由不等式公式知：(a+b+c)(1/a+1/b+1/c)>=9又 2(a+b+c)(1/(a+b)+1/(b+c)+1/(c+a))=[(a+b)+(b+c)+(c+a)][(a+b)+1/(b+c)+1/(c+a)]>=9a,b,c>01/(a+b)+1/(b+c)+1/(c+a)>=9/2(a+b+c)