sos-schur

TheSOS-SchurmethodBachNgocThanhCong,NguyenVuTuan,NguyenTrungKienTheSOS-Schurmethodconnectstwowell-knownresults,thesumofsquaresmethodandSchurinequality.TheideaoftheSOS-Schurmethodistoreduceathreevariableinequalitytoaninequalityofthefollowingtypef(a,b,c)=M(a−b)2+N(a−c)(b−c)≥0.Letusconsideraclassicalexample:Example1.Leta,b,cbepositiverealnumbers.Provethatab+bc+ca≥a+ca+b+b+cb+a+c+ac+b.MathlinksContestsSolution.Withoutlossofgenerality,assumethatc=min(a,b,c).Notethatforx,y,z0wehavexy+yz+zx−3=1xy(x−y)2+1xz(x−z)(y−z).Thereforethegiveninequalitycanberewrittenas1ab−1(a+c)(b+c)(a−b)2+1ac−1(a+b)(a+c)(a−c)(b−c)≥0.Thelastinequalityisclearlytrue,becausec=min(a,b,c),andwearedone.Theaboveexampleisprovedbyasimpletechnique:rewritingthedesiredin-equalityassumoftwononnegativenumbers.WetransformtheinequalityintothestandardSOS-Schurform:f(a,b,c)=M(a−b)2+N(a−c)(b−c)≥0.Afterthatwetrytoprovethat,ifc=min(a,b,c)orc=max(a,b,c),thenMandNarenonnegative.ThefollowingidentitiesareusefulinsolvinginequalitiesonthreevariablesusingtheSOS-Schurmethod.a2+b2+c2−ab−bc−ca=(a−b)2+(a−c)(b−c)a3+b3+c3−3abc=(a+b+c)(a−b)2+(a+b+c)(a−c)(b−c)(a+b)(b+c)(c+a)−8abc=2c(a−b)2+(a+b)(a−c)(b−c)ab2+bc2+ca2−3abc=c(a−b)2+b(a−c)(b−c)MathematicalReflections5(2007)1a2b2+b2c2+c2a2−abc(a+b+c)=c2(a−b)2+ab(a−c)(b−c)a4+b4+c4−a2b2−b2c2−c2a2=(a+b)2(a−b)2+(a+c)(b+c)(a−c)(b−c)a4+b4+c4−a3b−b3c−c3a=(a2+b2+ab)(a−b)2+(b2+bc+c2)(a−c)(b−c).ThenextstepistoestablishtherelationshipbetweentheSOS-SchurandSOSrepresentationforsymmetricpolynomials.Tofindthiscommonrepresentation,werelyonthestandardsumofsquaresrepresentation.Supposethatwehavethefollowingsymmetricexpression:f(a,b,c)=Sa(b−c)2+Sb(c−a)2+Sc(a−b)2.HowtotransformittotheSOS-Schurform?Theanswerissimple:observethat(a−b)2+(b−c)2+(c−a)2=2(a−b)2+2(a−c)(b−c).Hencef(a,b,c)=(Sa−Sc)(b−c)2+(Sb−Sc)(a−c)2+Sc(a−b)2+(b−c)2+(c−a)2=2Sc(a−b)2+(b−c)(Sa−Sc)a−c+(a−c)(Sb−Sc)b−c+2Sc(a−c)(b−c)Notethatf(a,b,c)isasymmetricpolynomial,whileSa,Sb,andScaresemi-symmetric.Thusf(a,b,c)=M(a−b)2+N(a−c)(b−c),whereM=2ScandN=(b−c)(Sa−Sc)a−c+(a−c)(Sb−Sc)b−c+2Sc.Alternatively,fromaSOSrepresentation,f(a,b,c)=Sa(b−c)2+Sb(c−a)2+Sc(a−b)2=(Sa+Sb)(a−c)(b−c)+(Sb+Sc)(b−a)(c−a)+(Sc+Sa)(c−b)(a−b)=c(Sa−Sb)(a−b)+aSb−bSa(a−b)+Sc(a−b)2+(Sa+Sb)(a−c)(b−c).Thusf(a,b,c)=M(a−b)2+N(a−c)(b−c),whereM=c(Sa−Sb)(a−b)+aSb−bSa(a−b)+ScandN=Sa+Sb.ItfollowsthatthemaindifficultyisactuallytotransformacyclicexpressionintoastandardSOS-Schurform.Allabovegivesusatheoreticalsenseofthismethod.ThefollowingappicationsillustratethespecialadvantageoftheSOS-Schurmethod.MathematicalReflections5(2007)2Example2.Leta,b,cbepositiverealnumberssuchthata≥b≥c.Provethata2b(a−b)+b2c(b−c)+c2a(c−a)≥0.Solution.Wehavef(a,b,c)=a2b(a−b)+b2c(b−c)+c2a(c−a)=a2b(a−b)−ab2(a−b)+b2c(b−c)−ab2(b−c)++c2a(c−a)−ab2(c−a)=ab(a−b)2+(ab+ac−b2)(a−c)(b−c).Clearlythelastexpressionispositive,andwearedone.Example3.Leta,b,cbepositiverealnumbers.Provethata+bb+c+b+cc+a+c+aa+b+3(ab+bc+ca)(a+b+c)2≥4.Solution.Withoutlossofgenerality,assumethatc=min(a,b,c).Wehavea+bb+c+b+cc+a+c+aa+b−3=1(a+c)(b+c)(a−b)2+1(a+b)(b+c)(a−c)(b−c)3(ab+bc+ca)(a+b+c)2−1=−1(a+b+c)2(a−b)2−1(a+b+c)2(a−c)(b−c)Hencethegiveninequalitycanberewrittenasf(a,b,c)=M(a−b)2+N(a−c)(b−c),whereM=1(a+c)(b+c)−1(a+b+c)2andN=1(a+b)(b+c)−1(a+b+c)2.ClearlyM,N≥0,andtheinequalityisproved.MathematicalReflections5(2007)3