Question

已知abc属于R*,且a+b+c=1,求证1/(a+b)+1/(b+c)+1/(c+a)大于等于9/2

请求详细过程 谢啦

Answer 1

因为a+b+c=1,
所以2(a+b+c)=2,
即(a+b)+(b+c)+(c+a)=2,
所以[2(a+b+c)]*[1/(a+b)+1/(b+c)+1/(c+a)]
=[(a+b)+(b+c)+(c+a)]*[1/(a+b)+1/(b+c)+1/(c+a)]
>=[(a+b)*1/(a+b)+(b+c)*1/(b+c)+(c+a)*1/(c+a)]^2
=(1+1+1)^2
=9,
所以1/(a+b)+1/(b+c)+1/(c+a)>=9/2.