A Best Possible Double Inequality for Power Mean

Abstract

We answer the question: for any 𝑝,𝑞∈ℝ with 𝑝≠𝑞 and 𝑝≠−𝑞, what are the greatest value 𝜆=𝜆(𝑝,𝑞) and the least value 𝜇=𝜇(𝑝,𝑞), such that the double inequality 𝑀_𝜆(𝑎,𝑏)<𝑀_𝑝(𝑎,𝑏)𝑀_𝑞(𝑎,𝑏)<𝑀_𝜇(𝑎,𝑏) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏? Where 𝑀_𝑝(𝑎,𝑏) is the 𝑝th power mean of two positive numbers 𝑎 and 𝑏.

1. Introduction

For 𝑝∈ℝ, the 𝑝th power mean 𝑀_𝑝(𝑎,𝑏) of two positive numbers 𝑎 and 𝑏 is defined by 𝑀_𝑝()=𝑎𝑎,𝑏^𝑝+𝑏^𝑝2^1/𝑝√,𝑝≠0,𝑎𝑏,𝑝=0. (1.1)

It is well known that 𝑀_𝑝(𝑎,𝑏) is continuous and strictly increasing with respect to 𝑝∈ℝ for fixed 𝑎, 𝑏>0 with 𝑎≠𝑏. Many classical means are special case of the power mean, for example, 𝑀₋₁(𝑎,𝑏)=𝐻(𝑎,𝑏)=2𝑎𝑏/(𝑎+𝑏), 𝑀₀(𝑎,𝑏)=𝐺(𝑎,𝑏)=√𝑎𝑏, and 𝑀₁(𝑎,𝑏)=𝐴(𝑎,𝑏)=(𝑎+𝑏)/2 are the harmonic, geometric, and arithmetic means of 𝑎 and 𝑏, respectively. Recently, the power mean has been the subject of intensive research. In particular, many remarkable inequalities and properties for the power mean can be found in literature [1–15].

Let 𝐿(𝑎,𝑏)=(𝑎−𝑏)/(log𝑎−log𝑏) and 𝐼(𝑎,𝑏)=1/𝑒(𝑎^𝑎/𝑏^𝑏)^{1/(𝑎−𝑏)} be the logarithmic and identric means of two positive numbers 𝑎 and 𝑏 with 𝑎≠𝑏, respectively. Then it is well known that 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) (1.2) for all 𝑎,𝑏>0 with 𝑎≠𝑏.

In [16–22], the authors presented the sharp power mean bounds for 𝐿, 𝐼, (𝐼𝐿)^1/2, and (𝐿+𝐼)/2 as follows: 𝑀₀(𝑎,𝑏)<𝐿(𝑎,𝑏)<𝑀_1/3(𝑎,𝑏),𝑀_2/3(𝑎,𝑏)<𝐼(𝑎,𝑏)<𝑀_log2(),𝑀𝑎,𝑏₀(𝑎,𝑏)<𝐿(𝑎,𝑏)𝐼(𝑎,𝑏)<𝑀_1/2(),1𝑎,𝑏2(𝐿(𝑎,𝑏)+𝐼(𝑎,𝑏))<𝑀_1/2(𝑎,𝑏) (1.3) for all 𝑎,𝑏>0 with 𝑎≠𝑏.

Alzer and Qiu [12] proved that the inequality 12(𝐿(𝑎,𝑏)+𝐼(𝑎,𝑏))>𝑀_𝑝(𝑎,𝑏) (1.4) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝑝≤log2/(1+log2)=0.40938….

The following sharp bounds for the sum 𝛼𝐴(𝑎,𝑏)+(1−𝛼)𝐿(𝑎,𝑏), and the products 𝐴^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏) and 𝐺^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏) in terms of power means were proved in [5, 8] as follows: 𝑀_{log2/(log2−log𝛼)}(𝑎,𝑏)<𝛼𝐴(𝑎,𝑏)+(1−𝛼)𝐿(𝑎,𝑏)<𝑀_(1+2𝛼)/3(),𝑀𝑎,𝑏₀(𝑎,𝑏)<𝐴^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏)<𝑀_(1+2𝛼)/3(),𝑀𝑎,𝑏₀(𝑎,𝑏)<𝐺^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏)<𝑀_(1−𝛼)/3(𝑎,𝑏) (1.5) for any 𝛼∈(0,1) and all 𝑎,𝑏>0 with 𝑎≠𝑏.

In [2, 7], the authors answered the question: for any 𝛼∈(0,1), what are the greatest values 𝑝₁=𝑝₁(𝛼),  𝑝₂=𝑝₂(𝛼),   𝑝₃=𝑝₃(𝛼), and 𝑝₄=𝑝₄(𝛼), and the least values 𝑞₁=𝑞₁(𝛼), 𝑞₂=𝑞₂(𝛼),  𝑞₃=𝑞₃(𝛼), and 𝑞₄=𝑞₄(𝛼), such that the inequalities 𝑀_𝑝1(𝑎,𝑏)<𝑃^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏)<𝑀_𝑞1(),𝑀𝑎,𝑏_𝑝2(𝑎,𝑏)<𝐴^𝛼(𝑎,𝑏)𝐺^1−𝛼(𝑎,𝑏)<𝑀_𝑞2(),𝑀𝑎,𝑏_𝑝3(𝑎,𝑏)<𝐺^𝛼(𝑎,𝑏)𝐻^1−𝛼(𝑎,𝑏)<𝑀_𝑞3(),𝑀𝑎,𝑏_𝑝4(𝑎,𝑏)<𝐴^𝛼(𝑎,𝑏)𝐻^1−𝛼(𝑎,𝑏)<𝑀_𝑞4(𝑎,𝑏) (1.6) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏?

In [4], the authors presented the greatest value 𝑝=𝑝(𝛼,𝛽) and the least value 𝑞=𝑞(𝛼,𝛽) such that the double inequality 𝑀_𝑝(𝑎,𝑏)<𝐴^𝛼(𝑎,𝑏)𝐺^𝛽(𝑎,𝑏)𝐻^{1−𝛼−𝛽}(𝑎,𝑏)<𝑀_𝑞(𝑎,𝑏) (1.7) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 and 𝛼,𝛽>0 with 𝛼+𝛽<1.

It is the aim of this paper to answer the question: for any 𝑝, 𝑞∈ℝ with 𝑝≠𝑞 and 𝑝≠−𝑞, what are the greatest value 𝜆=𝜆(𝑝,𝑞) and the least value 𝜇=𝜇(𝑝,𝑞), such that the double inequality 𝑀_𝜆(𝑎,𝑏)<𝑀_𝑝(𝑎,𝑏)𝑀_𝑞(𝑎,𝑏)<𝑀_𝜇(𝑎,𝑏) (1.8) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏?

2. Main Result

In order to establish our main result, we need a lemma which we present in this section.

Lemma 2.1. Let 𝑝, 𝑞≠0,  𝑝≠𝑞 and 𝑥>1. Then 𝑀_𝑝(𝑥,1)𝑀_𝑞(𝑥,1)<𝑀²_{(𝑝+𝑞)/2}(𝑥,1) (2.1) for 𝑝+𝑞>0, and 𝑀_𝑝(𝑥,1)𝑀_𝑞(𝑥,1)>𝑀²_{(𝑝+𝑞)/2}(𝑥,1) (2.2) for 𝑝+𝑞<0.

Proof. From (1.1), we have log𝑀_𝑝(𝑥,1)𝑀_𝑞(𝑥,1)−log𝑀²_{(𝑝+𝑞)/2}()=1𝑥,1𝑝log1+𝑥^𝑝2+1𝑞log1+𝑥^𝑞2−4𝑝+𝑞log1+𝑥^{(𝑝+𝑞)/2}2. (2.3)
Let 𝑓(𝑥)=1𝑝log1+𝑥^𝑝2+1𝑞log1+𝑥^𝑞2−4𝑝+𝑞log1+𝑥^{(𝑝+𝑞)/2}2, (2.4) then simple computations lead to 𝑓(1)𝑓=0,_′(𝑥)=1−𝑥^{(𝑝+𝑞)/2}𝑥^𝑝/2−𝑥^𝑞/22𝑥(1+𝑥^𝑝)(1+𝑥^𝑞)1+𝑥^{(𝑝+𝑞)/2}. (2.5)(2.6)
Equation (2.6) implies that 𝑓^′(𝑥)<0 (2.7) for 𝑝+𝑞>0, and 𝑓^′(𝑥)>0 (2.8) for 𝑝+𝑞<0.
Therefore, inequality (2.1) follows from (2.3)–(2.5) and inequality (2.7), and inequality (2.2) follows from (2.3)–(2.5) and inequality (2.8).
Let 𝐸₀=𝑝,𝑞∈ℝ²∶𝑝=𝑞,𝐸^′₀=𝑝,𝑞∈ℝ²,𝐸∶𝑝=−𝑞₁=𝑝,𝑞∈ℝ²∶𝑝,𝑞>0,𝑝>𝑞,𝐸^′₁=𝑝,𝑞∈ℝ²,𝐸∶𝑝,𝑞>0,𝑝<𝑞₂=𝑝,𝑞∈ℝ²∶𝑝,𝑞<0,𝑝>𝑞,𝐸^′₂=𝑝,𝑞∈ℝ²,𝐸∶𝑝,𝑞<0,𝑝<𝑞₃=𝑝,𝑞∈ℝ²∶𝑝>0,𝑞=0,𝐸^′₃=𝑝,𝑞∈ℝ²,𝐸∶𝑝=0,𝑞>0₄=𝑝,𝑞∈ℝ²∶𝑝>0,𝑞<0,𝑝+𝑞>0,𝐸^′₄=𝑝,𝑞∈ℝ²,𝐸∶𝑝<0,𝑞>0,𝑝+𝑞>0₅=𝑝,𝑞∈ℝ²∶𝑝=0,𝑞<0,𝐸^′₅=𝑝,𝑞∈ℝ²,𝐸∶𝑝<0,𝑞=0₆=𝑝,𝑞∈ℝ²∶𝑝>0,𝑞<0,𝑝+𝑞<0,𝐸^′₆=𝑝,𝑞∈ℝ²∶𝑝<0,𝑞>0,𝑝+𝑞<0. (2.9)
Then we clearly see that ℝ²=⋃⁶_𝑖=0𝐸_𝑖⋃⁶_𝑖=0𝐸^′_𝑖, and it is not difficult to verify that the identity 𝑀_𝑝(𝑎,𝑏)𝑀_𝑞(𝑎,𝑏)=𝑀_{(𝑝+𝑞)/2}(𝑎,𝑏) holds for all 𝑎, 𝑏>0 if (𝑝,𝑞)∈𝐸₀⋃𝐸^′₀. Let 𝜆=2𝑝𝑞,𝑝+𝑞𝑝,𝑞∈𝐸₁⋃𝐸^′₁,𝑝+𝑞2,𝑝,𝑞∈𝐸₂⋃𝐸^′₂⋃𝐸₅⋃𝐸^′₅⋃𝐸₆⋃𝐸^′₆,0,𝑝,𝑞∈𝐸₃⋃𝐸^′₃⋃𝐸₄⋃𝐸^′₄,𝜇=2𝑝𝑞,𝑝+𝑞𝑝,𝑞∈𝐸₂⋃𝐸^′₂,𝑝+𝑞2,𝑝,𝑞∈𝐸₁⋃𝐸^′₁⋃𝐸₃⋃𝐸^′₃⋃𝐸₄⋃𝐸^′₄,0,𝑝,𝑞∈𝐸₅⋃𝐸^′₅⋃𝐸₆⋃𝐸^′₆. (2.10)

Then we have Theorem 2.2 as follows.

Theorem 2.2. The double inequality 𝑀_𝜆(𝑎,𝑏)<𝑀_𝑝(𝑎,𝑏)𝑀_𝑞(𝑎,𝑏)<𝑀_𝜇(𝑎,𝑏) (2.11) holds for all 𝑎, 𝑏>0 with 𝑎≠𝑏, and 𝑀_𝜆(𝑎,𝑏) and 𝑀_𝜇(𝑎,𝑏) are the best possible lower and upper power mean bounds for the geometric mean of 𝑀_𝑝(𝑎,𝑏) and 𝑀_𝑞(𝑎,𝑏).

Proof. From (1.1), we clearly see that 𝑀_𝑝(𝑎,𝑏) is symmetric and homogenous of degree 1. Without loss of generality, we assume that 𝑏=1, 𝑎=𝑥>1 and 𝑝>𝑞. We divide the proof of inequality (2.11) into three cases.
Case 1. (𝑝,𝑞)∈𝐸₁⋃𝐸₂. Then from Lemma 2.1, we clearly see that 𝑀_𝑝(𝑥,1)𝑀_𝑞(𝑥,1)<𝑀_{(𝑝+𝑞)/2}(𝑥,1) (2.12) for (𝑝,𝑞)∈𝐸₁, and 𝑀_𝑝(𝑥,1)𝑀_𝑞(𝑥,1)>𝑀_{(𝑝+𝑞)/2}(𝑥,1) (2.13) for (𝑝,𝑞)∈𝐸₂.
From (1.1), we get log𝑀_𝑝(𝑥,1)𝑀_𝑞(𝑥,1)−log𝑀²_{2𝑝𝑞/(𝑝+𝑞)}()=1𝑥,1𝑝log1+𝑥^𝑝2+1𝑞log1+𝑥^𝑞2−𝑝+𝑞𝑝𝑞log1+𝑥^{2𝑝𝑞/(𝑝+𝑞)}2. (2.14)
Let 𝐹(𝑥)=1𝑝log1+𝑥^𝑝2+1𝑞log1+𝑥^𝑞2−𝑝+𝑞𝑝𝑞log1+𝑥^{2𝑝𝑞/(𝑝+𝑞)}2, (2.15) then simple computations lead to 𝐹(1)𝐹=0,_′(𝑥)=𝑥^𝑞𝐺(𝑥)𝑥(1+𝑥^𝑝)(1+𝑥^𝑞)1+𝑥^{2𝑝𝑞/(𝑝+𝑞)}, (2.16)(2.17) where 𝐺(𝑥)=𝑥^𝑝−𝑞−𝑥^{(2𝑝𝑞+𝑝2−𝑞2)/(𝑝+𝑞)}+2𝑥^𝑝−𝑥^{2𝑝𝑞/(𝑝+𝑞)}−2𝑥^{𝑞(𝑝−𝑞)/(𝑝+𝑞)}𝐺(1)𝐺+1,=0,_′(𝑥)=𝑥^{(𝑝𝑞−𝑞2−𝑝−𝑞)/(𝑝+𝑞)}𝐻(𝑥), (2.18)(2.19)(2.20) where 𝐻(𝑥)=𝑝−𝑞𝑥^{𝑝(𝑝−𝑞)/(𝑝+𝑞)}−2𝑝𝑞+𝑝²−𝑞²𝑥𝑝+𝑞^𝑝+2𝑝𝑥^{(𝑝2+𝑞2)/(𝑝+𝑞)}−2𝑝𝑞𝑥𝑝+𝑞^𝑞−2𝑞𝑝−𝑞,𝐻(1)=2𝑝+𝑞𝑝−𝑞²,𝐻𝑝+𝑞_′(𝑥)=𝑝𝑥𝑝+𝑞^𝑝−1𝐼(𝑥), (2.21)(2.22)(2.23) where 𝐼(𝑥)=𝑝−𝑞²𝑥^{−2𝑝𝑞/(𝑝+𝑞)}+2𝑝²+𝑞²𝑥^{−𝑞(𝑝−𝑞)/(𝑝+𝑞)}−2𝑞²𝑥^{−(𝑝−𝑞)}−2𝑝𝑞−𝑝²+𝑞²,𝐼(1)=2𝑝−𝑞²,𝐼^′(𝑥)=2𝑞𝑝−𝑞𝑥𝑝+𝑞^{(𝑞2−𝑝𝑞−𝑝−𝑞)/(𝑝+𝑞)}𝐽(𝑥), (2.24)(2.25)(2.26) where 𝐽(𝑥)=−𝑝𝑝−𝑞𝑥^−𝑞+𝑞𝑝+𝑞𝑥^{−𝑝(𝑝−𝑞)/(𝑝+𝑞)}−𝑝²−𝑞²,𝐽(1),𝐽=−2𝑝𝑝−𝑞_′(𝑥)=𝑝𝑞𝑝−𝑞𝑥^−𝑞−11−𝑥^{(𝑞2−𝑝2+2𝑝𝑞)/(𝑝+𝑞)}. (2.27)(2.28)(2.29)
If (𝑝,𝑞)∈𝐸₁, then (2.15), (2.18), (2.21), (2.22), (2.24), (2.25), (2.27), and (2.28) lead to lim_𝑥→+∞𝐹(𝑥)=0,lim_𝑥→+∞𝐺(𝑥)=−∞,lim_𝑥→+∞𝐻(𝑥)𝐻(1)=−∞,>0,lim_𝑥→+∞𝐼(𝑥)=−2𝑝𝑞−𝑝²+𝑞²𝐼(1)<0,>0,lim_𝑥→+∞𝐽(𝑥)=−𝑝²+𝑞²𝐽(1)<0,<0. (2.30)(2.31)(2.32)(2.33)(2.34)(2.35)(2.36)(2.37)
We divide the discussion into two subcases.
Subcase 1.1. (𝑝,𝑞)∈𝐸₁. Then (2.26) and (2.29) together with inequalities (2.36) and (2.37) imply that 𝐼(𝑥) is strictly decreasing in [1,+∞). In fact, if (𝑞²−𝑝²+2𝑝𝑞)/(𝑝+𝑞)≥0, then (2.29) and inequality (2.37) imply that 𝐽(𝑥)<0 for 𝑥∈[1,+∞). If (𝑞²−𝑝²+2𝑝𝑞)/(𝑝+𝑞)<0, then (2.29) and inequality (2.36) lead to the conclusion that 𝐽(𝑥)<0 for 𝑥∈[1,+∞).
From inequalities (2.34) and (2.35) together with the monotonicity of 𝐼(𝑥), we know that there exists 𝜆₁>1 such that 𝐼(𝑥)>0 for 𝑥∈[1,𝜆₁) and 𝐼(𝑥)<0 for 𝑥∈(𝜆₁,+∞). Then (2.23) leads to the conclusion that 𝐻(𝑥) is strictly increasing in [1,𝜆₁] and strictly decreasing in [𝜆₁,+∞).
It follows from (2.32) and (2.33) together with the piecewise monotonicity of 𝐻(𝑥) that there exists 𝜆₂>𝜆₁>1 such that 𝐻(𝑥)>0 for [1,𝜆₂] and 𝐻(𝑥)<0 for (𝜆₂,+∞). Then (2.20) leads to the conclusion that 𝐺(𝑥) is strictly increasing in [1,𝜆₂] and strictly decreasing in [𝜆₂,+∞).
From (2.17), (2.19) and (2.31) together with the piecewise monotonicity of 𝐺(𝑥), we clearly see that there exists 𝜆₃>𝜆₂>1 such that 𝐹(𝑥) is strictly increasing in [1,𝜆₃] and strictly decreasing in [𝜆₃,+∞).
Therefore, 𝑀_𝑝(𝑥,1)𝑀_𝑞(𝑥,1)>𝑀_{2𝑝𝑞/(𝑝+𝑞)}(𝑥,1) follows from (2.14)–(2.16) and (2.30) together with the piecewise monotonicity of 𝐹(𝑥).
Subcase 1.2.   (𝑝,𝑞)∈𝐸₂. Then (2.30) and (2.35) again hold, and (2.18), (2.21), (2.22), and (2.28) lead to lim_𝑥→+∞𝐺(𝑥)=+∞,lim_𝑥→+∞𝐻(𝑥)𝐻(1)𝐽(1)=+∞,<0,>0. (2.38)(2.39)(2.40)(2.41)
It follows from (2.29) and inequalities (𝑞²−𝑝²+2𝑝𝑞)/(𝑝+𝑞)<0 and (2.41) that 𝐽(𝑥)>0 for 𝑥∈[1,+∞). Then (2.26) and inequality (2.35) lead to the conclusion that 𝐼(𝑥)>0 for 𝑥∈[1,+∞). Therefore, 𝐻(𝑥) is strictly increasing in [1,+∞) follows from (2.23).
It follows from (2.20) and (2.39) together with inequality (2.40) and the monotonicity of 𝐻(𝑥) that there exists 𝜇₁>1 such that 𝐺(𝑥) is strictly decreasing in [1,𝜇₁] and strictly increasing in [𝜇₁,+∞).
From (2.17), (2.19) and (2.38) together with the piecewise monotonicity of 𝐺(𝑥), we clearly see that there exists 𝜇₂>𝜇₁>1 such that 𝐹(𝑥) is strictly decreasing in [1,𝜇₂] and strictly increasing in [𝜇₂,+∞).
Therefore, 𝑀_𝑝(𝑥,1)𝑀_𝑞(𝑥,1)<𝑀_{2𝑝𝑞/(𝑝+𝑞)}(𝑥,1) follows from (2.14)–(2.16) and (2.30) together with the piecewise monotonicity of 𝐹(𝑥).
Case 2. (𝑝,𝑞)∈𝐸₃⋃𝐸₅. Clearly, we have 𝑀₀(𝑥,1)<𝑀_𝑝(𝑥,1)𝑀_𝑞(𝑥,1) for (𝑝,𝑞)∈𝐸₃ and 𝑀₀(𝑥,1)>𝑀_𝑝(𝑥,1)𝑀_𝑞(𝑥,1) for (𝑝,𝑞)∈𝐸₅. Therefore, we need only to prove that 𝑀₀(𝑥,1)𝑀_𝑟(𝑥,1)<𝑀_𝑟/2(𝑥,1) (2.42) for 𝑟>0, and 𝑀₀(𝑥,1)𝑀_𝑟(𝑥,1)>𝑀_𝑟/2(𝑥,1) (2.43) for 𝑟<0.
From (1.1), one has log𝑀₀(𝑥,1)𝑀_𝑟(𝑥,1)−log𝑀²_𝑟/2()=1𝑥,121log𝑥+𝑟log1+𝑥^𝑟2−4𝑟log1+𝑥^𝑟/22. (2.44)
Let 𝑓(𝑥)=121log𝑥+𝑟log1+𝑥^𝑟2−4𝑟log1+𝑥^𝑟/22, (2.45) then simple computations lead to 𝑓(1)𝑓=0,_′(𝑥)𝑥=−^𝑟/2−1³2𝑥(1+𝑥^𝑟)1+𝑥^𝑟/2. (2.46)(2.47)
If 𝑟>0 (or 𝑟<0, resp.), then (2.47) leads to the conclusion that 𝑓(𝑥) is strictly decreasing (or increasing, resp.) in [1,+∞). Therefore, inequalities (2.42) and (2.43) follow from (2.44)–(2.46) and the monotonicity of 𝑓(𝑥).
Case 3.(𝑝,𝑞)∈𝐸₄⋃𝐸₆. Then from Lemma 2.1, we clearly see that 𝑀_{(𝑝+𝑞)/2}(𝑥,1)>𝑀_𝑝(𝑥,1)𝑀_𝑞(𝑥,1) for (𝑝,𝑞)∈𝐸₄ and 𝑀_𝑝(𝑥,1)𝑀_𝑞(𝑥,1)>𝑀_{(𝑝+𝑞)/2}(𝑥,1) for (𝑝,𝑞)∈𝐸₆. Therefore, we need only to prove that 𝑀_𝑝(𝑥,1)𝑀_𝑞(𝑥,1)>𝑀₀(𝑥,1) (2.48) for (𝑝,𝑞)∈𝐸₄, and 𝑀_𝑝(𝑥,1)𝑀_𝑞(𝑥,1)<𝑀₀(𝑥,1) (2.49) for (𝑝,𝑞)∈𝐸₆.
From (1.1), we get log𝑀_𝑝(𝑥,1)𝑀_𝑞(𝑥,1)−log𝑀²₀()=1𝑥,1𝑝log1+𝑥^𝑝2+1𝑞log1+𝑥^𝑞2−log𝑥. (2.50)
Let 𝑓(𝑥)=1𝑝log1+𝑥^𝑝2+1𝑞log1+𝑥^𝑞2−log𝑥, (2.51) then simple computations lead to 𝑓(1)𝑓=0,_′(𝑥)=𝑥^𝑝+𝑞−1𝑥(1+𝑥^𝑝)(1+𝑥^𝑞). (2.52)(2.53)
If (𝑝,𝑞)∈𝐸₄ (or 𝐸₆, resp.), then (2.53) implies that 𝑓(𝑥) is strictly increasing (or decreasing, resp.) in [1,+∞). Therefore, inequalities (2.48) and (2.49) follow from (2.50)–(2.52) and the monotonicity of 𝑓(𝑥).
Next, we prove that 𝑀_𝜆(𝑎,𝑏) and 𝑀_𝜇(𝑎,𝑏) are the best possible lower and upper power mean bounds for the geometric mean of 𝑀_𝑝(𝑎,𝑏) and 𝑀_𝑞(𝑎,𝑏). We divide the proof into six cases.
Case 𝐴. (𝑝,𝑞)∈𝐸₁. Then for any 𝜖∈(0,(𝑝+𝑞)/2) and 𝑥>0, from (1.1), one has 𝑀_𝑝(1+𝑥,1)𝑀_𝑞(1+𝑥,1)−𝑀²_{(𝑝+𝑞)/2−𝜖}()=1+𝑥,11+(1+𝑥)^𝑝2^1/𝑝1+(1+𝑥)^𝑞2^1/𝑞−1+(1+𝑥)^{(𝑝+𝑞)/2−𝜖}2^{4/(𝑝+𝑞−2𝜖)},lim_𝑥→+∞𝑀²_{2𝑝𝑞/(𝑝+𝑞)+𝜖}(𝑥,1)𝑀_𝑝(𝑥,1)𝑀_𝑞(𝑥,1)=2^{𝜖(𝑝+𝑞)2/𝑝𝑞[2𝑝𝑞+𝜖(𝑝+𝑞)]}>1. (2.54)(2.55)
Letting 𝑥→0 and making use of Taylor expansion, we get 1+(1+𝑥)^𝑝2^1/𝑝1+(1+𝑥)^𝑞2^1/𝑞−1+(1+𝑥)^{(𝑝+𝑞)/2−𝜖}2^{4/(𝑝+𝑞−2𝜖)}=𝜖4𝑥²+𝑜𝑥². (2.56)
Equations (2.54) and (2.56) together with inequality (2.55) imply that for any 𝜖∈(0,𝑝+𝑞/2), there exist 𝛿₁=𝛿₁(𝜖)>0 and 𝑋₁=𝑋₁(𝑝,𝑞,𝜖)>1 such that 𝑀_𝑝(1+𝑥,1)𝑀_𝑞(1+𝑥,1)>𝑀_{(𝑝+𝑞)/2−𝜖}(1+𝑥,1) for 𝑥∈(0,𝛿₁) and 𝑀_𝑝(𝑥,1)𝑀_𝑞(𝑥,1)<𝑀_{2𝑝𝑞/(𝑝+𝑞)+𝜖}(𝑥,1) for 𝑥∈(𝑋₁,+∞).
Case 𝐵. (𝑝,𝑞)∈𝐸₂. Then for 𝜖∈(0,−(𝑝+𝑞)/2) and 𝑥>0, making use of (1.1) and Taylor expansion, we have 𝑀²_{(𝑝+𝑞)/2+𝜖}(1+𝑥,1)−𝑀_𝑝(1+𝑥,1)𝑀_𝑞()=𝜖1+𝑥,14𝑥²+𝑜𝑥²(𝑥→0),lim_𝑥→+∞𝑀_𝑝(𝑥,1)𝑀_𝑞(𝑥,1)𝑀²_{2𝑝𝑞/(𝑝+𝑞)−𝜖}(𝑥,1)=2^{𝜖(𝑝+𝑞)2/𝑝𝑞[2𝑝𝑞−𝜖(𝑝+𝑞)]}>1. (2.57)(2.58)
Equation (2.57) and inequality (2.58) imply that for any 𝜖∈(0,−(𝑝+𝑞)/2), there exist 𝛿₂=𝛿₂(𝜖)>0 and 𝑋₂=𝑋₂(𝑝,𝑞,𝜖)>1 such that 𝑀_{(𝑝+𝑞)/2+𝜖}(1+𝑥,1)>𝑀_𝑝(1+𝑥,1)𝑀_𝑞(1+𝑥,1) for 𝑥∈(0,𝛿₂) and 𝑀_𝑝(𝑥,1)𝑀_𝑞(𝑥,1)>𝑀_{2𝑝𝑞/(𝑝+𝑞)−𝜖}(𝑥,1) for 𝑥∈(𝑋₂,+∞).
Case 𝐶. (𝑝,𝑞)∈𝐸₃. Then for 𝜖∈(0,𝑝/2) and 𝑥>0, making use of (1.1) and Taylor expansion, we have 𝑀_𝑝(1+𝑥,1)𝑀₀(1+𝑥,1)−𝑀²_{𝑝/2−𝜖}()=𝜖1+𝑥,14𝑥²+𝑜𝑥²(𝑥→0),lim_𝑥→+∞𝑀²_𝜖(𝑥,1)𝑀_𝑝(𝑥,1)𝑀₀(𝑥,1)=+∞. (2.59)
Equation (2.59) leads to the conclusion that for any 𝜖∈(0,𝑝/2), there exist 𝛿₃=𝛿₃(𝜖)>0 and 𝑋₃=𝑋₃(𝑝,𝜖)>1 such that 𝑀_𝑝(1+𝑥,1)𝑀₀(1+𝑥,1)>𝑀_{𝑝/2−𝜖}(1+𝑥,1) for 𝑥∈(0,𝛿₃) and 𝑀_𝜖(𝑥,1)>𝑀_𝑝(𝑥,1)𝑀₀(𝑥,1) for 𝑥∈(𝑋₃,+∞).
Case 𝐷. (𝑝,𝑞)∈𝐸₄. Then for 𝜖∈(0,(𝑝+𝑞)/2) and 𝑥>0, making use of (1.1) and Taylor expansion, we have 𝑀_𝑝(1+𝑥,1)𝑀_𝑞(1+𝑥,1)−𝑀²_{(𝑝+𝑞)/2−𝜖}()=𝜖1+𝑥,14𝑥²+𝑜𝑥²(𝑥→0),lim_𝑥→+∞𝑀²_𝜖(𝑥,1)𝑀_𝑝(𝑥,1)𝑀_𝑞(𝑥,1)=+∞. (2.60)
Equation (2.60) implies that for any 𝜖∈(0,(𝑝+𝑞)/2), there exist 𝛿₄=𝛿₄(𝜖)>0 and 𝑋₄=𝑋₄(𝑝,𝑞,𝜖)>1 such that 𝑀_{(𝑝+𝑞)/2−𝜖}(1+𝑥,1)<𝑀_𝑝(1+𝑥,1)𝑀_𝑞(1+𝑥,1) for 𝑥∈(0,𝛿₄) and 𝑀_𝜖(𝑥,1)>𝑀_𝑝(𝑥,1)𝑀_𝑞(𝑥,1) for 𝑥∈(𝑋₄,+∞).
Case 𝐸. (𝑝,𝑞)∈𝐸₅. Then for any 𝜖∈(0,−𝑞/2) and 𝑥>0, making use of (1.1) and Taylor expansion, one has 𝑀²_𝑞/2+𝜖(1+𝑥,1)−𝑀₀(1+𝑥,1)𝑀_𝑞()=𝜖1+𝑥,14𝑥²+𝑜𝑥²(𝑥→0),lim_𝑥→+∞𝑀₀(𝑥,1)𝑀_𝑞(𝑥,1)𝑀²_−𝜖(𝑥,1)=+∞. (2.61)
Equation (2.61) leads to the conclusion that for any 𝜖∈(0,−𝑞/2), there exist 𝛿₅=𝛿₅(𝜖)>0 and 𝑋₅=𝑋₅(𝑞,𝜖)>1 such that 𝑀_𝑞/2+𝜖(1+𝑥,1)>𝑀₀(1+𝑥,1)𝑀_𝑞(1+𝑥,1) for 𝑥∈(0,𝛿₅) and 𝑀_−𝜖(𝑥,1)<𝑀₀(𝑥,1)𝑀_𝑞(𝑥,1) for 𝑥∈(𝑋₅,+∞).
Case 𝐹. (𝑝,𝑞)∈𝐸₆. Then for any 𝜖∈(0,−(𝑝+𝑞)/2) and 𝑥>0, making use of (1.1) and Taylor expansion, one has 𝑀²_{(𝑝+𝑞)/2+𝜖}(1+𝑥,1)−𝑀_𝑝(1+𝑥,1)𝑀_𝑞()=𝜖1+𝑥,14𝑥²+𝑜𝑥²(𝑥→0),lim_𝑥→+∞𝑀_𝑝(𝑥,1)𝑀_𝑞(𝑥,1)𝑀²_−𝜖(𝑥,1)=+∞. (2.62)
Equation (2.62) shows that for any 𝜖∈(0,−(𝑝+𝑞)/2), there exist 𝛿₆=𝛿₆(𝜖)>0 and 𝑋₆=𝑋₆(𝑝,𝑞,𝜖)>1 such that 𝑀_{(𝑝+𝑞)/2+𝜖}(1+𝑥,1)>𝑀_𝑝(1+𝑥,1)𝑀_𝑞(1+𝑥,1) for 𝑥∈(0,𝛿₆) and 𝑀_𝑝(𝑥,1)𝑀_𝑞(𝑥,1)>𝑀²_−𝜖(𝑥,1) for 𝑥∈(𝑋₆,+∞).

Acknowledgments

This research was supported by the Natural Science Foundation of China under Grants 11071069 and 11171307 and the Innovation Team Foundation of the Department of Education of Zhejiang Province under Grant T200924.