ta co a+b+c=3<=>3cb3(abc)=3=>abc=1(ap dung cosi de a+b+c min)bieu thuc>=2abc(1/a+1/b+1/c)+1/abc(a+b+c)>=2(3/cb3(abc)+3>=9(dpcm)
ta co a+b+c=3
ma a+b+c>
=3cb3(abc)=3=>abc
>=1 de a+b+c min
=>abc=1bieu thuc>=2abc(1/a+1/b+1/c)+1/abc(a+b+c)>=2(3/cb3(abc)+3>=9(dpcm)