Chinese-Dumbass-Notation(不等式)

TheArtofDumbassingBrianHamrickJuly2,20101IntroductionTheeldofinequalitiesisfarfromtrivial.Manytechniques,includingCauchy,Holder,isolatedfudging,Jensen,andsmoothingexistandleadtosimple,shortsolutions.ThatisnotthepurposeofthetechniquethatIwillbedescribingbelow.Thistechniqueisverystraightforward,widely(butnotuniversally)applicable,andcanleadtoasolutionquickly,althoughitisalmostneverthenicestone.Thistechniqueisknownasdumbassing,andisquiteaptlynamed,asverylittlecreativityisrequired.Thecruxofthismethodishomogenizing,clearingdenominators,multiplyingeverythingout,thenndingthecorrectapplicationsofSchurand/orAM-GM.Occaisionally,moreadvancedtechniquesmayberequired.2ChineseDumbassNotationTherstandforemostpurposeofthislectureistointroducearelativelyobscurebutextremelyusefulnotation.BeforeIdescribeit,Iwantyoutobesuretorealizethatthisnotationisnotstandard.Ifyouaregoingtouseitonacontest,besuretodeneit.Withthatwarninginmind,allowmetodescribethenotation.ChineseDumbassNotationisaconciseandconvenientwaytowritedownthreevariablehomogeneousexpressions.Additionally,itprovidesaconvenientwaytomultiplyouttwoexpressionswithverylittlechanceofdroppingterms.Unlikecyclicandsymmetricsumnotation,ChineseDumbassNotationallowsonetokeepliketermscombinedwhilenotrequiringanysortofsymmetryfromtheexpression.However,let'srstlookatanexamplethatissymmetric.088202820828288082080Therstthingthatoneshouldnoteaboutthisnotationisthatitisnotlinear.ChineseDumbassNotationusesatriangletoholdthecoecientsofathreevariablesymmetricinequalityinawaythatiseasytovisualize.TheabovetrianglerepresentstheexpressionXsym8a3b+Xsym10a2b2+Xsym14a2bc.Ingeneral,afourthdegreeexpressionwouldbewrittenasfollows.[a4][a3b][a3c][a2b2][a2bc][a2c2][ab3][ab2c][abc2][ac3][b4][b3c][b2c2][bc3][c4]Here[x]representsthecoecientofx.Ingeneral,adegreedexpressionisrepresentedbyatrianglewithsidelengthd+1andthecoecientofabcisplacedatBarycentriccoordinates(;;).Multiplyingthesetrianglesismuchlikemultiplyingpolynomials:youdosobyshiftingandadding.Forexample,1112121=10@1120001A+20@0101201A+10@0010121A=133252Itispossibletoshortcutthisprocessmuchlikehowyoushortcutpolynomialmultiplications.Themainideaisthatwhenyouwanttocomputeacoecientintheanswertriangle,youlookatallthewaysitcanbemadeinthefactortriangles.Practicingthismethodwillallowyoutolearnwhatformofmultiplicationworksbestforyou.Nowthatwehavethenotationdown,let'slookatsomebasicinequalities.3AM-GMTheAM-GMinequalityfortwovariablesstatesthata2+b22ab.Inotherwords,1201000.Ofcourse,thisworksanywhereinthetriangle,evenforspacesthataren'tconsecutive.Ingeneral,weightedAM-GMallowsustotakesomepositivecoecientsand,thinkingofthemasweights,\slidethemtotheircenterofmasswhilenotincreasingthetotalsumoftheexpression.Oneextremelyimportantapplicationofthisistotakeasymmetricdistributionofweightsandslidetheminwardtoanothersymmetricdistributionofweights.ThefactthattheseinequalitiesfollowfromAM-GMisknownasMuirhead'sInequality.Forexample,011000100101010000020022000000followseasilyfromMuirhead'sInequality.WecanseethatinthiscaseitalsofollowsfromasymmetricsumofthefollowingapplicationofweightedAM-GM:02300000000000130000010000000000Ingeneral,citingMuirhead'sinequalityislookeddownupon,butitsimpliesmanydumbassingargu-mentsconsiderablyastheyareoftenasumofmanyapplicationsofAM-GMandndingtheweightsforeachonetakesconsiderabletime.4Schur'sInequalitySchur'sinequalitystatesthatXcycar(ab)(ac)0forallnonnegativenumbersa;b;c;r.TheprototypicalapplicationofSchur'sinequalitylooksasfollows:21110101111110110AswithAM-GM,thevariablescanbetweakedsothatSchur'sinequalitygivesusXcycar(asbs)(ascs)0.InChineseDumbassNotation,thiscorrespondstomanipulatingthecharacteristicdiamondsshownabove,withrcontrollingthedistancefromthecenterandscontrollingthesizeofeachofthediamonds.Theinequalityshownaboveiswithr=2ands=1.5ExamplesExample1.Provethatbc2a+b+c+aca+2b+c+aba+b+2c14(a+b+c).Proof.Wecleardenominatorstoobtainthatthisisequivalentto4Xcycbc(a+2b+c)(a+b+2c)(a+2b+c)(a+b+2c)(2a+b+c)(a+b+c),4Xcyc0@0000101A112121112121211111We'llexpandthelefthandsideas4Xcyc0@0000101A112121=4Xcyc0@0000101A0@1332521A=4Xcyc0BBBB@0000100330025201CCCCA=40BBBB@0225752772025201CCCCA=0882028208282880820803Wethenexpandtherighthandsideas112121211111=0@1332521A211111=0BB@277716727721CCA111=299143014930309291492Theinequalityisthusequivalenttoshowingthatthedierenceisnonnegative.Butthedierenceis2116261221216120336063003036300bySchurandAM-GM.Example2.Leta;b;cbepositiverealnumberssuchthat1a+1b+1c=a+b+c:Showthat1(2a+b+c)2+1(2b+c+a)2+1(2c+a+b)2316:Proof.Noticethattheconditionisequivalenttoab+ac+bc=a2bc+ab2c+abc2:Additionally,bymultiplyingourdesiredinequalitythroughby16(2a+b+c)2(2b+a+c)2(2c+a+b)2,weseethatitisequivalentto16Xcyc(2b+c+a)2(2c+a+b)23(2b+c+a)2(2c+a+b)2(2a+b+c)2whichisequivalentto16Xcyc(2b+c+a)2(2c+a+b)2(a2bc+ab2c+abc2)3(2b+c+a)2(2c+a+b)2(2a+b+c)2(ab+bc+ac):4Wewillprovethisinequalityforallpositiverealsa;b;c.First,werewritethelefthandsideas16Xcyc12121122!0BBBB@00001