑏+1−𝜆<𝑀𝑎,𝑏<𝐶𝜇𝑎+1−𝜇𝑏,𝜇𝑏+1−𝜇 (37) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝜆≤[1+1/log(1+√2)−1]/2 and 𝜇≥(6+√6)/12.

Proof. Corollary 7 follows easily from Theorem 5 with 𝛼=0.

Acknowledgments

This research was supported by the Natural Science Foundation of China (Grants no.11171307, 61173123), the Natural Science Foundation of Zhejiang Province (Grants no. Z1110551, LY12F02012), and the Natural science Foundation of Huzhou City (Grant no. 2012YZ06).