已知a,b,c为正实数,求证(a+b+c)(1/a+1/b+1/c)大于等于9,没有a+b+c=1这一条件
问题描述：
已知a,b,c为正实数,求证(a+b+c)(1/a+1/b+1/c)大于等于9,没有a+b+c=1这一条件
答
(a+b+c)(1/a+1/b+1/c)
=(a+b+c)/a+(a+b+c)/b+(a+b+c)/c
=1+(b+c)/a+1(a+c)/b+1(a+b)/c
=3+b/c+c/b+a/c+c/a+a/b+b/a
>=3+2+2+2=9=3+b/c+c/b+a/c+c/a+a/b+b/a这一步怎么出来哒？(a+b+c)(1/a+1/b+1/c) =(a+b+c)/a+(a+b+c)/b+(a+b+c)/c=1+(b+c)/a+1+(a+c)/b+1+(a+b)/c =3+b/a+c/a+a/b+c/b+a/c+b/c =3+b/c+c/b+a/c+c/a+a/b+b/a3+b/c+c/b+a/c+c/a+a/b+b/a怎么变成3+2+2+2？不理解……3+b/c+c/b+a/c+c/a+a/b+b/a >=3+2√(b/c*c/b)+2√(a/c*c/a)+2√(b/a*a/b)=3+2+2+2=9