Abstract

We find the greatest value 𝑝 and the least value 𝑞 in (0,1/2) such that the double inequality 𝐻(𝑝𝑎+(1−𝑝)𝑏,𝑝𝑏+(1−𝑝)𝑎)<𝐼(𝑎,𝑏)<𝐻(𝑞𝑎+(1−𝑞)𝑏,𝑞𝑏+(1−𝑞)𝑎) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏. Here, 𝐻(𝑎,𝑏), and 𝐼(𝑎,𝑏) denote the harmonic and identric means of two positive numbers 𝑎 and 𝑏, respectively.

1. Introduction

The classical harmonic mean 𝐻(𝑎,𝑏) and identric mean 𝐼(𝑎,𝑏) of two positive numbers 𝑎 and 𝑏 are defined by𝐻(𝑎,𝑏)=2𝑎𝑏,⎧⎪⎨⎪⎩1𝑎+𝑏(1.1)𝐼(𝑎,𝑏)=𝑒𝑏𝑏𝑎𝑎1/(𝑏−𝑎),𝑎≠𝑏,𝑎,𝑎=𝑏,(1.2) respectively. Recently, both mean values have been the subject of intensive research. In particular, many remarkable inequalities for 𝐻 and 𝐼 can be found in the literature [1–17].

Let 𝑀𝑝(𝑎,𝑏)=[(𝑎𝑝+𝑏𝑝)/2]1/𝑝, 𝐿(𝑎,𝑏)=(𝑎−𝑏)/(log𝑎−log𝑏), √𝐺(𝑎,𝑏)=𝑎𝑏, 𝐴(𝑎,𝑏)=(𝑎+𝑏)/2, and √𝑃(𝑎,𝑏)=(𝑎−𝑏)/[4arctan(𝑎/𝑏)−𝜋] be the 𝑝th power, logarithmic, geometric, arithmetic, and Seiffert means of two positive numbers 𝑎 and 𝑏 with 𝑎≠𝑏, respectively. Then it is well-known that min{𝑎,𝑏}<𝐻(𝑎,𝑏)=𝑀−1(𝑎,𝑏)<𝐺(𝑎,𝑏)=𝑀0(𝑎,𝑏)<𝐿(𝑎,𝑏)<𝑃(𝑎,𝑏)<𝐼(𝑎,𝑏)<𝐴(𝑎,𝑏)=𝑀1(𝑎,𝑏)<max{𝑎,𝑏}(1.3) for all 𝑎,𝑏>0 with 𝑎≠𝑏.

Long and Chu [18] answered the question: what are the greatest value 𝑝 and the least value 𝑞 such that 𝑀𝑝(𝑎,𝑏)<𝐴𝛼(𝑎,𝑏)𝐺𝛽(𝑎,𝑏)𝐻1−𝛼−𝛽(𝑎,𝑏)<𝑀𝑞(𝑎,𝑏) for all 𝑎,𝑏>0 with 𝑎≠𝑏 and 𝛼,𝛽>0 with 𝛼+𝛽<1.

In [19], the authors proved that the double inequality 𝛼𝐴(𝑎,𝑏)+(1−𝛼)𝐻(𝑎,𝑏)<𝑃(𝑎,𝑏)<𝛽𝐴(𝑎,𝑏)+(1−𝛽)𝐻(𝑎,𝑏)(1.4) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if 𝛼≤2/𝜋 and 𝛽≥5/6.

The following sharp bounds for 𝐼,(𝐿𝐼)1/2, and (𝐿+𝐼)/2 in terms of power means are presented in [20]: 𝑀2/3(𝑎,𝑏)<𝐼(𝑎,𝑏)<𝑀log2(𝑎,𝑏),𝑀0√(𝑎,𝑏)<𝐿(𝑎,𝑏)𝐼(𝑎,𝑏)<𝑀1/2𝑀(𝑎,𝑏),log2/(1+log2)(𝑎,𝑏)<𝐿(𝑎,𝑏)+𝐼(𝑎,𝑏)2<𝑀1/2(𝑎,𝑏)(1.5)

for all 𝑎,𝑏>0 with 𝑎≠𝑏.

Alzer and Qiu [21] proved that the inequalities 𝛼𝐴(𝑎,𝑏)+(1−𝛼)𝐺(𝑎,𝑏)<𝐼(𝑎,𝑏)<𝛽𝐴(𝑎,𝑏)+(1−𝛽)𝐺(𝑎,𝑏)(1.6) hold for all positive real numbers 𝑎 and 𝑏 with 𝑎≠𝑏 if and only if 𝛼≤2/3 and 𝛽≥2/𝑒=0.73575, and so forth.

For fixed 𝑎,𝑏>0 with 𝑎≠𝑏 and 𝑥∈[0,1/2], let 𝑓(𝑥)=𝐻(𝑥𝑎+(1−𝑥)𝑏,𝑥𝑏+(1−𝑥)𝑎).(1.7)

Then it is not difficult to verify that 𝑓(𝑥) is continuous and strictly increasing in [0,1/2]. Note that 𝑓(0)=𝐻(𝑎,𝑏)<𝐼(𝑎,𝑏) and 𝑓(1/2)=𝐴(𝑎,𝑏)>𝐼(𝑎,𝑏). Therefore, it is natural to ask what are the greatest value 𝑝 and the least value 𝑞 in (0,1/2) such that the double inequality 𝐻(𝑝𝑎+(1−𝑝)𝑏,𝑝𝑏+(1−𝑝)𝑎)<𝐼(𝑎,𝑏)<𝐻(𝑞𝑎+(1−𝑞)𝑏,𝑞𝑏+(1−𝑞)𝑎) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏. The main purpose of this paper is to answer these questions. Our main result is Theorem 1.1.

Theorem 1.1. If 𝑝,𝑞∈(0,1/2), then the double inequality 𝐻(𝑝𝑎+(1−𝑝)𝑏,𝑝𝑏+(1−𝑝)𝑎)<𝐼(𝑎,𝑏)<𝐻(𝑞𝑎+(1−𝑞)𝑏,𝑞𝑏+(1−𝑞)𝑎)(1.8) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 if and only if √𝑝≤(1−1−2/𝑒)/2 and √𝑞≥(6−6)/12.

2. Proof of Theorem 1.1

Proof of Theorem 1.1. Let √𝜆=(6−6)/12 and √𝜇=(1−1−2/𝑒)/2. Then from the monotonicity of the function 𝑓(𝑥)=𝐻(𝑥𝑎+(1−𝑥)𝑏,𝑥𝑏+(1−𝑥)𝑎) in [0,1/2] we know that to prove inequality (1.8) we only need to prove that inequalities 𝐼(𝑎,𝑏)<𝐻(𝜆𝑎+(1−𝜆)𝑏,𝜆𝑏+(1−𝜆)𝑎),(2.1)𝐼(𝑎,𝑏)>𝐻(𝜇𝑎+(1−𝜇)𝑏,𝜇𝑏+(1−𝜇)𝑎),(2.2) hold for all 𝑎,𝑏>0 with 𝑎≠𝑏.
Without loss of generality, we assume that 𝑎>𝑏. Let 𝑡=𝑎/𝑏>1 and 𝑟∈(0,1/2), then from (1.1) and (1.2) one has log𝐻(𝑟𝑎+(1−𝑟)𝑏,𝑟𝑏+(1−𝑟)𝑎)−log𝐼(𝑎,𝑏)=log𝑟(1−𝑟)𝑡2+𝑟2+(1−𝑟)2𝑡+𝑟(1−𝑟)−log(𝑡+1)−𝑡log𝑡𝑡−1+1+log2.(2.3)
Let 𝑔(𝑡)=log𝑟(1−𝑟)𝑡2+𝑟2+(1−𝑟)2−𝑡+𝑟(1−𝑟)−log(𝑡+1)𝑡log𝑡𝑡−1+1+log2.(2.4) Then simple computations lead to 𝑔(1)=0,(2.5)lim𝑡→+∞[]𝑔𝑔(𝑡)=log𝑟(1−𝑟)+1+log2,(2.6)(𝑔𝑡)=1(𝑡)(𝑡−1)2,(2.7) where 𝑔1(𝑡)=log𝑡−(𝑡−1)2𝑟2𝑡−2𝑟+12+4𝑟(1−𝑟)𝑡+2𝑟2−2𝑟+1(𝑡+1)𝑟(1−𝑟)𝑡2+2𝑟2,𝑔−2𝑟+1𝑡+𝑟(1−𝑟)(2.8)1(1)=0,(2.9)lim𝑡→+∞𝑔1𝑔(𝑡)=+∞,(2.10)1𝑔(𝑡)=2(𝑡)𝑡(𝑡+1)2𝑟(1−𝑟)𝑡2+2𝑟2−2𝑟+1𝑡+𝑟(1−𝑟)2,(2.11) where 𝑔2(𝑡)=𝑟2(1−𝑟)2𝑡6+2𝑟4−4𝑟3−2𝑟2𝑡+4𝑟−15−17𝑟4−34𝑟3+25𝑟2𝑡−8𝑟+14+47𝑟4−14𝑟3+13𝑟2𝑡−6𝑟+13−17𝑟4−34𝑟3+25𝑟2𝑡−8𝑟+12+2𝑟4−4𝑟3−2𝑟2+4𝑟−1𝑡+𝑟2(1−𝑟)2,𝑔(2.12)2(1)=0,(2.13)lim𝑡→+∞𝑔2𝑔(𝑡)=+∞,(2.14)2(𝑡)=6𝑟2(1−𝑟)2𝑡5+52𝑟4−4𝑟3−2𝑟2𝑡+4𝑟−14−417𝑟4−34𝑟3+25𝑟2𝑡−8𝑟+13+127𝑟4−14𝑟3+13𝑟2𝑡−6𝑟+12−217𝑟4−34𝑟3+25𝑟2𝑡−8𝑟+1+2𝑟4−4𝑟3−2𝑟2𝑔+4𝑟−1,(2.15)2(1)=0,(2.16)lim𝑡→+∞𝑔2𝑔(𝑡)=+∞,(2.17)2(𝑡)=30𝑟2(1−𝑟)2𝑡4+202𝑟4−4𝑟3−2𝑟2𝑡+4𝑟−13−1217𝑟4−34𝑟3+25𝑟2𝑡−8𝑟+12+247𝑟4−14𝑟3+13𝑟2−6𝑟+1𝑡−217𝑟4−34𝑟3+25𝑟2,𝑔−8𝑟+1(2.18)2(1)=−224𝑟2−24𝑟+5,(2.19)lim𝑡→+∞𝑔2𝑔(𝑡)=+∞,(2.20)2(𝑡)=120𝑟2(1−𝑟)2𝑡3+602𝑟4−4𝑟3−2𝑟2𝑡+4𝑟−12−2417𝑟4−34𝑟3+25𝑟2−8𝑟+1𝑡+247𝑟4−14𝑟3+13𝑟2,𝑔−6𝑟+1(2.21)2(1)=−1224𝑟2−24𝑟+5,(2.22)lim𝑡→∞𝑔2(𝑔𝑡)=∞,(2.23)2(4)(𝑡)=360𝑟2(1−𝑟)2𝑡2+1202𝑟4−4𝑟3−2𝑟2𝑡+4𝑟−1−2417𝑟4−34𝑟3+25𝑟2,𝑔−8𝑟+1(2.24)2(4)(1)=484𝑟4−8𝑟3−10𝑟2+14𝑟−3,(2.25)lim𝑡→+∞𝑔2(4)𝑔(𝑡)=+∞,(2.26)2(5)(𝑡)=720𝑟2(1−𝑟)2𝑡+1202𝑟4−4𝑟3−2𝑟2𝑔+4𝑟−1,(2.27)2(5)(1)=1208𝑟4−16𝑟3+4𝑟2.+4𝑟−1(2.28)
We divide the proof into two cases. Case 1 (√𝑟=𝜆=(6−6)/12). Then (2.19), (2.22), (2.25), and (2.28) lead to 𝑔2𝑔(1)=0,(2.29)2𝑔(1)=0,(2.30)2(4)(1)=133𝑔>0,(2.31)2(5)(1)=653>0.(2.32)
From (2.27) we clearly see that 𝑔2(5)(𝑡) is strictly increasing in [1,+∞), then inequality (2.32) leads to the conclusion that 𝑔2(5)(𝑡)>0 for 𝑡∈[1,+∞), hence 𝑔2(4)(𝑡) is strictly increasing in [1,+∞).
It follows from inequality (2.31) and the monotonicity of 𝑔2(4)(𝑡) that 𝑔2(𝑡) is strictly increasing in [1,+∞). Then (2.30) implies that 𝑔2(𝑡)>0 for 𝑡∈[1,+∞), so 𝑔2(𝑡) is strictly increasing in [1,+∞).
From (2.29) and the monotonicity of 𝑔2(𝑡) we clearly see that 𝑔2(𝑡) is strictly increasing in [1,+∞).
From (2.5), (2.7), (2.9), (2.11), (2.13), (2.16), and the monotonicity of 𝑔2(𝑡) we conclude that 𝑔(𝑡)>0(2.33) for 𝑡∈(1,+∞).
Therefore, inequality (2.1) follows from (2.3) and (2.4) together with inequality (2.33).
Case 2 (√𝑟=𝜇=(1−1−2/𝑒)/2). Then (2.19), (2.22), (2.25), and (2.28) lead to 𝑔22(1)=−𝑒𝑔(5𝑒−12)<0,(2.34)2(1)=−12𝑒𝑔(5𝑒−12)<0,(2.35)2(4)(1)=−48𝑒23𝑒2𝑔−7𝑒−1<0,(2.36)2(5)(1)=120𝑒22+2𝑒−𝑒2>0.(2.37) From (2.27) and (2.37) we know that 𝑔2(4)(𝑡) is strictly increasing in [1,+∞). Then (2.26) and (2.36) lead to the conclusion that there exists 𝑡1>1 such that 𝑔2(4)(𝑡)<0 for 𝑡∈[1,𝑡1) and 𝑔2(4)(𝑡)>0 for 𝑡∈(𝑡1,+∞), hence 𝑔2(𝑡) is strictly decreasing in [1,𝑡1] and strictly increasing in [𝑡1,+∞).
It follows from (2.23) and (2.35) together with the piecewise monotonicity of 𝑔2(𝑡) that there exists 𝑡2>𝑡1>1 such that 𝑔2(𝑡) is strictly decreasing in [1,𝑡2] and strictly increasing in [𝑡2,+∞). Then (2.20) and (2.34) lead to the conclusion that there exists 𝑡3>𝑡2>1 such that 𝑔2(𝑡) is strictly decreasing in [1,𝑡3] and strictly increasing in [𝑡3,+∞).
From (2.16) and (2.17) together with the piecewise monotonicity of 𝑔2(𝑡) we clearly see that there exists 𝑡4>𝑡3>1 such that 𝑔2(𝑡)<0 for 𝑡∈(1,𝑡4) and 𝑔2(𝑡)>0 for 𝑡∈(𝑡4,+∞). Therefore, 𝑔2(𝑡) is strictly decreasing in [1,𝑡4] and strictly increasing in [𝑡4,+∞). Then (2.11)–(2.14) lead to the conclusion that there exists 𝑡5>𝑡4>1 such that 𝑔1(𝑡) is strictly decreasing in [1,𝑡5] and strictly increasing in [𝑡5,+∞).
It follows from (2.7)–(2.10) and the piecewise monotonicity of 𝑔1(𝑡) that there exists 𝑡6>𝑡5>1 such that 𝑔(𝑡) is strictly decreasing in [1,𝑡6] and strictly increasingin [𝑡6,+∞).
Note that (2.6) becomes lim𝑡→+∞[]𝑔(𝑡)=log𝑟(1−𝑟)+1+log2=0(2.38) for √𝑟=𝜇=(1−1−2/𝑒)/2.
From (2.5) and (2.38) together with the piecewise monotonicity of 𝑔(𝑡) we clearly see that 𝑔(𝑡)<0(2.39) for 𝑡∈(1,+∞).
Therefore, inequality (2.2) follows from (2.3) and (2.4) together with inequality (2.39).
Next, we prove that the parameter √𝜆=(6−6)/12 is the best possible parameter in (0,1/2) such that inequality (2.1) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏. In fact, if √𝑟<𝜆=(6−6)/12, then (2.19) leads to 𝑔2(1)=−2(24𝑟2−24𝑟+5)<0. From the continuity of 𝑔2(𝑡) we know that there exists 𝛿>0 such that 𝑔2(𝑡)<0(2.40) for 𝑡∈(1,1+𝛿).
It follows from (2.3)–(2.5), (2.7), (2.9), (2.11), (2.13), and (2.16) that 𝐼(𝑎,𝑏)>𝐻(𝑟𝑎+(1−𝑟)𝑏,𝑟𝑏+(1−𝑟)𝑎) for 𝑎/𝑏∈(1,1+𝛿).
Finally, we prove that the parameter √𝜇=(1−1−2/𝑒)/2 is the best possible parameter in (0,1/2) such that inequality (2.2) holds for all 𝑎,𝑏>0 with 𝑎≠𝑏. In fact, if √(1−1−2/𝑒)/2=𝜇<𝑟<1/2, then (2.6) leads to lim𝑡→+∞𝑔(𝑡)>0. Hence, there exists 𝑇>1 such that 𝑔(𝑡)>0(2.41) for 𝑡∈(𝑇,+∞).
Therefore, 𝐻(𝑟𝑎+(1−𝑟)𝑏,𝑟𝑏+(1−𝑟)𝑎)>𝐼(𝑎,𝑏) for 𝑎/𝑏∈(𝑇,+∞), follows from (2.3) and (2.4) together with inequality (2.41).

Acknowledgments

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