Abstract

For 1<𝑟<+, we find the least value 𝛼 and the greatest value 𝛽 such that the inequality 𝐻𝛼(𝑎,𝑏)<𝐴𝑟(𝑎,𝑏)<𝐻𝛽(𝑎,𝑏) holds for all 𝑎,𝑏>0 with 𝑎𝑏. Here, 𝐻𝜔(𝑎,𝑏) and 𝐴𝑟(𝑎,𝑏) are the generalized Heronian and the power means of two positive numbers 𝑎 and 𝑏, respectively.

1. Introduction and Statement of Result

For 𝑎,𝑏>0 with 𝑎𝑏, the generalized Heronian mean of 𝑎 and 𝑏 is defined by Janous [1] as 𝐻𝜔(𝑎,𝑏)=𝑎+𝜔𝑎𝑏+𝑏𝜔+2,0𝜔<+,𝑎𝑏,𝜔=+.(1.1) If we take 𝜔=1 in (1.1), then we arrive at the classical Heronian mean 𝐻𝑒(𝑎,𝑏)=𝑎+𝑎𝑏+𝑏3.(1.2) The domain of definition for the function 𝜔𝐻𝜔(𝑎,𝑏) can be extended to all 𝜔 with 𝜔(2,+), that is, 𝐻𝜔(𝑎,𝑏)=𝑎+𝜔𝑎𝑏+𝑏𝜔+2,2<𝜔<+,𝑎𝑏,𝜔=+.(1.3) For all fixed 𝑎,𝑏>0, it is easy to derive that 𝜔𝐻𝜔(𝑎,𝑏), 2<𝜔<+ is monotonically decreasing, and lim𝜔2+𝐻𝜔(𝑎,𝑏)=+.(1.4)

Let 𝐴𝑟𝑎(𝑎,𝑏)=𝑟+𝑏𝑟21/𝑟,𝑟0,𝑎𝑏,𝑟=0,max{𝑎,𝑏},𝑟=+,min{𝑎,𝑏},𝑟=,(1.5) denote the power mean of order 𝑟. In particular, the harmonic, geometric, square-root, arithmetic, and root-square means of 𝑎 and 𝑏 are 𝐻(𝑎,𝑏)=𝐴1(𝑎,𝑏)=2𝑎,𝑎+𝑏𝐺(𝑎,𝑏)=𝐴0(𝑎,𝑏)=𝑁𝑎𝑏,1(𝑎,𝑏)=𝐴1/2(𝑎,𝑏)=𝑎+𝑏22,𝐴(𝑎,𝑏)=𝐴1(𝑎,𝑏)=𝑎+𝑏2,𝑆(𝑎,𝑏)=𝐴2(𝑎,𝑏)=𝑎2+𝑏22.(1.6) It is well known that the power mean of order 𝑟 given in (1.5) is monotonically increasing in 𝑟, then we can write min{𝑎,𝑏}<𝐻(𝑎,𝑏)<𝐺(𝑎,𝑏)<𝑁1(𝑎,𝑏)<𝐴(𝑎,𝑏)<𝑆(𝑎,𝑏)<max{𝑎,𝑏}.(1.7)

Recently, the inequalities for means have been the subject of intensive research [115]. In particular, many remarkable inequalities for the generalized Heronian and power means can be found in the literature [49].

In [4], the authors established two sharp inequalities 231𝐺(𝑎,𝑏)+3𝐻(𝑎,𝑏)𝐴1/31(𝑎,𝑏),32𝐺(𝑎,𝑏)+3𝐻(𝑎,𝑏)𝐴2/3(𝑎,𝑏).(1.8)

In [5], Long and Chu found the greatest value 𝑝 and the least value 𝑞 such that the double inequality 𝐴𝑝(𝑎,𝑏)𝐴(𝑎,𝑏)𝛼𝐺(𝑎,𝑏)𝛽𝐻(𝑎,𝑏)1𝛼𝛽𝐴𝑞(𝑎,𝑏)(1.9) holds for all 𝑎,𝑏>0 and 𝛼,𝛽>0 with 𝛼+𝛽<1.

In [6], Shi et al. gave two optimal inequalities 𝐴𝛼(𝑎,𝑏)𝐿1𝛼(𝑎,𝑏)𝐴(1+2𝛼)/3𝐺(𝑎,𝑏),𝛼(𝑎,𝑏)𝐿1𝛼(𝑎,𝑏)𝐴(1𝛼)/3(𝑎,𝑏),(1.10) for 0<𝛼<1, where 𝐿(𝑎,𝑏)=𝑎𝑏log𝑎log𝑏,𝑎𝑏,(1.11) is the logarithmic mean for 𝑎,𝑏>0.

In [7], Guan and Zhu obtained sharp bounds for the generalized Heronian mean in terms of the power mean with 𝜔>0. The optimal values 𝛼 and 𝛽 such that 𝐴𝛼(𝑎,𝑏)𝐻𝜔(𝑎,𝑏)𝐴𝛽(𝑎,𝑏)(1.12) holds in general are (1)in case of 𝜔(0,2],𝛼max=log2/log(𝜔+2) and 𝛽min=2/(𝜔+2),(2) in case of 𝜔[2,+),𝛼max=2/(𝜔+2) and 𝛽min=log2/log(𝜔+2).

In this paper, we find the least value 𝛼 and the greatest value 𝛽, such that for any fixed 1<𝑟<+, the inequality 𝐻𝛼(𝑎,𝑏)<𝐴𝑟(𝑎,𝑏)<𝐻𝛽(𝑎,𝑏)(1.13) holds for all 𝑎,𝑏>0 with 𝑎𝑏.

Theorem 1.1. For 1<𝑟<+, the optimal numbers 𝛼 and 𝛽 such that 𝐻𝛼(𝑎,𝑏)<𝐴𝑟(𝑎,𝑏)<𝐻𝛽(𝑎,𝑏)(1.14) is valid for all 𝑎,𝑏>0 with 𝑎𝑏, are 𝛼min=21/𝑟2 and 𝛽max=2(1𝑟)/𝑟.

Notice that in our case 𝑟>1; the two numbers 𝛼min and 𝛽max are all negative see Corollary 2.2 below. Thus, the result in this paper is different from [7, Theorem A].

2. Preliminary Lemmas

The following lemma will be repeatedly used in the proof of Theorem 1.1.

Lemma 2.1. For 1<𝑟<+, one has 𝑟21/𝑟1>1.(2.1)

Proof. We show that 𝑚(𝑟)=(1𝑟)log2+𝑟log𝑟>0,(2.2) which is clearly equivalent to the claim. Equation (2.2) follows from the facts lim𝑟1+𝑚(𝑟)=0,𝑚(𝑟)=log2+log𝑟+1>0.(2.3)

Corollary 2.2. If 1<𝑟<+, then 2<2(1𝑟)𝑟<21/𝑟2<0.(2.4)

Proof. Since for 1<𝑟<+, the two functions 𝜑1(𝑟)=2(1𝑟)𝑟,𝜑2(𝑟)=21/𝑟2(2.5) are strictly decreasing, then one has 2=lim𝑟+𝜑1(𝑟)<𝜑1(𝑟),𝜑2(𝑟)<lim𝑟1+𝜑2(𝑟)=0.(2.6) It suffices to show that 22𝑟<𝑟21/𝑟2𝑟,(2.7) which is equivalent to (2.1).

Lemma 2.3. For 𝑥>1 and 𝑟>1, let 𝑥(𝑥)=2𝑟+11/𝑟2𝑥2(𝑟1)𝑥2𝑟.+2𝑟1(2.8) Then, (𝑥) is strictly decreasing for 𝑥>1, and lim𝑥1+(𝑥)=𝑟21/𝑟1,lim𝑥+(𝑥)=1.(2.9)

Proof. The fact (𝑥)>0 for 𝑥>1 and 𝑟>1 is obvious, which allows us to take the logarithmic function of (𝑥), 1log(𝑥)=𝑟𝑥2log2𝑟𝑥+1+2(𝑟1)log𝑥+log2𝑟.+2𝑟1(2.10) Some tedious, but not difficult calculations lead to log(𝑥)=1𝑟22𝑟𝑥2𝑟1𝑥2𝑟++12(𝑟1)𝑥+2𝑟𝑥2𝑟1𝑥2𝑟=+2𝑟1𝑚(𝑥)𝑥𝑥2𝑟𝑥+12𝑟,+2𝑟1(2.11) where 𝑚(𝑥)=2(12𝑟)𝑥2𝑟𝑥2𝑟𝑥+2𝑟1+(2𝑟1)2𝑟𝑥+12𝑟+2𝑟1+2𝑟𝑥2𝑟𝑥2𝑟+1=2(𝑟1)(2𝑟1)1𝑥2𝑟.(2.12) It is easy to see that lim𝑥1+𝑚𝑚(𝑥)=0,(2.13)(𝑥)=4𝑟(𝑟1)(2𝑟1)𝑥2𝑟1<0.(2.14) Equation (2.14) implies that 𝑚(𝑥) is strictly decreasing for 𝑥>1, which together with (2.13) implies 𝑚(𝑥)<0 for 𝑥>1. Thus, by (2.11), log(𝑥)<0,(2.15) which implies (𝑥)=log(𝑥)(𝑥)<0.(2.16) Hence, (𝑥) is strictly decreasing.
It remains to show (2.9). The first equality in (2.9) is obvious. The second one follows from lim𝑥+(𝑥)=lim𝑥+𝑥2𝑟+11/𝑟2𝑥2(𝑟1)𝑥2𝑟+2𝑟1=lim𝑡0+(2𝑟1)𝑡2𝑟+11+𝑡2𝑟(2𝑟1)/𝑟=1.(2.17) This ends the proof of Lemma 2.3.

Lemma 2.4. For 𝑥>1, 𝑟>1, and 𝜔=21/𝑟2, let 𝑓𝑟(𝑥)=21/𝑟𝑥2𝑥+𝜔𝑥+1(𝜔+2)2𝑟+11/𝑟.(2.18) Then, lim𝑥+𝑓𝑟(𝑥)=,lim𝑥+𝑓𝑟(𝑥)=21/𝑟21/𝑟.2(2.19)

Proof. Simple calculations lead to lim𝑥+𝑓𝑟(𝑥)=lim𝑥+21/𝑟𝑥2𝑥+𝜔𝑥+1(𝜔+2)2𝑟+11/𝑟=lim𝑡0+21/𝑟𝑡2𝑡+𝜔𝑡+1(𝜔+2)2𝑟+11/𝑟𝑡2=,lim𝑥+𝑓𝑟(𝑥)=lim𝑥+21/𝑟𝑥(2𝑥+𝜔)2(𝜔+2)2𝑟+11/𝑟=lim𝑡0+21/𝑟(2+𝜔𝑡)2(𝜔+2)1+𝑡2𝑟(1𝑟)/𝑟𝑡=lim𝑡0+21/𝑟(2+𝜔𝑡)1+𝑡2𝑟(𝑟1)/𝑟2(𝜔+2)𝑡1+𝑡2𝑟(𝑟1)/𝑟=lim𝑡0+21/𝑟𝜔1+𝑡2𝑟(𝑟1)/𝑟+2(1/𝑟)+1(𝑟1)(2+𝜔𝑡)1+𝑡2𝑟1/𝑟𝑡2𝑟11+𝑡2𝑟(𝑟1)/𝑟+2(𝑟1)1+𝑡2𝑟1/𝑟𝑡2𝑟=21/𝑟𝜔=21/𝑟𝜔1/𝑟2<0,(2.20) where we have used L'Hospital's law. This ends the proof of Lemma 2.4.

3. Proof of Theorem 1.1

Proof. Firstly, we prove that for 1<𝑟<+, 𝐻2(1𝑟)/𝑟(𝑎,𝑏)>𝐴𝑟𝐻(𝑎,𝑏),(3.1)21/𝑟2(𝑎,𝑏)<𝐴𝑟(𝑎,𝑏)(3.2) hold true for all 𝑎,𝑏>0 with 𝑎𝑏. It is no loss of generality to assume that 𝑎>𝑏>0. Let 𝑥=𝑏/𝑎>1 and 𝜔{2(1𝑟)/𝑟,21/𝑟2}. In view of Corollary 2.2, 2<𝜔<0. Equations (1.3) and (1.5) lead to 1𝑎𝐻𝜔(𝑎,𝑏)𝐴𝑟(𝑎,𝑏)=𝐻𝜔𝑥2,1𝐴𝑟𝑥2=𝑥,12+𝜔𝑥+1𝑥𝜔+22𝑟+121/𝑟=21/𝑟𝑥2𝑥+𝜔𝑥+1(𝜔+2)2𝑟+11/𝑟21/𝑟=𝑓(𝜔+2)𝑟(𝑥)21/𝑟,(𝜔+2)(3.3) where 𝑓𝑟(𝑥) is defined by (2.18). It is easy to see that lim𝑥1+𝑓𝑟𝑓(𝑥)=0,(3.4)𝑟(𝑥)=21/𝑟𝑥(2𝑥+𝜔)2(𝜔+2)2𝑟+11/𝑟1𝑥2𝑟1,(3.5)lim𝑥1+𝑓𝑟(𝑥)=0.(3.6) By Lemma 2.3, 𝑓𝑟2(𝑥)=21/𝑟𝑥(𝜔+2)2(1𝑟)2𝑟+11/𝑟2𝑥4𝑟2𝑥+(2𝑟1)2𝑟+11/𝑟1𝑥2(𝑟1)2=21/𝑟2(𝜔+2)(𝑥)>21/𝑟(𝜔+2)𝑟21/𝑟1=21/𝑟[],2(𝜔+2)𝑟(3.7)lim𝑥1+𝑓𝑟2(𝑥)=21/𝑟(𝜔+2)2(1𝑟)21/𝑟2+(2𝑟1)21/𝑟1=21/𝑟[].2(𝜔+2)𝑟(3.8)
We now distinguish between two cases.
Case 1 (𝜔=2(1𝑟)/𝑟). Since 2(𝜔+2)𝑟=0, then by (3.7), 𝑓𝑟(𝑥)>0. Thus, 𝑓𝑟(𝑥) is strictly increasing for 𝑥>1, which together with (3.6) implies 𝑓𝑟(𝑥)>0. Hence, 𝑓𝑟(𝑥) is strictly increasing for 𝑥>1. Since (3.4), then 𝑓𝑟(𝑥)>0. Equation (3.1) follows from (3.3).Case 2 (𝜔=21/𝑟2). By (3.5) and (2.11), 𝑓𝑟(𝑥)=2(𝜔+2)(𝑥)=2(𝜔+2)log(𝑥)(𝑥)=2(𝜔+2)𝑚(𝑥)(𝑥)𝑥𝑥2𝑟+1(𝑥𝑟)+2𝑟1>0.(3.9) Thus, 𝑓𝑟(𝑥) is strictly increasing. Equations (3.8) and (2.1) imply lim𝑥1+𝑓𝑟(𝑥)=21/𝑟[]2(𝜔+2)𝑟=21/𝑟+11𝑟21/𝑟1<0.(3.10) Equations (3.7) and (2.9) imply lim𝑥+𝑓𝑟(𝑥)=lim𝑥+221/𝑟2(𝜔+2)(𝑥)=21/𝑟1>0.(3.11) Combining (3.10) with (3.11), we obviously know that there exists 𝜆1>1 such that 𝑓𝑟(𝑥)<0 for 𝑥(1,𝜆1) and 𝑓𝑟(𝑥)>0 for 𝑥(𝜆1,+). This implies that 𝑓𝑟(𝑥) is strictly decreasing for 𝑥(1,𝜆1) and strictly increasing for 𝑥(𝜆1,+). By (3.6) and Lemma 2.4, we know that 𝑓𝑟(𝑥)<0 for 𝑥>1. Therefore, 𝑓𝑟(𝑥) is strictly decreasing. By (3.4) and Lemma 2.4 again, we derive that 𝑓𝑟(𝑥)<0 for 𝑥>1. Equation (3.2) follows from (3.3).Secondly, we prove that 𝐻21/𝑟2(𝑎,𝑏) is the best lower bound for the power mean 𝐴𝑟(𝑎,𝑏) for 1<𝑟<+. For any 𝛼<21/𝑟2, lim𝑥+𝐻𝛼(𝑥,1)𝐴𝑟(𝑥,1)=lim𝑥+21/𝑟𝑥+𝛼𝑥+1(𝛼+2)(𝑥𝑟+1)𝑟=21/𝑟𝛼+2>1.(3.12) Hence, there exists 𝑋=𝑋(𝛼)>1 such that 𝐻𝛼(𝑥,1)>𝐴𝑟(𝑥,1) for 𝑥(𝑋,+).
Finally, we prove that 𝐻2(1𝑟)/𝑟(𝑎,𝑏) is the best upper bound for the power mean 𝐴𝑟(𝑎,𝑏) for 1<𝑟<+. For any 𝛽>2(1𝑟)/𝑟, by (3.7) (with 𝛽 in place of 𝜔), we have lim𝑥1+𝑓𝑟(𝑥)=21/𝑟[]2(𝛽+2)𝑟<0.(3.13) Hence, by the continuity of 𝑓𝑟(𝑥), there exists 𝛿=𝛿(𝛽)>0 such that 𝑓𝑟(𝑥)<0 for 𝑥(1,1+𝛿). Thus 𝑓𝑟(𝑥) is strictly decreasing for 𝑥(1,1+𝛿). From (3.6), 𝑓𝑟(𝑥)<0 for 𝑥(1,1+𝛿). This result together with (3.4) implies that 𝑓𝑟(𝑥)<0 for 𝑥(1,1+𝛿). Hence, by (3.3), 𝐻𝛽𝑥2,1<𝐴𝑟𝑥2,,1(3.14) for 𝑥(1,1+𝛿).

Acknowledgments

This work was supported by NSFC (10971224) and NSF of Hebei Province (A2011201011).