note on an open problem of feng qi

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1、JournalofInequalitiesinPureandAppliedMathematicshttp://jipam.vu.edu.au/Volume7,Issue1,Article4,2006NOTEONANOPENPROBLEMOFFENGQIYINCHENANDJOHNKIMBALLDEPARTMENTOFMATHEMATICALSCIENCESLAKEHEADUNIVERSITYTHUNDERBAY,ONTARIOCANADAP7B5E1yin.chen@lakeheadu.cajfkimbal@lakeheau.caRec

2、eived04July,2005;accepted24August,2005CommunicatedbyF.QiABSTRACT.Inthispaper,anintegralinequalityisstudied.AnanswertoanopenproblemproposedbyFengQiisgiven.Keywordsandphrases:Integralinequality,Cauchy’sMeanValueTheorem.2000MathematicsSubjectClassification.26D15.In[5],Qistud

3、iedaveryinterestingintegralinequalityandprovedthefollowingresultTheorem1.Letf(x)becontinuouson[a,b],differentiableon(a,b)andf(a)=0.Iff0(x)≥1forx∈(a,b),thenZZ2bb3(1)[f(x)]dx≥f(x)dx.aaIf0≤f0(x)≤1,thentheinequality(1)reverses.Qiextendedthisresulttoamoregeneralcase[5],ando

4、btainedthefollowinginequality(2).Theorem2.Letnbeapositiveinteger.Supposef(x)hascontinuousderivativeofthen-thorderontheinterval[a,b]suchthatf(i)(a)≥0where0≤i≤n−1,andf(n)(x)≥n!,thenZZn+1bbn+2(2)[f(x)]dx≥f(x)dx.aaQithenproposedanopenproblem:Underwhatconditionistheinequali

5、ty(2)stilltrueifnisreplacedbyanypositiverealnumberr?Somenewresultsonthissubjectcanbefoundin[1],[2],[3],and[4].WenowgiveananswertoQi’sopenproblem.ThefollowingresultisageneralizationofTheorem1.ISSN(electronic):1443-5756c2006VictoriaUniversity.Allrightsreserved.202-052YINCH

6、ENANDJOHNKIMBALLTheorem3.Letpbeapositivenumberandf(x)becontinuouson[a,b]anddifferentiableon11−1(a,b)suchthatf(a)=0.If[fp]0(x)≥(p+1)pforx∈(a,b),thenZZp+1bbp+2(3)[f(x)]dx≥f(x)dx.aa11−1If0≤[fp]0(x)≤(p+1)pforx∈(a,b),thentheinequality(3)reverses.11Proof.Supposethat[fp]0(x)≥

7、0,x∈(a,b).Thenfp(x)isanon-decreasingfunction.Itfollowsthatf(x)≥0forallx∈(a,b].11−1If[fp]0(x)≥(p+1)pforx∈(a,b),thenf(x)>0forx∈(a,b].Thusbothsidesof(3)arenot0.Nowconsiderthequotientofbothsidesof(3).ByusingCauchy’sMeanValueTheoremtwice,wehaveRb[f(x)]p+2dx[f(b)]p+1ahR1i(4)hR

8、ip+1=bp(a