数学不等式的3variable

Onaclassofthree-variableinequalitiesVoQuocBaCan1TheoremLeta,b,cberealnumberssatisfyinga+b+c=1.BytheAM-GMinequality,wehaveab+bc+ca≤13,thereforesettingab+bc+ca=1−q23(q≥0),wewillfindthemaximumandminimumvaluesofabcintermsofq.Ifq=0,thena=b=c=13,thereforeabc=127.Ifq6=0,then(a−b)2+(b−c)2+(c−a)20.Considerthefunctionf(x)=(x−a)(x−b)(x−c)=x3−x2+1−q23x−abc.Wehavef0(x)=3x2−2x+1−q23whosezerosarex1=1+q3,andx2=1−q3.Wecanseethatf0(x)0forx2xx1andf0(x)0forxx2orxx1.Furthermore,f(x)hasthreezeros:a,b,andc.Thenf1−q3=(1−q)2(1+2q)27−abc≥0andf1+q3=(1+q)2(1−2q)27−abc≤0.Hence(1+q)2(1−2q)27≤abc≤(1−q)2(1+2q)27andweobtainTheorem1.1Ifa,b,carearbitraryrealnumberssuchthata+b+c=1,thensettingab+bc+ca=1−q23(q≥0),thefollowinginequalityholds(1+q)2(1−2q)27≤abc≤(1−q)2(1+2q)27.Or,moregeneral,Theorem1.2Ifa,b,carearbitraryrealnumberssuchthata+b+c=p,thensettingab+bc+ca=p2−q23(q≥0)andr=abc,wehave(p+q)2(p−2q)27≤r≤(p−q)2(p+2q)27.Thisisapowerfultoolsincetheequalityholdsifandonlyif(a−b)(b−c)(c−a)=0.MathematicalReflections2(2007)1Herearesomeidentitieswhichwecanusewiththistheorema2+b2+c2=p2+2q23a3+b3+c3=pq2+3rab(a+b)+bc(b+c)+ca(c+a)=p(p2−q2)3−3r(a+b)(b+c)(c+a)=p(p2−q2)3−ra2b2+b2c2+c2a2=(p2−q2)29−2prab(a2+b2)+bc(b2+c2)+ca(c2+a2)=(p2+2q2)(p2−q2)9−pra4+b4+c4=−p4+8p2q2+2q49+4pr2Applications2.1Leta,b,cbepositiverealnumberssuchthata+b+c=1.Provethat1a+1b+1c+48(ab+bc+ca)≥25.Solution.Wecaneasilycheckthatq∈[0,1],byusingthetheoremwehaveLHS=1−q23r+16(1−q2)≥9(1+q)(1−q)(1+2q)+16(1−q2)=2q2(4q−1)2(1−q)(1+2q)+25≥25.Theinequalityisproved.Equalityholdsifandonlyifa=b=c=13ora=12,b=c=14andtheirpermutations.2.2[Vietnam2002]Leta,b,cberealnumberssuchthata2+b2+c2=9.Provethat2(a+b+c)−abc≤10.Solution.Theconditioncanberewrittenasp2+2q2=9.Usingourtheorem,wehaveLHS=2p−r≤2p−(p+q)2(p−2q)27=p(5q2+27)+2q327.Weneedtoprovethatp(5q2+27)≤270−2q3.Thisfollowsfrom(270−2q3)2≥p2(5q2+27)2,or,equivalently,27(q−3)2(2q4+12q3+49q2+146q+219)≥0.Theinequalityisproved.Equalityholdsifandonlyifa=b=2,c=−1andtheirpermutations.2.3[VoQuocBaCan]Forallpositiverealnumbersa,b,c,wehavea+bc+b+ca+c+ab+11rab+bc+caa2+b2+c2≥17.MathematicalReflections2(2007)2Solution.Becausetheinequalityishomogeneous,withoutlossofgenerality,wemayassumethatp=1.Thenq∈[0,1]andtheinequalitycanberewrittenas1−q23r+11s1−q21+2q2≥20.Usingourtheorem,itsufficestoprove11s1−q21+2q2≥20−9(1+q)(1−q)(1+2q)=−40q2+11+11(1−q)(1+2q).If−40q2+11q+11≤0,orq≥11+3√20980,itistrivial.Ifq≤11+3√2098023,wehave121(1−q2)(1+2q2)−(−40q2+11q+11)2(1−q)2(1+2q)2=3q2(11−110q+255q2+748q3−1228q4)(1+2q2)(1−q)2(1+2q)2.Ontheotherhand,11−110q+255q2+748q3−1228q4=q411q4−110q3+255q2+748q−1228≥q411(2/3)4−110(2/3)3+255(2/3)2+7482/3−1228=243516q4≥0.Theinequalityisproved.Equalityoccursifandonlyifa=b=c.2.4[VietnamTST1996]Provethatforanya,b,c∈R,thefollowinginequalityholds(a+b)4+(b+c)4+(c+a)4≥47(a4+b4+c4).Solution.Ifp=0theinequalityistrivial,sowewillconsiderthecasep6=0.Withoutlossofgenerality,wemayassumep=1.Theinequalitybecomes3q4+4q2+10−108r≥0Usingourtheorem,wehave3q4+4q2+10−108r≥3q4+4q2+10−4(1−q)2(1+2q)=q2(q−4)2+2q4+6≥0.Theinequalityisproved.Equalityholdsonlyfora=b=c=0.2.5[PhamHuuDuc,MR1/2007]Provethatforanypositiverealnumbersa,b,andc,rb+ca+rc+ab+ra+bc≥s6·a+b+c3√abcSolution.ByHolder’sinequality,wehaveXcycrb+ca!2Xcyc1a2(b+c)!≥Xcyc1a!3ItsufficestoprovethatXcyc1a!3≥6(a+b+c)3√abcXcyc1a2(b+c)MathematicalReflections2(2007)3Settingx=1a,y=1b,z=1c,theinequalitybecomes(x+y+z)3≥63√xyz(xy+yz+zx)Xcycxy+z,or(x+y+z)3≥63√xyz(xy+yz+zx)((x+y+z)3−2(x+y+z)(xy+yz+zx)+3xyz)(x+y)(y+z)(z+x).BytheAM-GMinequality,(x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)−xyz≥89(x+y+z)(xy+yz+zx).Itremainstoprovethat4(x+y+z)4≥273√xyz((x+y+z)3−2(x+y+z)(xy+yz+zx)+3xyz).Settingp=x+y+z,xy+yz+zx=p2−q23(p≥q≥0),theinequalitybecomes4p4≥93√xyz(p3+2pq2+9xyz).Applyingourtheorem,itsufficestoprovethat4p4≥93r(p−q)2(p+2q)27p3+2pq2+(p−q)2(p+2q)3,4p4≥3p(p−q)2(p+2q)(3p3+6pq2+(p−q)2(p+2q)).Settingu=3qp−qp+2q≤1,theinequalityisequivalentto4(2u3+1)4≥27u2(4u9+5u6+2u3+1),orf(u)=(2u3+1)4u2(4u9+5u6+2u3+1)≥274Wehavef0(u)=2(2u3+1)3(u3−1)(2u3−1)(2u6+2u3−1)u3(u3+1)2(4u6+u3+1)2f0(u)=0⇔u=3s√3−12,oru=13√3,oru=1.Now,wecaneasilyverifythatf(u)≥minf3s√3−12,f(1)=274,whichistrue.Theinequalityisproved.Equalityholdsifandonlyifa=b=c.2.6[DarijGrinberg]Ifa,b,c≥0,thena2+b2+c2+2abc+1≥2(ab+bc+ca).MathematicalReflections2(2007)4Solution.Rewritetheinequalityas6r+3+4q2−p2≥0.If2q≥p,itistrivial.Ifp≥2q,usingthetheorem,itsufficestoprovethat2(p−2q)(p+q)29+3+4q2−p2≥0,or(p−3)2(2p+3)≥2q2(2q+3p−18).If2p≤9,wehave2q+3p≤4p≤18,thereforetheinequalityistrue.If2p≥9,wehave2q2(2q+3p−18)≤4q2(2p−9)≤p2(2p−9)=(p−3)2(2p+3)−27(p−3)2(2p+3).Theinequalityisproved.Equalityholdsifandonlyifa=b=c=1.2.7[Schur’sinequality]Foranynonnegativerealnumbersa,b,c,a3+b3+c3+3abc≥ab(a+b)+bc(b+c)+ca(c+a).Solution.Becausetheinequalityishomogeneous,wecanassumethata+b+c=1.Thenq∈[0,1]andtheinequalityisequivalentto27r+4q2−1≥0.Ifq≥12,itistrivial.Ifq≤12,bythetheoremweneedtoprovethat(1+q)2(1−2q)+4q2−1≥0,orq2(1−2q)≥0,whichistrue.Equalityholdsifandonlyifa=b=cora=b,c=0andtheirpermutations.2.8[PhamKimHung]Findthegreatestconstantksucht