TWO SHARP INEQUALITIES FOR BOUNDING THE SEIFFERT MEAN BY THE ARITHMETIC, CENTROIDAL, AND CONTRA-HARMONIC MEANS 2 1 0 WEI-DONGJIANG,JIANCAO,ANDFENGQI 2 n Abstract. Inthepaper,theauthorsfindthebestpossibleconstantsappeared a intwoinequalities forbounding the Seiffertmeanbythe linearcombinations J ofthearithmetic,centroidal,andcontra-harmonicmeans. 1 3 ] A 1. Introduction C Fora,b>0witha=b,theSeiffertmeanT(a,b)andthecentroidalmeanC(a,b) . 6 h are defined respectively by t a a b m T(a,b)= 2arcta−n a−b (1.1) a+b [ and (cid:0) (cid:1) 1 2 a2+ab+b2 v C(a,b)= . (1.2) 3(a+b) 2 (cid:0) (cid:1) 3 It is well known that 4 a+b a2+b2 6 A(a,b)= , G(a,b)=√ab, S(a,b)= , 1. 2 r 2 0 C(a,b)= a2+b2, M (a,b)= p ap+bp 2 a+b p 2 r 1 for p = 0 are respectively the arithmetic, geometric, root-square, contra-harmonic : 6 v andp-th power means of two positive numbers a and b, that the p-th power means Xi Mp(a,b)iscontinuousandstrictlyincreasingwithrespecttop Rforfixeda,b>0 ∈ with a=b, and that the inequalities in r 6 a G(a,b)=M (a,b)<A(a,b)=M (a,b)<C(a,b) 0 1 <S(a,b)=M (a,b)<C(a,b) (1.3) 2 holdfora,b>0witha=b. Formoreinformationonresultsofmeanvalues,please 6 refer to, for example, [11, 12, 19, 20, 21] and closely related references therein. In [22], Seiffert proved the double inequality A(a,b)=M (a,b)<T(a,b)<M (a,b)=S(a,b) (1.4) 1 2 2010 Mathematics Subject Classification. Primary 26E60; Secondary 11H60, 26A48, 26D05, 33B10. Key words and phrases. Seiffert mean; Arithmetic mean; Centroidal mean; Contra-harmonic mean;Inequality; Bestconstant. Thefirstauthor waspartiallysupportedbytheProjectofShandong ProvinceHigherEduca- tionalScienceandTechnologyProgramundergrantNo. J11LA57. 1 2 W.-D.JIANG,J.CAO,ANDF.QI fora,b>0witha=b. In[13],Ha¨sto¨showedthatthefunction T(1,x) isincreasing 6 Mp(1,x) with respect to x (0, ) if p 1. In [3, 5], the authors demonstrated that the ∈ ∞ ≤ double inequalities α S(a,b)+(1 α )A(a,b)<T(a,b)<β S(a,b)+(1 β )A(a,b) (1.5) 1 1 1 1 − − and C α a+(1 α )b,α b+(1 α )a <T(a,b) 2 2 2 2 − − (cid:0) <(cid:1) C β2a+(1 β2)b,β2b+(1 β2)a (1.6) − − hold for a,b>0 with a=b if and only if (cid:0) (cid:1) 6 4 π 2 1 4 3+√3 α − , β , α 1+ 1 , β . (1.7) 1 ≤ √2 1 π 1 ≥ 3 2 ≤ 2 rπ − ! 2 ≥ 6 − For more(cid:0)informa(cid:1)tion on this topic, please refer to recently published papers [4, 6, 7, 8, 9, 10, 14, 15, 16, 18, 23, 24] and cited references therein. For positive numbers a,b>0 with a=b, let 6 J(x)=C xa+(1 x)b,xb+(1 x)a (1.8) − − on 1,1 . It is not difficult to(cid:0)directly verify that J(x) is c(cid:1)ontinuous and strictly 2 increasing on 1,1 and to notice that (cid:2) (cid:3) 2 (cid:2)1 (cid:3) J =A(a,b)<T(a,b) and J(1)=C(a,b)>T(a,b). (1.9) 2 (cid:18) (cid:19) Therefore,itismuchnaturaltoaskaquestion: Whatarethebestconstantsα 1 ≥ 2 and β 1 such that the double inequality ≤ C αa+(1 α)b,αb+(1 α)a <T(a,b)<C βa+(1 β)b,βb+(1 β)a (1.10) − − − − hol(cid:0)ds for a,b>0 with a=b? (cid:1) (cid:0) (cid:1) 6 ThefollowingTheorem1.1,thefirstmainresultofthispaper,givesanaffirmative answer to this question. Theorem 1.1. For positive numbers a,b > 0 with a = b, the double inequal- 6 ity (1.10) is valid if and only if 1 12 α 1+ 3 and β =1. (1.11) ≤ 2 r π − ! In [17] the author posed an unsolved problem: Find the greatest value α and 1 the least value β such that the double inequality 1 α C(a,b)+(1 α )A(a,b)<T(a,b)<β C(a,b)+(1 β )A(a,b) (1.12) 1 1 1 1 − − holds for a,b>0 with a=b. 6 The following Theorem 1.2, the second main result of this paper, solves this problem. Theorem 1.2. for a,b > 0 with a = b, the double inequality (1.12) holds if and 6 only if α 4 1 and β 1. 1 ≤ π − 1 ≥ 3 TWO SHARP INEQUALITIES FOR BOUNDING SEIFFERT MEAN 3 2. Proof of Theorem 1.1 In this section, we supply a proof of Theorem 1.1. For simplicity, we denote two numbers in (1.11) by λ and µ respectively. It is clearthat, in orderto provethe double inequality (1.10), it suffices to show T(a,b)>C λa+(1 λ)b,λb+(1 λ)a (2.1) − − and (cid:0) (cid:1) T(a,b)<C µa+(1 µ)b,µb+(1 µ)a . (2.2) − − Fromdefinitions(1.1)and(1.2)(cid:0)weseethatbothT(a,b)and(cid:1)C(a,b)aresymmetric and homogenous of degree 1. Hence, without loss of generality, we assume that a>b. If replacing a >1 by t>1 and letting p 1,1 , then b ∈ 2 C pa+(1 p)b,pb+(1 p)a T(a,b) (cid:0) (cid:1) − − − (cid:0) [pt+(1 p)]2+[pt+(cid:1)(1 p)][p+(1 p)t]+[p+(1 p)t]2 = − − − − bf(t), (2.3) 6(1+t)arctant−1 t+1 where t 1 f(t)=4arctan − t+1 (2.4) 3(t2 1) − . − [pt+(1 p)]2+[pt+(1 p)][p+(1 p)t]+[p+(1 p)t]2 − − − − Standard computations lead to f(1)=0, (2.5) 3 lim f(t)=π , (2.6) t→∞ − p2 p+1 − and f (t) f′(t)= 1 , (2.7) h (t) 1 where f (t)= 4p4 8p3+18p2 14p+1 t4 4 4p4 8p3+9p2 5p+1 t3 1 − − − − − +(cid:0) 6 4p4 8p3+6p2 2p+1(cid:1) t2 4(cid:0) 4p4 8p3+9p2 5p+1(cid:1) t (2.8) − − − − − +4p(cid:0)4 8p3+18p2 14p+1(cid:1), (cid:0) (cid:1) − − f (1)=0, (2.9) 1 and h (t)= [pt+(1 p)]2+[pt+(1 p)][p+(1 p)t]+[p+(1 p)t]2 2 1+t2 . 1 − − − − Let (cid:8) (cid:9) (cid:0) (cid:1) f′(t) f′(t) f′(t) f (t)= 1 , f (t)= 2 , and f (t)= 3 . 2 3 4 4 3 2 Then, by standard argument, we have f (t)= 4p4 8p3+18p2 14p+1 t3 3 4p4 8p3+9p2 5p+1 t2 2 − − − − − (2.10) +(cid:0) 3 4p4 8p3+6p2 2p+1(cid:1) t 4(cid:0)p4 8p3+9p2 5p+1 ,(cid:1) − − − − − f (1)=0, (2.11) 2 (cid:0) (cid:1) (cid:0) (cid:1) 4 W.-D.JIANG,J.CAO,ANDF.QI f (t)= 4p4 8p3+18p2 14p+1 t2 2 4p4 8p3+9p2 5p+1 t 3 − − − − − (2.12) +(cid:0) 4p4 8p3+6p2 2p+1,(cid:1) (cid:0) (cid:1) − − f (1)=6p2 6p, (2.13) 3 − f (t)= 4p4 8p3+18p2 14p+1 t 4p4 8p3+9p2 5p+1 , (2.14) 4 − − − − − f4(1)=(cid:0)9p2 9p. (cid:1) (cid:0) (cid:1) (2.15) − If p=λ, then the quantities (2.6), (2.13), and (2.15) become lim f(t)=0, (2.16) t→∞ 18 f (1)= 6<0, (2.17) 3 π − 27 f (1)= 9<0, (2.18) 4 π − and 36+18π 9π2 4p4 8p3+18p2 14p+1= − >0. (2.19) − − π2 Thus, from (2.8), (2.10), (2.12), (2.14), and (2.19), it is very easy to obtain that lim f (t)= , (2.20) t→∞ 1 ∞ lim f (t)= , (2.21) t→∞ 2 ∞ lim f (t)= , (2.22) t→∞ 3 ∞ lim f (t)= . (2.23) t→∞ 4 ∞ From (2.14) and (2.19), it is clear that the function f (t) is strictly increasing on 4 [1, ), and so, by virtue of (2.18) and (2.23), there exists a point t > 1 such 0 ∞ that f (t) < 0 on [1,t ) and f (t) > 0 on (t , ). Hence, the function f (t) is 4 0 4 0 3 ∞ strictly decreasing on [1,t ] and strictly increasing on [t , ). Similarly, by (2.17) 0 0 ∞ and (2.22), there exists a point t > t > 1 such that f (t) is strictly decreasing 1 0 2 on [1,t ] and strictly increasing on [t , ), and, by (2.11) and (2.21), there exists 1 1 ∞ a point t > t > 1 such that f (t) is strictly decreasing on [1,t ] and strictly 2 1 1 2 increasing on [t , ). Further, by (2.7), (2.9), and (2.20), there exists a point 2 ∞ t >t >1 such that f(t) is strictly decreasing on [1,t ] and strictly increasing on 3 2 3 [t , ). Finally,by (2.3)and(2.16),itisdeducedthatthefunctionf(t)isnegative 3 ∞ on (1, ). The inequality (2.1) is thus proved. ∞ If p=µ=1, then the function (2.8) becomes f (t)=(t 1)4 >0 (2.24) 1 − for t > 1. Combining this with (2.7) and (2.5) results in that f(t) is strictly increasing and positive on (1, ). Therefore, the inequality (2.2) is obtained. ∞ Combiningtheinequalities(2.1)and(2.2)withthemonotonicityofJ(x)defined by (1.8), the double inequality (1.10) is established for all α λ and β 1. ≤ ≥ For any given number p satisfying 1 > p > λ, it is obvious that the limit (2.6) is positive. This positivity together with (2.3) and (2.4) implies that for 1>p>λ there exists T =T (p)>1 such that the inequality 0 0 C pa+(1 p)b,pb+(1 p)a >T(a,b) − − holds for ab ∈(T0,∞). (cid:0)This tells us that the consta(cid:1)nt λ is the best possible. TWO SHARP INEQUALITIES FOR BOUNDING SEIFFERT MEAN 5 For 1 < p < µ = 1, the quantity (2.13) is positive. Accordingly, there exists a 2 number δ = δ(p) > 0 such that the function f (t) is negative on (1,1+δ). This 3 negativity together with (2.3), (2.5), (2.7) and (2.9) implies that for any 1 < p < 2 µ=1, there exists δ =δ(p)>0 such that the inequality T(a,b)>C pa+(1 p)b,pb+(1 p)a − − is valid for ab ∈ (T0,∞). Conse(cid:0)quently, the number µ is th(cid:1)e best possible. The proof of Theorem 1.1 is complete. 3. Proof of Theorem 1.2 In order to prove Theorem 1.2, we need the following Lemmas. Lemma 3.1. The Bernoulli numbers B for n N have the property 2n ∈ ( 1)n−1B = B , (3.1) 2n 2n − | | where the Bernoulli numbers B for i 0 are defined by i ≥ ∞ ∞ x B x x2i = ixi =1 + B , x <2π. (3.2) ex 1 n! − 2 2i(2i)! | | − i=0 i=1 X X Proof. In [2, p. 16 and p. 56], it is listed that for q 1 ≥ (2π)2q B ζ(2q)=( 1)q−1 2q, (3.3) − (2q)! 2 where ζ is the Riemann zeta function defined by ∞ 1 ζ(s)= . (3.4) ns n=1 X From (3.3), the formula (3.1) follows. (cid:3) Lemma 3.2. For 0< x <π, | | ∞ 1 22n B cotx= | 2n|x2n−1. (3.5) x − (2n)! n=1 X Proof. This may be derived readily from combining the formula [1, p. 75, 4.3.70] with the identity (3.1). (cid:3) Lemma 3.3. For 0< x <π, | | ∞ 1 1 22n(2n 1)B = + − | 2n|x2(n−1). (3.6) sin2x x2 (2n)! n=1 X Proof. Since 1 d =csc2x= (cotx), sin2x −dx the formula (3.6) follows from differentiating (3.5). (cid:3) Now we are ready to prove Theorem 1.2. It is easy to see that the double inequality (1.12) is equivalent to T(a,b) A(a,b) α < − <β . (3.7) 1 C(a,b) A(a,b) 1 − 6 W.-D.JIANG,J.CAO,ANDF.QI Without loss of generality, we assume a>b>0 and let x= a. Then x>1 and b x−1 x+1 T(a,b) A(a,b) 2arctanx−1 − 2 − = x+1 . C(a,b) A(a,b) x2+1 x+1 − x+1 − 2 Let t= x−1. Then t (0,1) and x+1 ∈ T(a,b) A(a,b) t 1 − = arctant − . C(a,b) A(a,b) t2 − Let t=tanθ for θ 0,π . Then ∈ 4 T(a,(cid:0)b) A(cid:1)(a,b) tanθ 1 cotθ 1 − = θ − = +1. C(a,b) A(a,b) (tanθ)2 θ − sin2θ − By Lemmas 3.2 and 3.3, we have ∞ ∞ cotθ 1 22n (2n 1)22n +1=1 B θ2n−2 − B θ2n−2 θ − sin2θ − (2n)!| 2n| − (2n)! | 2n| n=1 n=1 X X ∞ n22n+1 =1 B θ2n−2 − (2n)! | 2n| n=1 X which is strictly decreasing on 0,π . 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Math.Forum5(2010), no.26,1297–1302. 2 (Jiang)DepartmentofInformationEngineering,WeihaiVocationalUniversity,Wei- haiCity, ShandongProvince,264210,China E-mail address: [email protected] (Cao) Department of Mathematics, Hangzhou Normal University, Hangzhou City, ZhejiangProvince,310036,China E-mail address: [email protected], [email protected] (Qi)SchoolofMathematicsandInformatics,HenanPolytechnicUniversity,Jiaozuo City, Henan Province, 454010, China; Department of Mathematics, School of Science, TianjinPolytechnic University,Tianjin City, 300387,China E-mail address: [email protected], [email protected], [email protected] URL:http://qifeng618.wordpress.com