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Abstract and Applied Analysis
VolumeΒ 2011Β (2011), Article IDΒ 679201, 9 pages
Research Article

Sharp Bounds for Power Mean in Terms of Generalized Heronian Mean

College of Mathematics and Computer Science, Hebei University, Baoding 071002, China

Received 2 March 2011; Accepted 7 April 2011

Academic Editor: MarciaΒ Federson

Copyright Β© 2011 Hongya Gao et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


For 1<π‘Ÿ<+∞, we find the least value 𝛼 and the greatest value 𝛽 such that the inequality 𝐻𝛼(π‘Ž,𝑏)<π΄π‘Ÿ(π‘Ž,𝑏)<𝐻𝛽(π‘Ž,𝑏) holds for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘. Here, π»πœ”(π‘Ž,𝑏) and π΄π‘Ÿ(π‘Ž,𝑏) are the generalized Heronian and the power means of two positive numbers π‘Ž and 𝑏, respectively.

1. Introduction and Statement of Result

For π‘Ž,𝑏>0 with π‘Žβ‰ π‘, the generalized Heronian mean of π‘Ž and 𝑏 is defined by Janous [1] as π»πœ”βŽ§βŽͺ⎨βŽͺ⎩√(π‘Ž,𝑏)=π‘Ž+πœ”π‘Žπ‘+π‘βˆšπœ”+2,0β‰€πœ”<+∞,π‘Žπ‘,πœ”=+∞.(1.1) If we take πœ”=1 in (1.1), then we arrive at the classical Heronian mean π»π‘’βˆš(π‘Ž,𝑏)=π‘Ž+π‘Žπ‘+𝑏3.(1.2) The domain of definition for the function πœ”β†¦π»πœ”(π‘Ž,𝑏) can be extended to all πœ” with πœ”βˆˆ(βˆ’2,+∞), that is, π»πœ”βŽ§βŽͺ⎨βŽͺ⎩√(π‘Ž,𝑏)=π‘Ž+πœ”π‘Žπ‘+π‘βˆšπœ”+2,βˆ’2<πœ”<+∞,π‘Žπ‘,πœ”=+∞.(1.3) For all fixed π‘Ž,𝑏>0, it is easy to derive that πœ”β†¦π»πœ”(π‘Ž,𝑏), βˆ’2<πœ”<+∞ is monotonically decreasing, and limπœ”β†’βˆ’2+π»πœ”(π‘Ž,𝑏)=+∞.(1.4)

Let π΄π‘ŸβŽ§βŽͺβŽͺβŽͺ⎨βŽͺβŽͺβŽͺβŽ©ξ‚΅π‘Ž(π‘Ž,𝑏)=π‘Ÿ+π‘π‘Ÿ2ξ‚Ά1/π‘Ÿβˆš,π‘Ÿβ‰ 0,π‘Žπ‘,π‘Ÿ=0,max{π‘Ž,𝑏},π‘Ÿ=+∞,min{π‘Ž,𝑏},π‘Ÿ=βˆ’βˆž,(1.5) denote the power mean of order π‘Ÿ. In particular, the harmonic, geometric, square-root, arithmetic, and root-square means of π‘Ž and 𝑏 are 𝐻(π‘Ž,𝑏)=π΄βˆ’1(π‘Ž,𝑏)=2π‘Ž,π‘Ž+𝑏𝐺(π‘Ž,𝑏)=𝐴0√(π‘Ž,𝑏)=π‘π‘Žπ‘,1(π‘Ž,𝑏)=𝐴1/2ξƒ©βˆš(π‘Ž,𝑏)=βˆšπ‘Ž+𝑏2ξƒͺ2,𝐴(π‘Ž,𝑏)=𝐴1(π‘Ž,𝑏)=π‘Ž+𝑏2,𝑆(π‘Ž,𝑏)=𝐴2ξ‚™(π‘Ž,𝑏)=π‘Ž2+𝑏22.(1.6) It is well known that the power mean of order π‘Ÿ given in (1.5) is monotonically increasing in π‘Ÿ, then we can write min{π‘Ž,𝑏}<𝐻(π‘Ž,𝑏)<𝐺(π‘Ž,𝑏)<𝑁1(π‘Ž,𝑏)<𝐴(π‘Ž,𝑏)<𝑆(π‘Ž,𝑏)<max{π‘Ž,𝑏}.(1.7)

Recently, the inequalities for means have been the subject of intensive research [1–15]. In particular, many remarkable inequalities for the generalized Heronian and power means can be found in the literature [4–9].

In [4], the authors established two sharp inequalities 231𝐺(π‘Ž,𝑏)+3𝐻(π‘Ž,𝑏)β‰₯π΄βˆ’1/31(π‘Ž,𝑏),32𝐺(π‘Ž,𝑏)+3𝐻(π‘Ž,𝑏)β‰₯π΄βˆ’2/3(π‘Ž,𝑏).(1.8)

In [5], Long and Chu found the greatest value 𝑝 and the least value π‘ž such that the double inequality 𝐴𝑝(π‘Ž,𝑏)≀𝐴(π‘Ž,𝑏)𝛼𝐺(π‘Ž,𝑏)𝛽𝐻(π‘Ž,𝑏)1βˆ’π›Όβˆ’π›½β‰€π΄π‘ž(π‘Ž,𝑏)(1.9) holds for all π‘Ž,𝑏>0 and 𝛼,𝛽>0 with 𝛼+𝛽<1.

In [6], Shi et al. gave two optimal inequalities 𝐴𝛼(π‘Ž,𝑏)𝐿1βˆ’π›Ό(π‘Ž,𝑏)≀𝐴(1+2𝛼)/3𝐺(π‘Ž,𝑏),𝛼(π‘Ž,𝑏)𝐿1βˆ’π›Ό(π‘Ž,𝑏)≀𝐴(1βˆ’π›Ό)/3(π‘Ž,𝑏),(1.10) for 0<𝛼<1, where 𝐿(π‘Ž,𝑏)=π‘Žβˆ’π‘logπ‘Žβˆ’log𝑏,π‘Žβ‰ π‘,(1.11) is the logarithmic mean for π‘Ž,𝑏>0.

In [7], Guan and Zhu obtained sharp bounds for the generalized Heronian mean in terms of the power mean with πœ”>0. The optimal values 𝛼 and 𝛽 such that 𝐴𝛼(π‘Ž,𝑏)β‰€π»πœ”(π‘Ž,𝑏)≀𝐴𝛽(π‘Ž,𝑏)(1.12) holds in general are (1)in case of πœ”βˆˆ(0,2],𝛼max=log2/log(πœ”+2) and 𝛽min=2/(πœ”+2),(2) in case of πœ”βˆˆ[2,+∞),𝛼max=2/(πœ”+2) and 𝛽min=log2/log(πœ”+2).

In this paper, we find the least value 𝛼 and the greatest value 𝛽, such that for any fixed 1<π‘Ÿ<+∞, the inequality 𝐻𝛼(π‘Ž,𝑏)<π΄π‘Ÿ(π‘Ž,𝑏)<𝐻𝛽(π‘Ž,𝑏)(1.13) holds for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘.

Theorem 1.1. For 1<π‘Ÿ<+∞, the optimal numbers 𝛼 and 𝛽 such that 𝐻𝛼(π‘Ž,𝑏)<π΄π‘Ÿ(π‘Ž,𝑏)<𝐻𝛽(π‘Ž,𝑏)(1.14) is valid for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘, are 𝛼min=21/π‘Ÿβˆ’2 and 𝛽max=2(1βˆ’π‘Ÿ)/π‘Ÿ.

Notice that in our case π‘Ÿ>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<π‘Ÿ<+∞, one has π‘Ÿ21/π‘Ÿβˆ’1>1.(2.1)

Proof. We show that π‘š(π‘Ÿ)=(1βˆ’π‘Ÿ)log2+π‘Ÿlogπ‘Ÿ>0,(2.2) which is clearly equivalent to the claim. Equation (2.2) follows from the facts limπ‘Ÿβ†’1+π‘š(π‘Ÿ)=0,π‘šξ…ž(π‘Ÿ)=βˆ’log2+logπ‘Ÿ+1>0.(2.3)

Corollary 2.2. If 1<π‘Ÿ<+∞, then βˆ’2<2(1βˆ’π‘Ÿ)π‘Ÿ<21/π‘Ÿβˆ’2<0.(2.4)

Proof. Since for 1<π‘Ÿ<+∞, the two functions πœ‘1(π‘Ÿ)=2(1βˆ’π‘Ÿ)π‘Ÿ,πœ‘2(π‘Ÿ)=21/π‘Ÿβˆ’2(2.5) are strictly decreasing, then one has βˆ’2=limπ‘Ÿβ†’+βˆžπœ‘1(π‘Ÿ)<πœ‘1(π‘Ÿ),πœ‘2(π‘Ÿ)<limπ‘Ÿβ†’1+πœ‘2(π‘Ÿ)=0.(2.6) It suffices to show that 2βˆ’2π‘Ÿ<π‘Ÿ21/π‘Ÿβˆ’2π‘Ÿ,(2.7) which is equivalent to (2.1).

Lemma 2.3. For π‘₯>1 and π‘Ÿ>1, let ξ€·π‘₯β„“(π‘₯)=2π‘Ÿξ€Έ+11/π‘Ÿβˆ’2π‘₯2(π‘Ÿβˆ’1)ξ€·π‘₯2π‘Ÿξ€Έ.+2π‘Ÿβˆ’1(2.8) Then, β„“(π‘₯) is strictly decreasing for π‘₯>1, and limπ‘₯β†’1+β„“(π‘₯)=π‘Ÿ21/π‘Ÿβˆ’1,limπ‘₯β†’+βˆžβ„“(π‘₯)=1.(2.9)

Proof. The fact β„“(π‘₯)>0 for π‘₯>1 and π‘Ÿ>1 is obvious, which allows us to take the logarithmic function of β„“(π‘₯), ξ‚€1logβ„“(π‘₯)=π‘Ÿξ‚ξ€·π‘₯βˆ’2log2π‘Ÿξ€Έξ€·π‘₯+1+2(π‘Ÿβˆ’1)logπ‘₯+log2π‘Ÿξ€Έ.+2π‘Ÿβˆ’1(2.10) Some tedious, but not difficult calculations lead to ξ€Ίξ€»logβ„“(π‘₯)ξ…ž=ξ‚€1π‘Ÿξ‚βˆ’22π‘Ÿπ‘₯2π‘Ÿβˆ’1π‘₯2π‘Ÿ++12(π‘Ÿβˆ’1)π‘₯+2π‘Ÿπ‘₯2π‘Ÿβˆ’1π‘₯2π‘Ÿ=+2π‘Ÿβˆ’1π‘š(π‘₯)π‘₯ξ€·π‘₯2π‘Ÿπ‘₯+1ξ€Έξ€·2π‘Ÿξ€Έ,+2π‘Ÿβˆ’1(2.11) where π‘š(π‘₯)=2(1βˆ’2π‘Ÿ)π‘₯2π‘Ÿξ€·π‘₯2π‘Ÿξ€Έξ€·π‘₯+2π‘Ÿβˆ’1+(2π‘Ÿβˆ’1)2π‘Ÿπ‘₯+1ξ€Έξ€·2π‘Ÿξ€Έ+2π‘Ÿβˆ’1+2π‘Ÿπ‘₯2π‘Ÿξ€·π‘₯2π‘Ÿξ€Έξ€·+1=2(π‘Ÿβˆ’1)(2π‘Ÿβˆ’1)1βˆ’π‘₯2π‘Ÿξ€Έ.(2.12) It is easy to see that limπ‘₯β†’1+π‘šπ‘š(π‘₯)=0,(2.13)ξ…ž(π‘₯)=βˆ’4π‘Ÿ(π‘Ÿβˆ’1)(2π‘Ÿβˆ’1)π‘₯2π‘Ÿβˆ’1<0.(2.14) Equation (2.14) implies that π‘š(π‘₯) is strictly decreasing for π‘₯>1, which together with (2.13) implies π‘š(π‘₯)<0 for π‘₯>1. Thus, by (2.11), ξ€Ίξ€»logβ„“(π‘₯)ξ…ž<0,(2.15) which implies β„“ξ…žξ€Ίξ€»(π‘₯)=logβ„“(π‘₯)ξ…žβ„“(π‘₯)<0.(2.16) Hence, β„“(π‘₯) is strictly decreasing.
It remains to show (2.9). The first equality in (2.9) is obvious. The second one follows from limπ‘₯β†’+βˆžβ„“(π‘₯)=limπ‘₯β†’+βˆžξ€·π‘₯2π‘Ÿξ€Έ+11/π‘Ÿβˆ’2π‘₯2(π‘Ÿβˆ’1)ξ€·π‘₯2π‘Ÿξ€Έ+2π‘Ÿβˆ’1=lim𝑑→0+(2π‘Ÿβˆ’1)𝑑2π‘Ÿ+1ξ€·1+𝑑2π‘Ÿξ€Έ(2π‘Ÿβˆ’1)/π‘Ÿ=1.(2.17) This ends the proof of Lemma 2.3.

Lemma 2.4. For π‘₯>1, π‘Ÿ>1, and πœ”=21/π‘Ÿβˆ’2, let π‘“π‘Ÿ(π‘₯)=21/π‘Ÿξ€·π‘₯2ξ€Έξ€·π‘₯+πœ”π‘₯+1βˆ’(πœ”+2)2π‘Ÿξ€Έ+11/π‘Ÿ.(2.18) Then, limπ‘₯β†’+βˆžπ‘“π‘Ÿ(π‘₯)=βˆ’βˆž,limπ‘₯β†’+βˆžπ‘“ξ…žπ‘Ÿ(π‘₯)=21/π‘Ÿξ€·21/π‘Ÿξ€Έ.βˆ’2(2.19)

Proof. Simple calculations lead to limπ‘₯β†’+βˆžπ‘“π‘Ÿ(π‘₯)=limπ‘₯β†’+∞21/π‘Ÿξ€·π‘₯2ξ€Έξ€·π‘₯+πœ”π‘₯+1βˆ’(πœ”+2)2π‘Ÿξ€Έ+11/π‘Ÿ=lim𝑑→0+21/π‘Ÿξ€·π‘‘2𝑑+πœ”π‘‘+1βˆ’(πœ”+2)2π‘Ÿξ€Έ+11/π‘Ÿπ‘‘2=βˆ’βˆž,limπ‘₯β†’+βˆžπ‘“ξ…žπ‘Ÿ(π‘₯)=limπ‘₯β†’+∞21/π‘Ÿξ€·π‘₯(2π‘₯+πœ”)βˆ’2(πœ”+2)2π‘Ÿξ€Έ+11/π‘Ÿ=lim𝑑→0+21/π‘Ÿξ€·(2+πœ”π‘‘)βˆ’2(πœ”+2)1+𝑑2π‘Ÿξ€Έ(1βˆ’π‘Ÿ)/π‘Ÿπ‘‘=lim𝑑→0+21/π‘Ÿξ€·(2+πœ”π‘‘)1+𝑑2π‘Ÿξ€Έ(π‘Ÿβˆ’1)/π‘Ÿβˆ’2(πœ”+2)𝑑1+𝑑2π‘Ÿξ€Έ(π‘Ÿβˆ’1)/π‘Ÿ=lim𝑑→0+21/π‘Ÿπœ”ξ€·1+𝑑2π‘Ÿξ€Έ(π‘Ÿβˆ’1)/π‘Ÿ+2(1/π‘Ÿ)+1ξ€·(π‘Ÿβˆ’1)(2+πœ”π‘‘)1+𝑑2π‘Ÿξ€Έβˆ’1/π‘Ÿπ‘‘2π‘Ÿβˆ’1ξ€·1+𝑑2π‘Ÿξ€Έ(π‘Ÿβˆ’1)/π‘Ÿξ€·+2(π‘Ÿβˆ’1)1+𝑑2π‘Ÿξ€Έβˆ’1/π‘Ÿπ‘‘2π‘Ÿ=21/π‘Ÿπœ”=21/π‘Ÿξ€·πœ”1/π‘Ÿξ€Έβˆ’2<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<π‘Ÿ<+∞, 𝐻2(1βˆ’π‘Ÿ)/π‘Ÿ(π‘Ž,𝑏)>π΄π‘Ÿπ»(π‘Ž,𝑏),(3.1)21/π‘Ÿβˆ’2(π‘Ž,𝑏)<π΄π‘Ÿ(π‘Ž,𝑏)(3.2) hold true for all π‘Ž,𝑏>0 with π‘Žβ‰ π‘. It is no loss of generality to assume that π‘Ž>𝑏>0. Let √π‘₯=𝑏/π‘Ž>1 and πœ”βˆˆ{2(1βˆ’π‘Ÿ)/π‘Ÿ,21/π‘Ÿβˆ’2}. In view of Corollary 2.2, βˆ’2<πœ”<0. Equations (1.3) and (1.5) lead to 1π‘Žξ€Ίπ»πœ”(π‘Ž,𝑏)βˆ’π΄π‘Ÿξ€»(π‘Ž,𝑏)=π»πœ”ξ€·π‘₯2ξ€Έ,1βˆ’π΄π‘Ÿξ€·π‘₯2ξ€Έ=π‘₯,12+πœ”π‘₯+1βˆ’ξ‚΅π‘₯πœ”+22π‘Ÿ+12ξ‚Ά1/π‘Ÿ=21/π‘Ÿξ€·π‘₯2ξ€Έξ€·π‘₯+πœ”π‘₯+1βˆ’(πœ”+2)2π‘Ÿξ€Έ+11/π‘Ÿ21/π‘Ÿ=𝑓(πœ”+2)π‘Ÿ(π‘₯)21/π‘Ÿ,(πœ”+2)(3.3) where π‘“π‘Ÿ(π‘₯) is defined by (2.18). It is easy to see that limπ‘₯β†’1+π‘“π‘Ÿπ‘“(π‘₯)=0,(3.4)ξ…žπ‘Ÿ(π‘₯)=21/π‘Ÿξ€·π‘₯(2π‘₯+πœ”)βˆ’2(πœ”+2)2π‘Ÿξ€Έ+11/π‘Ÿβˆ’1π‘₯2π‘Ÿβˆ’1,(3.5)limπ‘₯β†’1+π‘“ξ…žπ‘Ÿ(π‘₯)=0.(3.6) By Lemma 2.3, π‘“π‘Ÿξ…žξ…žξ‚†2(π‘₯)=21/π‘Ÿξ‚ƒξ€·π‘₯βˆ’(πœ”+2)2(1βˆ’π‘Ÿ)2π‘Ÿξ€Έ+11/π‘Ÿβˆ’2π‘₯4π‘Ÿβˆ’2ξ€·π‘₯+(2π‘Ÿβˆ’1)2π‘Ÿξ€Έ+11/π‘Ÿβˆ’1π‘₯2(π‘Ÿβˆ’1)ξ€Ί2=21/π‘Ÿξ€»ξ€Ί2βˆ’(πœ”+2)β„“(π‘₯)>21/π‘Ÿβˆ’(πœ”+2)π‘Ÿ21/π‘Ÿβˆ’1ξ€»=21/π‘Ÿ[],2βˆ’(πœ”+2)π‘Ÿ(3.7)limπ‘₯β†’1+π‘“π‘Ÿξ…žξ…žξ€½2(π‘₯)=21/π‘Ÿξ€Ίβˆ’(πœ”+2)2(1βˆ’π‘Ÿ)21/π‘Ÿβˆ’2+(2π‘Ÿβˆ’1)21/π‘Ÿβˆ’1ξ€»ξ€Ύ=21/π‘Ÿ[].2βˆ’(πœ”+2)π‘Ÿ(3.8)
We now distinguish between two cases.
Case 1 (πœ”=2(1βˆ’π‘Ÿ)/π‘Ÿ). Since 2βˆ’(πœ”+2)π‘Ÿ=0, then by (3.7), π‘“π‘Ÿξ…žξ…ž(π‘₯)>0. Thus, π‘“ξ…žπ‘Ÿ(π‘₯) is strictly increasing for π‘₯>1, which together with (3.6) implies π‘“ξ…žπ‘Ÿ(π‘₯)>0. Hence, π‘“π‘Ÿ(π‘₯) is strictly increasing for π‘₯>1. Since (3.4), then π‘“π‘Ÿ(π‘₯)>0. Equation (3.1) follows from (3.3).Case 2 (πœ”=21/π‘Ÿβˆ’2). By (3.5) and (2.11), π‘“π‘Ÿξ…žξ…žξ…ž(π‘₯)=βˆ’2(πœ”+2)β„“ξ…žξ€Ίξ€»(π‘₯)=βˆ’2(πœ”+2)logβ„“(π‘₯)ξ…žβ„“(π‘₯)=βˆ’2(πœ”+2)π‘š(π‘₯)β„“(π‘₯)π‘₯ξ€·π‘₯2π‘Ÿξ€Έ+1(π‘₯π‘Ÿ)+2π‘Ÿβˆ’1>0.(3.9) Thus, π‘“π‘Ÿξ…žξ…ž(π‘₯) is strictly increasing. Equations (3.8) and (2.1) imply limπ‘₯β†’1+π‘“π‘Ÿξ…žξ…ž(π‘₯)=21/π‘Ÿ[]2βˆ’(πœ”+2)π‘Ÿ=21/π‘Ÿ+1ξ€·1βˆ’π‘Ÿ21/π‘Ÿβˆ’1ξ€Έ<0.(3.10) Equations (3.7) and (2.9) imply limπ‘₯β†’+βˆžπ‘“π‘Ÿξ…žξ…ž(π‘₯)=limπ‘₯β†’+∞2ξ€Ί21/π‘Ÿβˆ’ξ€»ξ€·2(πœ”+2)β„“(π‘₯)=21/π‘Ÿξ€Έβˆ’1>0.(3.11) Combining (3.10) with (3.11), we obviously know that there exists πœ†1>1 such that π‘“π‘Ÿξ…žξ…ž(π‘₯)<0 for π‘₯∈(1,πœ†1) and π‘“π‘Ÿξ…žξ…ž(π‘₯)>0 for π‘₯∈(πœ†1,+∞). This implies that π‘“ξ…žπ‘Ÿ(π‘₯) is strictly decreasing for π‘₯∈(1,πœ†1) and strictly increasing for π‘₯∈(πœ†1,+∞). By (3.6) and Lemma 2.4, we know that π‘“ξ…žπ‘Ÿ(π‘₯)<0 for π‘₯>1. Therefore, π‘“π‘Ÿ(π‘₯) is strictly decreasing. By (3.4) and Lemma 2.4 again, we derive that π‘“π‘Ÿ(π‘₯)<0 for π‘₯>1. Equation (3.2) follows from (3.3).Secondly, we prove that 𝐻21/π‘Ÿβˆ’2(π‘Ž,𝑏) is the best lower bound for the power mean π΄π‘Ÿ(π‘Ž,𝑏) for 1<π‘Ÿ<+∞. For any 𝛼<21/π‘Ÿβˆ’2, limπ‘₯β†’+βˆžπ»π›Ό(π‘₯,1)π΄π‘Ÿ(π‘₯,1)=limπ‘₯β†’+∞21/π‘Ÿξ‚€βˆšπ‘₯+𝛼π‘₯+1(𝛼+2)(π‘₯π‘Ÿ+1)π‘Ÿ=21/π‘Ÿπ›Ό+2>1.(3.12) Hence, there exists 𝑋=𝑋(𝛼)>1 such that 𝐻𝛼(π‘₯,1)>π΄π‘Ÿ(π‘₯,1) for π‘₯∈(𝑋,+∞).
Finally, we prove that 𝐻2(1βˆ’π‘Ÿ)/π‘Ÿ(π‘Ž,𝑏) is the best upper bound for the power mean π΄π‘Ÿ(π‘Ž,𝑏) for 1<π‘Ÿ<+∞. For any 𝛽>2(1βˆ’π‘Ÿ)/π‘Ÿ, by (3.7) (with 𝛽 in place of πœ”), we have limπ‘₯β†’1+π‘“π‘Ÿξ…žξ…ž(π‘₯)=21/π‘Ÿ[]2βˆ’(𝛽+2)π‘Ÿ<0.(3.13) Hence, by the continuity of π‘“π‘Ÿξ…žξ…ž(π‘₯), there exists 𝛿=𝛿(𝛽)>0 such that π‘“π‘Ÿξ…žξ…ž(π‘₯)<0 for π‘₯∈(1,1+𝛿). Thus π‘“π‘Ÿ(π‘₯) is strictly decreasing for π‘₯∈(1,1+𝛿). From (3.6), π‘“ξ…žπ‘Ÿ(π‘₯)<0 for π‘₯∈(1,1+𝛿). This result together with (3.4) implies that π‘“π‘Ÿ(π‘₯)<0 for π‘₯∈(1,1+𝛿). Hence, by (3.3), 𝐻𝛽π‘₯2ξ€Έ,1<π΄π‘Ÿξ€·π‘₯2ξ€Έ,,1(3.14) for π‘₯∈(1,1+𝛿).


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


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