问题描述: a>0.b>0,c>0 a+b+c=1,请证明1/(a+b)+1/(b+c)+1/(a+c)≥9/2 1个回答 分类:综合 2014-10-10 问题解答: 我来补答 解,法一:因为2(a+b+c)=2,所以由柯西不等式[(a+b)+(b+c)+(c+a)][1/(a+b)+1/(b+c)+1/(a+c)]>=(1+1+1))^2=9即2[1/(a+b)+1/(b+c)+1/(a+c)]>=9所以1/(a+b)+1/(b+c)+1/(a+c)>=9/2法二:把 a+b+c=1代入1/(a+b)+1/(b+c)+1/(a+c)>=9/2得2a/(b+c)+2b/(a+c)+2c/(a+b)>=3由对称性不妨设a 展开全文阅读