Sharp Bounds for the Weighted Geometric Mean of the First Seiffert and Logarithmic Means in terms of Weighted Generalized Heronian Mean

Abstract

Optimal bounds for the weighted geometric mean of the first Seiffert and logarithmic means by weighted generalized Heronian mean are proved. We answer the question: for 𝛼∈(0,1), what the greatest value 𝑝(𝛼) and the least value 𝑞(𝛼) such that the double inequality, 𝐻_𝑝(𝛼)(𝑎,𝑏)<𝑃^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏)<𝐻_𝑞(𝛼)(𝑎,𝑏), holds for all 𝑎,𝑏>0 with 𝑎≠𝑏 are. Here, 𝑃(𝑎,𝑏),𝐿(𝑎,𝑏), and 𝐻_𝜔(𝑎,𝑏) denote the first Seiffert, logarithmic, and weighted generalized Heronian means of two positive numbers 𝑎 and 𝑏, respectively.

1. Introduction

Recently, means has been the subject of intensive research. In particular, many remarkable inequalities for the Seiffert, logarithmic, and Heronian mean can be found in the literature [1–11]. In the paper [1], authors proved the following optimal inequalities:

Let 𝑎>0, 𝑏>0, 𝑎≠𝑏 then 𝐻_𝛿(𝑎,𝑏)<𝑃(𝑎,𝑏)<𝐻_𝛽()𝐻𝑎,𝑏for𝛿≥𝜋−2,𝛽≤1,𝛿=𝜋−2,𝛽=1arethebestconstants._𝛾(𝑎,𝑏)<𝐿(𝑎,𝑏)<𝐻_𝜏(𝑎,𝑏)for𝛾=+∞,𝜏≤4,𝛾=+∞,𝜏=4arethebestconstants. (1)𝑃(𝑎,𝑏) is the first Seiffert mean, which was introduced by Seiffert in [9] 𝑃(𝑎,𝑏)=𝑎−𝑏4arctan=𝑎/𝑏−𝜋𝑎−𝑏2arcsin((𝑎−𝑏)/(𝑎+𝑏))for𝑎,𝑏>0,𝑎≠𝑏. (2) In [9], Seiffert proved that 𝐿(𝑎,𝑏)<𝑃(𝑎,𝑏)<𝐼(𝑎,𝑏), where 𝐼(𝑎,𝑏) is the identric mean 𝐼()=1𝑎,𝑏𝑒𝑎^𝑎𝑏^{𝑏1/(𝑏−𝑎)}if𝑎≠𝑏,𝐼(𝑎,𝑎)=𝑎. (3)𝐿(𝑎,𝑏) is the logarithmic mean 𝐿(𝑎,𝑏)=𝑎−𝑏log𝑎−log𝑏for𝑎,𝑏>0,𝑎≠𝑏. (4)𝐺_𝛼(𝑎,𝑏) is the weighted geometric mean 𝐺_𝛼(𝑎,𝑏)=𝑎^𝛼𝑏^1−𝛼for𝑎,𝑏>0,0≤𝛼≤1. (5)𝐻_𝜔(𝑎,𝑏) is the weighted generalized Heronian mean introduced by Janous [7] 𝐻_𝜔()=√𝑎,𝑏𝑎+𝜔𝑎𝑏+𝑏=√𝜔+2for0≤𝜔<+∞𝑎𝑏for𝜔=+∞. (6) It is well known, that 𝐻_𝜔(𝑎,𝑏) is a strictly decreasing continuous function of the argument 𝜔. From this and from results of [1], it is natural to assume that there exist optimal functions 𝑝(𝛼), 𝑞(𝛼), 0≤𝛼≤1 such that 𝐻_𝑝(𝛼)(𝑎,𝑏)<𝑃^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏)<𝐻_𝑞(𝛼)(𝑎,𝑏). (7) The purpose of this paper is to find the optimal functions. For some other details about means, see [1–11] and the related references cited there in.

2. Main Results

The main result of this paper is the following theorem.

Theorem 1. Let 𝑎,𝑏>0, 𝑎≠𝑏, 𝛼∈(0,1). Then 𝐻_𝑝(𝑎,𝑏)<𝑃^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏)<𝐻_𝑞(),(𝛼),𝑎,𝑏𝑓𝑜𝑟𝑝=+∞,𝑞≤𝑞 (8) where 𝑝=𝑝(𝛼)=+∞, 𝑞(𝛼)=2(2−𝛼)/(1+𝛼) are the best possible functions.

Proof. First, we prove the left inequality of (8). The inequalities (1) imply that 𝐻_+∞(𝑎,𝑏)<𝑃^𝛼(𝑎,𝑏)𝐿^1−𝛼(𝑎,𝑏)for𝑎,𝑏>0,𝑎≠𝑏,0<𝛼<1. (9) From lim_𝑡→0⁺𝐺(𝑡,𝛼)=+∞ for 𝛼∈(0,1) (see (14)) we obtain that 𝑝(𝛼)=+∞ is the optimal function.
Without loss of generality, we assume that 0<𝑎<𝑏. Let 𝑡=𝑎/𝑏; then 0<𝑡<1. The right inequality of (8) can be rewritten as 1𝑏^𝛼𝑃^𝛼()1𝑎,𝑏𝑏^1−𝛼𝐿^1−𝛼()<1𝑎,𝑏𝑏𝐻_𝑞(𝑎,𝑏)for𝑎,𝑏>0,𝑎≠𝑏,0<𝛼<1. (10) Simple computations lead to 1−𝑡²(𝜋−4arctan𝑡)^𝛼(−2ln𝑡)^1−𝛼−𝑡²+𝑞𝑡+1𝑞+2<0for0<𝑡<1,0<𝛼<1. (11) Then the inequality (11) is equivalent to 𝑞1−𝑡²−𝑡(𝜋−4arctan𝑡)^𝛼(−2ln𝑡)^1−𝛼<1+𝑡²(𝜋−4arctan𝑡)^𝛼(−2ln𝑡)^1−𝛼−21−𝑡². (12) Denote 𝑠(𝑡,𝛼)=1−𝑡²−𝑡(𝜋−4arctan𝑡)^𝛼(−2ln𝑡)^1−𝛼,𝑟(𝑡)=𝜋−4arctan𝑡+2ln𝑡,𝑣(𝑡)=𝑡²−2𝑡ln𝑡−1. (13) From 𝑟(1)=0 and 𝑟^′(𝑡)=(2−4𝑡+2𝑡²)/(𝑡+𝑡³)>0 we have 𝑟(𝑡)<0 for 𝑡∈(0,1). It implies ((𝜋−4arctan𝑡)/(−2ln𝑡))^𝛼<1. From 𝑣(1)=0, 𝑣^′(1)=0, 𝑣^′′(𝑡)=2−2/𝑡<0 we obtain 𝑣^′(𝑡)>0 and so 𝑣(𝑡)<0. It implies that 𝑠(𝑡,𝛼)>0 for 𝑡, 𝛼∈(0,1). This leads to ()=𝑞<𝐺𝑡,𝛼1+𝑡²(𝜋−4arctan𝑡)^𝛼(−2ln𝑡)^1−𝛼−21−𝑡²1−𝑡²−𝑡(𝜋−4arctan𝑡)^𝛼(−2ln𝑡)^1−𝛼. (14) If we show 𝐺^′_𝑡(𝑡,𝛼)<0 for 𝑡,𝛼∈(0,1), then 𝑞(𝛼)=lim_𝑡→1⁻𝐺(𝑡,𝛼) will be the best function in (8). Simple computations lead to 𝐺^′_𝑡(𝑡,𝛼)<0 which is equivalent to 𝐻()()()𝑡,𝛼=21−𝑡ln𝑡+4𝛼1+𝑡)(1−𝑡²ln𝑡−()𝜋−4arctan𝑡1−𝛼)(1+𝑡)(1−𝑡²𝑡+2(1+𝑡)ln²𝑡𝜋−4arctan𝑡−2ln𝑡^𝛼<0. (15) Using the inequality 𝑡^𝛼<1−𝛼(1−𝑡) for 𝑡,𝛼∈(0,1) it suffices to show that 𝑅()()()𝑡,𝛼=21−𝑡ln𝑡+4𝛼1+𝑡)(1−𝑡²ln𝑡−()𝜋−4arctan𝑡1−𝛼)(1+𝑡)(1−𝑡²𝑡+(2ln𝑡+𝛼(−2ln𝑡−𝜋+4arctan𝑡))(1+𝑡)ln𝑡<0. (16) It will be done, if we show 𝑅(𝑡,0)<0 and 𝑅(𝑡,1)<0. It follows from 𝑅(𝑡,𝛼) being a linear continuous function in the argument 𝛼𝑅()()()𝑡,0=21−𝑡ln𝑡−1+𝑡)(1−𝑡²𝑡+2(1+𝑡)ln²𝑡<0 (17) is equivalent to 𝑠(𝑡)=2(1−𝑡)ln𝑡−()1+𝑡1−𝑡²𝑡+2ln²𝑡<0. (18) From 𝑠(1)=0 it suffices to show that 𝑠^′(𝑡)>0 which is equivalent to 𝑣(𝑡)=ln𝑡+1+3𝑡−3𝑡²−𝑡³(1+𝑡)4𝑡1+𝑡+𝑡²>0. (19) It follows from 𝑣(1)=0 and 𝑣^′(𝑡)=𝑤(𝑡)4𝑡²1+𝑡+𝑡²²<0, (20) where 𝑤(𝑡)=−(1−𝑡)²(1−𝑡²)².
Next, we show that 𝑅()()8()𝑡,1=41−𝑡ln𝑡+1+𝑡)(1−𝑡²ln𝑡()𝜋−4arctan𝑡−2𝜋−4arctan𝑡)(1+𝑡ln𝑡<0. (21) The inequality (21) is equivalent to 𝜋−4arctan𝑡−1−𝑡1+𝑡²<1−𝑡1+𝑡²−4(1−𝑡)². (22) So, it suffices to show that 𝑔(𝑡)=−𝜋+4arctan𝑡+1−𝑡1+𝑡1+1+4(1+𝑡)²>0. (23) It is easy to see that 1+1+4(1+𝑡)²()+1>1+21+𝑡5()=1+𝑡16+25𝑡+10𝑡²5(1+𝑡)for0≤𝑡≤1. (24) Because of 𝑔(𝑡)>𝑔₁(𝑡)+=−𝜋+4arctan𝑡1−𝑡1+𝑡16+25𝑡+10𝑡²5(1+𝑡)>0, (25) it suffices to prove 𝑔₁(𝑡)>0 for 0<𝑡<1. From 𝑔₁(𝑡)=15𝑔₂(𝑡)=15()−5𝜋+20arctan𝑡+1−𝑡16+25𝑡+10𝑡²(1+𝑡)², (26)arctan(𝑡)>𝑡−𝑡³/3, for 𝑡∈(0,1), 𝑔₂(1)=0 we have done it, if we show 𝑔₃(𝑡)𝑡=−5𝜋+20𝑡−³3+(1−𝑡)16+25𝑡+10𝑡²(1+𝑡)²>0, (27) on (0,0.67⟩ and 𝑔^′₂(𝑡)<0 on ⟨0.67,1).
Simple computation gives 𝑔^′₂(𝑡)=201+𝑡²−23+39𝑡+30𝑡²+10𝑡³(1+𝑡)³. (28) The inequality 𝑔^′₂(𝑡)<0 is equivalent to 𝑐ℎ(𝑡)=−3+21𝑡+7𝑡²−29𝑡³−30𝑡⁴−10𝑡⁵<0. (29) From 𝑐ℎ^′′′(𝑡)<0 we get 𝑐ℎ^′′(𝑡) is a decreasing function. 𝑐ℎ^′′(0.67)=−324.3366 implies 𝑐ℎ^′′(𝑡)<0 on ⟨0.67,1). So, we obtain 𝑐ℎ^′(𝑡) is a decreasing function. From 𝑐ℎ^′(0.67)=−54.8414 we have 𝑐ℎ^′(𝑡)<0 on ⟨0.67,1). It implies that 𝑐ℎ(𝑡) is a decreasing function. From 𝑐ℎ(0.67)=−1.9053 we get 𝑐ℎ(𝑡)<0 on ⟨0.67,1). So 𝑔^′₂(𝑡)<0 on ⟨0.67,1).
Next, we show 𝑔₃(𝑡)>0 on (0,0.67⟩.
Simple computation gives 𝑔₃(𝑡)=16−5𝜋+𝑡(29−10𝜋)+𝑡²(25−5𝜋)+𝑡³103−𝑡⁴403−𝑡⁵203(1+𝑡)²⁻¹. (30) The inequality 𝑔₃(𝑡)>0 is equivalent to ℎ(𝑡)=16−5𝜋+𝑡(29−10𝜋)+𝑡²(25−5𝜋)+𝑡³103−𝑡⁴403−𝑡⁵203>0, (31) on (0,0.67⟩. From ℎ(0)=16−5𝜋>0, ℎ(0.1)=0.1453, ℎ(0.15)=0.1427, ℎ(0.67)=0.2602, ℎ^′(0.15)=0.3998 it suffices to show that ℎ^′(𝑡)<0 on (0,0.1⟩; ℎ(𝑡)>0 on ⟨0.1,0.15⟩ and ℎ^′(𝑡) has only one root in (0.15,0.67).
First, we show ℎ(𝑡)>0 on ⟨0.1,0.15⟩. From 𝑡³>0.1𝑡², 𝑡⁴<0.15²𝑡², 𝑡⁵<0.15³𝑡² we have ℎ(𝑡)>16−5𝜋+𝑡(29−10𝜋)+𝑡²25−5𝜋+0.1103−0.15²403−0.15³203>𝑙(𝑡), (32) where 𝑙(𝑡)=16−5𝜋+𝑡(29−10𝜋)+𝑡²9.3. (33) It is easy to see that 𝑙^′(𝑡)=0 for 𝑡=(10𝜋−29)/18.6=0.1299. From 𝑙^′′(𝑡)>0 on ⟨0.1,0.15⟩ and 𝑙(0.1299)=0.1351 we have 𝑙(𝑡)>0. It implies ℎ(𝑡)>0 on ⟨0.1,0.15⟩.
Next, we show ℎ^′(𝑡)<0 on (0,0.1⟩. Simple computation gives ℎ^′(𝑡)=(29−10𝜋)+(50−10𝜋)𝑡+10𝑡²−1603𝑡³−1003𝑡⁴<𝑗(𝑡), (34) where 𝑗(𝑡)=(29−10𝜋)+(50−10𝜋)𝑡+10𝑡², (35)𝑗(0)=29−10𝜋<0, 𝑗^′′(𝑡)>0, 𝑗(0.1)=−0.45750 imply 𝑗(𝑡)<0 so ℎ^′(𝑡)<0 on (0,0.1⟩.
Finally, we show that ℎ^′(𝑡) has only one root on (0.15,0.67). From ℎ^′′′′(𝑡)<0 we obtain ℎ^′′′(𝑡) is a decreasing function. Because of ℎ^′′′(0.15)=−37 we have ℎ^′′′(𝑡)<0 on (0.15,0.67) so ℎ^′(𝑡) is a concave function. From ℎ^′(0.15)=0.3998 and ℎ^′(0.67)=−8.2333 we have that ℎ^′(𝑡) has only one root on (0.15,0.67). It implies ℎ(𝑡)>0 on ⟨0.15,0.67⟩. So, the proof of decreasing of 𝐺(𝑡,𝛼) is complete.
In what follows, we find the representation of the function 𝑞(𝛼).
It is easy to see that𝑞(𝛼)=lim_𝑡→1−1+𝑡²(𝜋−4arctan𝑡)^𝛼×(−2ln𝑡)^1−𝛼−21−𝑡²×1−𝑡²−𝑡(𝜋−4arctan𝑡)^𝛼×(−2ln𝑡)^1−𝛼−1. (36) Equation (36) can be rewritten as 𝑞(𝛼)=lim_𝑡→1−1+𝑡²𝑌(𝑡,𝛼)𝑈(𝑡,𝛼)−2(1+𝑡)()𝑈(),1+𝑡−𝑡𝑌𝑡,𝛼𝑡,𝛼 (37) where 𝑌(𝑡,𝛼)=𝜋−4arctan𝑡1−𝑡^𝛼,𝑈(𝑡,𝛼)=−2ln𝑡1−𝑡^1−𝛼. (38) Simple computations give 𝑌(𝑡,𝛼)=2^𝛼1+1−𝑡2+(1−𝑡)²6+𝑦(𝑡)(1−𝑡)³, (39) where 𝑦(𝑡) is a suitable function. Similarly we have 𝑈(𝑡,𝛼)=2^1−𝛼1+1−𝑡2+(1−𝑡)²3+𝑢(𝑡)(1−𝑡)³, (40) where 𝑢(𝑡) is a suitable function. Denote 𝑆(𝑡,𝛼)=𝑌(𝑡,𝛼)𝑈(𝑡,𝛼). Then 𝑆(𝑡,𝛼)=21+1−𝑡2+(2−𝛼)(1−𝑡)²6+𝑠(𝑡)(1−𝑡)³, (41) where 𝑠(𝑡) is a suitable function. Using the L’Hospital's rule we obtain 𝑞(𝛼)=lim_𝑡→1−1+𝑡²𝑆(𝑡,𝛼)−2(1+𝑡)1+𝑡−𝑡𝑆(𝑡,𝛼)=lim_𝑡→1−2𝑆(𝑡,𝛼)+4𝑡𝑆^′_𝑡(𝑡,𝛼)+1+𝑡²𝑆^′′_𝑡𝑡(𝑡,𝛼)−2𝑆^′_𝑡(𝑡,𝛼)−𝑡𝑆^′′_𝑡𝑡()=2()𝑡,𝛼2−𝛼.1+𝛼 (42) The proof is complete.

Acknowledgments

The work was supported by VEGA Grant no. 1/0530/11 and KEGA Grant no. 0007 TnUAD-4/2013. The author thanks to the faculty FPT TnUAD in Púchov, Slovakia for its kind support and the anonymous referees for their careful reading of the paper and fruitful comments and suggestions. The author would especially like to thank Prof. Walther Janous for his kind reading the manuscript and for his correction of the calculation 𝑞(𝛼).