Optimal Inequalities among Various Means of Two Arguments

Abstract

We establish two optimal inequalities among power mean 𝑀_𝑝(𝑎,𝑏)=(𝑎^𝑝/2+𝑏^𝑝/2)^1/𝑝, arithmetic mean 𝐴(𝑎,𝑏)=(𝑎+𝑏)/2, logarithmic mean 𝐿(𝑎,𝑏)=(𝑎−𝑏)/(log𝑎−log𝑏), and geometric mean 𝐺(𝑎,𝑏)=√𝑎𝑏.

1. Introduction

For 𝑝∈𝑅, the power mean 𝑀_𝑝(𝑎,𝑏) of order 𝑝 of two positive numbers 𝑎 and 𝑏 is defined by

𝑀_𝑝(𝑎(𝑎,𝑏)=^𝑝+𝑏^𝑝2)^1/𝑝√,𝑝≠0,𝑎𝑏,𝑝=0. (1.1)

In the recent past, the power mean 𝑀_𝑝(𝑎,𝑏) has been the subject of intensive research. In particular, many remarkable inequalities for the mean can be found in literature [1–11]. It is well-known that 𝑀_𝑝(𝑎,𝑏) is continuous and strictly increasing with respect to 𝑝∈𝑅 for fixed 𝑎 and 𝑏.

If we denote by

1𝐼(𝑎,𝑏)=𝑒(𝑎^𝑎𝑏^𝑏)^{1/(𝑎−𝑏)},𝑏≠𝑎,𝑎,𝑏=𝑎,𝐿(𝑎,𝑏)=𝑏−𝑎log𝑏−log𝑎,𝑏≠𝑎,𝑎,𝑏=𝑎, (1.2)𝐴(𝑎,𝑏)=(𝑎+𝑏)/2, 𝐺(𝑎,𝑏)=√𝑎𝑏 and 𝐻(𝑎,𝑏)=2𝑎𝑏/(𝑎+𝑏) the identric mean, logarithmic mean, arithmetic mean, geometric mean and the harmonic mean, of two positive real numbers 𝑎 and 𝑏, respectively, then

min{𝑎,𝑏}≤𝐻(𝑎,𝑏)=𝑀₋₁(𝑎,𝑏)≤𝐺(𝑎,𝑏)=𝑀₀(𝑎,𝑏)≤𝐿(𝑎,𝑏)≤𝐼(𝑎,𝑏)≤𝐴(𝑎,𝑏)=𝑀₁(𝑎,𝑏)≤max{𝑎,𝑏} (1.3) with equality if and only if 𝑏=𝑎 in each inequality.

In [12], Alzer and Janous established the following sharp double inequality (see also [13, page 350]):

𝑀_log2/log32(𝑎,𝑏)≤31𝐴(𝑎,𝑏)+3𝐺(𝑎,𝑏)≤𝑀_2/3(𝑎,𝑏) (1.4) for all real numbers 𝑎,𝑏>0.

In [14], Mao proved

𝑀_1/31(𝑎,𝑏)≤32𝐴(𝑎,𝑏)+3𝐺(𝑎,𝑏)≤𝑀_1/2(𝑎,𝑏) (1.5) for all real numbers 𝑎,𝑏>0, and the constant 1/3 in the left side inequality cannot be improved.

In [15–17], the authors presented the bounds for 𝐿 and 𝐼 in terms of 𝐴 and 𝐺 as follows:

𝐺^2/3(𝑎,𝑏)𝐴^1/31(𝑎,𝑏)<𝐿(𝑎,𝑏)<32𝐴(𝑎,𝑏)+31𝐺(𝑎,𝑏),32𝐺(𝑎,𝑏)+3𝐴(𝑎,𝑏)<𝐼(𝑎,𝑏) (1.6) for all 𝑎,𝑏>0 with 𝑎≠𝑏.

Alzer [18] proved

𝐺(𝑎,𝑏)𝐴(𝑎,𝑏)<1𝐿(𝑎,𝑏)𝐼(𝑎,𝑏)<21(𝐿(𝑎,𝑏)+𝐼(𝑎,𝑏))<2(𝐺(𝑎,𝑏)+𝐴(𝑎,𝑏)). (1.7)

In [5], Alzer and Qiu established

𝛼𝐴(𝑎,𝑏)+(1−𝛼)𝐺(𝑎,𝑏)<𝐼(𝑎,𝑏)<𝛽𝐴(𝑎,𝑏)+(1−𝛽)𝐺(𝑎,𝑏) (1.8) for 𝛼≤2/3, 𝛽≥2/𝑒=0.73575… and 𝑎,𝑏>0 with 𝑎≠𝑏.

The main purpose of this paper is to present the optimal bounds for 𝐴^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏) and 𝐺^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏) for all 𝛼∈(0,1) in terms of the power mean 𝑀_𝑝(𝑎,𝑏). Moreover, two optimal inequalities among 𝐴(𝑎,𝑏), 𝐺(𝑎,𝑏), 𝐿(𝑎,𝑏) and 𝑀_𝑝(𝑎,𝑏) are proved.

2. Lemmas

In order to establish our main results we need two inequalities, which we present in this section.

Lemma 2.1. If 𝑟∈(0,1), then 𝑔(𝑡)=−[𝑡²+𝑡^(1+2𝑟)/3+(1−2𝑟)(𝑡^{(1+2𝑟)/3+1}+𝑡)]log𝑡+(1−𝑟)(𝑡^{(1+2𝑟)/3+2}−𝑡^(1+2𝑟)/3+𝑡²−1)>0 (2.1) for 𝑡∈(1,+∞).

Proof. Let 𝑝=(1+2𝑟)/3, 𝑔₁(𝑡)=𝑡^1−𝑝𝑔^′(𝑡), 𝑔₂(𝑡)=𝑡^𝑝𝑔^′₁(𝑡), 𝑔₃(𝑡)=𝑡^1−𝑝𝑔^′₂(𝑡), 𝑔₄(𝑡)=𝑡^𝑝𝑔^′₃(𝑡), 𝑔₅(𝑡)=𝑡^3−𝑝𝑔^′₄(𝑡), 𝑔₆(𝑡)=𝑡^𝑝𝑔^′₅(𝑡), 𝑔₇(𝑡)=𝑡^1−𝑝𝑔^′₆(𝑡), and 𝑔₈(𝑡)=𝑡^𝑝𝑔^′₇(𝑡), then simple computation yields 𝑔𝑔(1)=0,₁(𝑡)=−[2𝑡^2−𝑝+(1−2𝑟)𝑡^1−𝑝+(1−2𝑟)(𝑝+1)𝑡+𝑝]log𝑡+(1−𝑟)(𝑝+2)𝑡²−(1−2𝑟)𝑡+(1−2𝑟)𝑡^2−𝑝−(1−2𝑟)𝑡^1−𝑝𝑔−𝑝(1−𝑟)−1,₁𝑔(1)=0,₂(𝑡)=−[2(2−𝑝)𝑡+(1−2𝑟)(𝑝+1)𝑡^𝑝+(1−2𝑟)(1−𝑝)]log𝑡+2(1−𝑟)(2+𝑝)𝑡^1+𝑝−(1−2𝑟)(𝑝+2)𝑡^𝑝−𝑝𝑡^𝑝−1𝑔+(2𝑟𝑝−4𝑟−𝑝)𝑡−(1−2𝑟)(2−𝑝),₂𝑔(1)=0,₃(𝑡)=−[2(2−𝑝)𝑡^1−𝑝+𝑝(𝑝+1)(1−2𝑟)]log𝑡+𝑝(1−𝑝)𝑡⁻¹+(2𝑟𝑝−4𝑟+𝑝−4)𝑡^1−𝑝−(1−2𝑟)(1−𝑝)𝑡^−𝑝+2(1−𝑟)(2+𝑝)(1+𝑝)𝑡−(1−2𝑟)(𝑝²𝑔+3𝑝+1),₃𝑔(1)=6𝑝−2−4𝑟=0,₄(𝑡)=−2(1−𝑝)(2−𝑝)log𝑡+2(1−𝑟)(2+𝑝)(1+𝑝)𝑡^𝑝−𝑝(1+𝑝)(1−2𝑟)𝑡^𝑝−1+𝑝(𝑝−1)𝑡^𝑝−2+𝑝(1−𝑝)(1−2𝑟)𝑡⁻¹−𝑝²−2𝑟𝑝²𝑔+6𝑟𝑝−4𝑟+7𝑝−8,₄(1)=12𝑝−4−8𝑟=0, (2.2)(2.3)(2.4)(2.5)(2.6)𝑔₅(𝑡)=2𝑝(1−𝑟)(𝑝+1)(𝑝+2)𝑡²+𝑝(1−𝑝²)(1−2𝑟)𝑡−2(2−𝑝)(1−𝑝)𝑡^2−𝑝−𝑝(1−𝑝)(1−2𝑟)𝑡^1−𝑝𝑔+𝑝(𝑝−1)(𝑝−2),₅(1)=2𝑝³+2(1−4𝑟)𝑝²=8+4(3−𝑟)𝑝−427(−10𝑟³−15𝑟²𝑔+24𝑟+1)>0,(2.7)₆(𝑡)=4𝑝(𝑝+1)(𝑝+2)(1−𝑟)𝑡^𝑝+1+𝑝(1−𝑝²)(1−2𝑟)𝑡^𝑝−2(2−𝑝)²(1−𝑝)𝑡−𝑝(1−𝑝)²𝑔(1−2𝑟),₆(1)=4𝑝³+4(1−4𝑟)𝑝²=+8(3−𝑟)𝑝−81627(−10𝑟³−15𝑟²𝑔+24𝑟+1)>0,(2.8)₇(𝑡)=4𝑝(2+𝑝)(1+𝑝)²(1−𝑟)𝑡−2(1−𝑝)(2−𝑝)²𝑡^1−𝑝+𝑝²(1−𝑝²𝑔)(1−2𝑟),₇(1)=(3−2𝑟)𝑝⁴+2(9−8𝑟)𝑝³+11(1−2𝑟)𝑝²=1+8(3−𝑟)𝑝−881(−32𝑟⁵−400𝑟⁴−888𝑟³−412𝑟²𝑔+1576𝑟+156)>0,(2.9)₈(𝑡)=4𝑝(1+𝑝)²(2+𝑝)(1−𝑟)𝑡^𝑝−2(1−𝑝)²(2−𝑝)²𝑔,(2.10)₈(1)=2(1−2𝑟)𝑝⁴+4(7−4𝑟)𝑝³−2(3+10𝑟)𝑝²=8+8(4−𝑟)𝑝−881(−8𝑟⁵−60𝑟⁴−82𝑟³−79𝑟²+198𝑟+31)>0.(2.11)
From (2.10) we clearly see that is strictly increasing in 𝑔₈(𝑡). Therefore, Lemma 2.1 follows from (2.2)–(2.9) and (2.11) together with the monotonicity of (1,+∞).

Lemma 2.2. If 𝑔₈(𝑡), then 𝑟∈(0,1) for 𝑔(𝑡)=[𝑟𝑡^{(1−𝑟)/3+1}+(𝑟−2)𝑡^(1−𝑟)/3+(𝑟−2)𝑡+𝑟]log𝑡+2(1−𝑟)(𝑡^{(1−𝑟)/3+1}−𝑡^(1−𝑟)/3+𝑡−1)>0(2.12).

Proof. Let 𝑡∈(1,+∞), 𝑝=(1−𝑟)/3, 𝑔₁(𝑡)=𝑡^1−𝑝𝑔^′(𝑡), 𝑔₂(𝑡)=𝑡^𝑝𝑔^′₁(𝑡), 𝑔₃(𝑡)=𝑡^1−𝑝𝑔^′₂(𝑡), and 𝑔₄(𝑡)=𝑡^𝑝+2𝑔^′₃(𝑡), then simple computation leads to 𝑔₅(𝑡)=𝑡^1−𝑝𝑔^′₄(𝑡)
From (2.18) we clearly see that 𝑔𝑔(1)=0,(2.13)₁(𝑡)=[𝑟(𝑝+1)𝑡+(𝑟−2)𝑡^1−𝑝+𝑝(𝑟−2)]log𝑡−𝑟𝑡^1−𝑝+𝑟𝑡^−𝑝𝑔+(2+2𝑝−2𝑝𝑟−𝑟)𝑡+2𝑝𝑟−2𝑝+𝑟−2,₁𝑔(1)=0,(2.14)₂(𝑡)=[𝑟(𝑝+1)𝑡^𝑝+(1−𝑝)(𝑟−2)]log𝑡+(2𝑝+2−𝑟𝑝)𝑡^𝑝+𝑝(𝑟−2)𝑡^𝑝−1−𝑟𝑝𝑡⁻¹𝑔+𝑝𝑟−2,₂𝑔(1)=0,(2.15)₃(𝑡)=𝑝𝑟(𝑝+1)log𝑡+(𝑟−2)(1−𝑝)𝑡^−𝑝+𝑟𝑝𝑡^−𝑝−1+𝑝(𝑝−1)(𝑟−2)𝑡⁻¹+2𝑝²−𝑟𝑝²𝑔+𝑝𝑟+2𝑝+𝑟,₃𝑔(1)=6𝑝+2𝑟−2=0,(2.16)₄(𝑡)=𝑝(𝑝+1)𝑟𝑡^𝑝+1+𝑝(1−𝑝)(𝑟−2)𝑡^𝑝𝑔−𝑝(1−𝑝)(𝑟−2)𝑡−𝑝(𝑝+1)𝑟,₄𝑔(1)=0,(2.17)₅(𝑡)=𝑟𝑝(1+𝑝)²𝑡+𝑝(1−𝑝)(2−𝑟)𝑡^1−𝑝+𝑝²𝑔(1−𝑝)(𝑟−2),(2.18)₅(1)=2𝑝(𝑝²=+2𝑟𝑝−2𝑝+1)2(1−𝑟)27(−5𝑟²+10𝑟+4)>0.(2.19) is strictly increasing in 𝑔₅(𝑡). Therefore, Lemma 2.2 follows from (2.13)–(2.17) and (2.19) together with the monotonicity of (1,+∞).

3. Main Results

Theorem 3.1. If 𝑔₅(𝑡), then 𝛼∈(0,1) with equality if and only if 𝐴^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏)≤𝑀_(1+2𝛼)/3(𝑎,𝑏)(3.1), and the parameter 𝑎=𝑏 cannot be improved.

Proof. If (1+2𝛼)/3, then we clearly see that 𝑎=𝑏
If 𝑎=𝐴^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏)=𝑀_(1+2𝛼)/3(𝑎,𝑏)=𝑏.(3.2), then without loss of generality we assume that 𝑎≠𝑏 and let 𝑎>𝑏; hence elementary calculation yields
𝑡=𝑎/𝑏>1
Let
𝑀_(1+2𝛼)/3(𝑎,𝑏)−𝐴^𝛼(𝑎,𝑏)𝐿^1−𝛼𝑡(𝑎,𝑏)=𝑏[(^(1+2𝛼)/3+12)^3/(1+2𝛼)−(1+𝑡2)^𝛼(𝑡−1)log𝑡^1−𝛼]. (3.3) then 3𝑓(𝑡)=1+2𝛼log(𝑡^(1+2𝛼)/33+1)−1+2𝛼log2−𝛼log𝑡+12+𝛼log2−(1−𝛼)log(𝑡−1)+(1−𝛼)log(log𝑡), (3.4) where lim_𝑡→1𝑓𝑓(𝑡)=0,_′(𝑡)=𝑔(𝑡)𝑡(𝑡+1)(𝑡−1)(𝑡^(1+2𝛼)/3,+1)log𝑡 (3.5)(3.6)
From Lemma 2.1 and (3.6) we know that
𝑔(𝑡)=−[𝑡²+𝑡^(1+2𝛼)/3+(1−2𝛼)(𝑡^{(1+2𝛼)/3+1}+𝑡)]log𝑡+(1−𝛼)(𝑡^{(1+2𝛼)/3+2}−𝑡^(1+2𝛼)/3+𝑡²−1). (3.7) for 𝑓^′(𝑡)>0(3.8).
Therefore, we get
𝑡∈(1,+∞) for 𝑀_(1+2𝛼)/3(𝑎,𝑏)>𝐴^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏)(3.9) that follows from (3.3)–(3.5) and (3.8).
Next, we prove that the parameter 𝑎>𝑏 cannot be improved.
For any (1+2𝛼)/3, let 0<𝜀<(1+2𝛼)/3, then (1.1) leads to
0<𝑡<1 where 𝐴^𝛼(𝑡+1,1)𝐿^1−𝛼(𝑡+1,1)−𝑀_{(1+2𝛼)/3−𝜀}𝑡(𝑡+1,1)=(1+2)^𝛼𝑡⋅[]log(𝑡+1)^1−𝛼−[(𝑡+1)^{(1+2𝛼)/3−𝜀}+12]^{1/((1+2𝛼)/3−𝜀)}=𝑓₁(𝑡)[log(1+𝑡)]^1−𝛼, (3.10)
Making use of the Taylor expansion we get
𝑓₁𝑡(𝑡)=(1+2)^𝛼⋅𝑡^1−𝛼−[(𝑡+1)^{(1+2𝛼−3𝜀)/3}+12]^{3/(1+2𝛼−3𝜀)}[log(1+𝑡)]^1−𝛼. (3.11)
Equations (3.10) and (3.12) imply that for any 𝑓₁𝜀(𝑡)=[8𝑡²+𝑜(𝑡²)]⋅𝑡^1−𝛼.(3.12) and 𝛼∈(0,1), there exists 0<𝜀<(1+2𝛼)/3, such that
0<𝛿₁(𝜀,𝛼)<1 for 𝐴^𝛼(1,1+𝑡)𝐿^1−𝛼(1,1+𝑡)>𝑀_{(1+2𝛼)/3−𝜀}(1,1+𝑡)(3.13).

Remark 3.2. For any 𝑡∈(0,𝛿₁) we have 0<𝛼<1 for all 𝑀₀(𝑎,𝑏)≤𝐴^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏)(3.14), with equality if and only if 𝑎,𝑏>0, and the parameter 𝑎=𝑏 in the lower bound cannot be improved.
In fact, if 0, then we clearly see that 𝑎=𝑏. If 𝑀₀(𝑎,𝑏)=𝐴^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏)=𝑎, then 𝑎≠𝑏 follows from 𝑀₀(𝑎,𝑏)<𝐴^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏).
Next, we prove that the parameter 𝑀₀(𝑎,𝑏)<𝐿(𝑎,𝑏)<𝐴(𝑎,𝑏) in the lower bound cannot be improved.
For any 0, we have
𝜀>0
Equation (3.15) implies that for any lim_𝑡→∞𝑀_𝜀(𝑡,1)𝐴^𝛼(𝑡,1)𝐿^1−𝛼(𝑡,1)=lim_𝑡→∞((𝑡^𝜀+1)/2)^1/𝜀((𝑡+1)/2)^𝛼((𝑡−1)/log𝑡)^1−𝛼=lim_𝑡→∞((1+𝑡^−𝜀)/2)^1/𝜀(log𝑡)^1−𝛼((1+𝑡⁻¹)/2)^𝛼(1−𝑡⁻¹)^1−𝛼=+∞.(3.15), there exists 𝜀>0, such that
𝑇₁=𝑇₁(𝜀)>1 for 𝐴^𝛼(𝑡,1)𝐿^1−𝛼(𝑡,1)<𝑀_𝜀(𝑡,1)(3.16).

Theorem 3.3. If 𝑡∈(𝑇₁,+∞), then 𝛼∈(0,1) for all 𝐺^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏)≤𝑀_(1−𝛼)/3(𝑎,𝑏)(3.17), with equality if and only if 𝑎,𝑏>0, and the parameter 𝑎=𝑏 cannot be improved.

Proof. If (1−𝛼)/3, then we clearly see that 𝑎=𝑏
If 𝑎=𝐺^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏)=𝑀_(1−𝛼)/3(𝑎,𝑏)=𝑏.(3.18), then without loss of generality, we assume that 𝑎≠𝑏, let 𝑎>𝑏; hence simple computation leads to
𝑡=𝑎/𝑏>1 Let 𝑀_(1−𝛼)/3(𝑎,𝑏)−𝐺^𝛼(𝑎,𝑏)𝐿^1−𝛼𝑡(𝑎,𝑏)=𝑏[(^(1−𝛼)/3+12)^3/(1−𝛼)−𝑡^𝛼/2(𝑡−1)log𝑡^1−𝛼]. (3.19) then 3ℎ(𝑡)=1−𝛼log(𝑡^(1−𝛼)/33+1)−1−𝛼log2−𝛼log𝑡+12+𝛼log2−(1−𝛼)log(𝑡−1)+(1−𝛼)log(log𝑡), (3.20) where lim_𝑡→1ℎℎ(𝑡)=0,_′(𝑡)=𝑔(𝑡)2𝑡(𝑡−1)(𝑡^(1−𝛼)/3,+1)log𝑡 (3.21)(3.22)
From Lemma 2.2 and (3.22) we know that
𝑔(𝑡)=[𝛼𝑡^{(1−𝛼)/3+1}+(𝛼−2)𝑡^(1−𝛼)/3+(𝛼−2)𝑡+𝛼]log𝑡+2(1−𝛼)(𝑡^{(1−𝛼)/3+1}−𝑡^(1−𝛼)/3+𝑡−1). (3.23) for ℎ^′(𝑡)>0(3.24).
Therefore, we get
𝑡∈(1,+∞) for 𝑀_(1−𝛼)/3(𝑎,𝑏)>𝐺^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏)(3.25) that follows from (3.19)–(3.21) and (3.24).
Next, we prove that the constants 𝑎>𝑏 cannot be improved.
For any (1−𝛼)/3, let 0<𝜀<(1−𝛼)/3, then (1.1) leads to
0<𝑡<1 where 𝐺^𝛼(𝑡+1,1)𝐿^1−𝛼(𝑡+1,1)−𝑀_{(1−𝛼)/3−𝜀}(𝑡+1,1)=(1+𝑡)^𝛼/2𝑡⋅[]log(𝑡+1)^1−𝛼−[(𝑡+1)^{(1−𝛼)/3−𝜀}+12]^{1/((1−𝛼)/3−𝜀)}=ℎ₁(𝑡)[log(1+𝑡)]^1−𝛼, (3.26)
Making use of the Taylor expansion we get
ℎ₁(𝑡)=(1+𝑡)^𝛼/2⋅𝑡^1−𝛼−[(𝑡+1)^{(1−𝛼−3𝜀)/3}+12]^{3/(1−𝛼−3𝜀)}[log(1+𝑡)]^1−𝛼. (3.27)
Equations (3.26) and (3.28) imply that for any ℎ₁𝜀(𝑡)=[8𝑡²+𝑜(𝑡²)]⋅𝑡^1−𝛼.(3.28) and 𝛼∈(0,1), there exists 0<𝜀<(1−𝛼)/3, such that
0<𝛿₂(𝜀,𝛼)<1 for 𝐴^𝛼(1,1+𝑡)𝐿^1−𝛼(1,1+𝑡)>𝑀_{(1−𝛼)/3−𝜀}(1,1+𝑡)(3.29).

Remark 3.4. For any 𝑡∈(0,𝛿₂), we have 𝜀>0 Therefore, (3.30) implies that inequality lim_𝑡→∞𝑀_𝜀(𝑡,1)𝐺^𝛼(𝑡,1)𝐿^1−𝛼(𝑡,1)=lim_𝑡→∞((𝑡^𝜀+1)/2)^1/𝜀𝑡^𝛼/2((𝑡−1)/log𝑡)^1−𝛼=lim_𝑡→∞((1+𝑡^−𝜀)/2)^1/𝜀𝑡^𝛼/2(log𝑡)^1−𝛼(1−𝑡⁻¹)^1−𝛼=+∞. (3.30) holds for all 𝑀₀(𝑎,𝑏)≤𝐺^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏)(3.31) and 𝛼∈(0,1), with equality if and only if 𝑎,𝑏>0 and the parameter 𝑎=𝑏 in the lower bound cannot be improved.

Acknowledgments

This work was supported by the National Natural Science Foundation of China (Grant no. 60850005) and the Natural Science Foundation of Zhejiang Province (Grant no. D7080080 and no. Y607128).