Question

一道高中数学不等式证明题。。急等答案

设a,b,c>0,求证1/(a+b)+1/(b+c)+1/(c+a)>=9/2(a+b+c)

Answer 1

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)]
>=9
a,b,c>0
1/(a+b)+1/(b+c)+1/(c+a)>=9/2(a+b+c)

Answer 2

证明：由柯西不等式可知[(a+b)+(b+c)+(c+a)]×[1/(a+b)+1/(b+c)+1/(c+a)]≥(1+1+1)²=9.即2(a+b+c)[1/(a+b)+1/(b+c)+1/(c+a)]≥9.====>1/(a+b)+1/(b+c)+1/(c+a)≥9/[2(a+b+c)]