On a class of three-variable inequalities.pdf
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1、On a class of three-variable inequalitiesVo Quoc Ba Can1TheoremLet a,b,c be real numbers satisfying a+b+c=1.By the AM-GM inequality,we have ab+bc+ca 13,therefore setting ab+bc+ca=1q23(q 0),we will find the maximum and minimum values of abc interms of q.If q=0,then a=b=c=13,therefore abc=127.If q 6=0
2、,then(ab)2+(bc)2+(ca)2 0.Considerthe function f(x)=(x a)(x b)(x c)=x3 x2+1q23x abc.We havef0(x)=3x2 2x+1 q23whose zeros are x1=1+q3,and x2=1 q3.We can see that f0(x)0 for x2 x 0 for x x1.Furthermore,f(x)hasthree zeros:a,b,and c.Thenf?1 q3?=(1 q)2(1+2q)27 abc 0andf?1+q3?=(1+q)2(1 2q)27 abc 0.Hence(1+
3、q)2(1 2q)27 abc(1 q)2(1+2q)27and we obtainTheorem 1.1 If a,b,c are arbitrary real numbers such that a+b+c=1,then settingab+bc+ca=1q23(q 0),the following inequality holds(1+q)2(1 2q)27 abc(1 q)2(1+2q)27.Or,more general,Theorem 1.2 If a,b,c are arbitrary real numbers such that a+b+c=p,then settingab+b
4、c+ca=p2q23(q 0)and r=abc,we have(p+q)2(p 2q)27 r(p q)2(p+2q)27.This is a powerful tool since the equality holds if and only if(a b)(b c)(c a)=0.Mathematical Reflections 2(2007)1Here are some identities which we can use with this theorema2+b2+c2=p2+2q23a3+b3+c3=pq2+3rab(a+b)+bc(b+c)+ca(c+a)=p(p2 q2)3
5、 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+4pr2Applications2.1 Let a,b,c be positive real numbers such that a+b+c=1.Prove that1a+1b+1c+48(ab+bc+ca)25.Solution.We can easily check that q 0,1,by using the theorem we
6、 haveLHS=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.The inequality is proved.Equality holds if and only if a=b=c=13or a=12,b=c=14and theirpermutations.2.2 Vietnam 2002 Let a,b,c be real numbers such that a2+b2+c2=9.Prove that2(a+b+c)abc 10.Solution.The condition can be rew
7、ritten as p2+2q2=9.Using our theorem,we haveLHS=2p r 2p(p+q)2(p 2q)27=p(5q2+27)+2q327.We need to prove thatp(5q2+27)270 2q3.This follows from(270 2q3)2 p2(5q2+27)2,or,equivalently,27(q 3)2(2q4+12q3+49q2+146q+219)0.The inequality is proved.Equality holds if and only if a=b=2,c=1 and their permutation
8、s.2.3 Vo Quoc Ba Can For all positive real numbers a,b,c,we havea+bc+b+ca+c+ab+11rab+bc+caa2+b2+c2 17.Mathematical Reflections 2(2007)2Solution.Because the inequality is homogeneous,without loss of generality,we may assume that p=1.Then q 0,1 and the inequality can be rewritten as1 q23r+11s1 q21+2q2
9、 20.Using our theorem,it suffices to prove11s1 q21+2q2 20 9(1+q)(1 q)(1+2q)=40q2+11+11(1 q)(1+2q).If 40q2+11q+11 0,or q 11+320980,it is trivial.If q 11+32098023,we have121(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.On the other hand,11 110q+255q2+748q
10、3 1228q4=q4?11q4110q3+255q2+748q 1228?q4?11(2/3)4110(2/3)3+255(2/3)2+7482/3 1228?=243516q4 0.The inequality is proved.Equality occurs if and only if a=b=c.2.4 Vietnam TST 1996 Prove that for any a,b,c R,the following inequality holds(a+b)4+(b+c)4+(c+a)447(a4+b4+c4).Solution.If p=0 the inequality is
11、trivial,so we will consider the case p 6=0.Without loss of generality,we may assume p=1.The inequality becomes3q4+4q2+10 108r 0Using our theorem,we have3q4+4q2+10 108r 3q4+4q2+10 4(1 q)2(1+2q)=q2(q 4)2+2q4+6 0.The inequality is proved.Equality holds only for a=b=c=0.2.5 Pham Huu Duc,MR1/2007 Prove t
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