先两边平方1/(a-b)+1/(b-c)+1/(c-a)=1
得[1/(a-b)]^2+[1/(b-c)]^2+[1/(c-a)]^2+2/(a-b)(b-c)+2/(a-b)(c-a)+2/(b-c)(c-a) (注意：2/(a-b)(b-c)=2除以（(a-b)×(b-c)）
因为2/(a-b)(b-c)+2/(a-b)(c-a)+2/(b-c)(c-a)=2/(a-b)(b-c)+(2/(a-b)+2/(b-c))/c-a=2/(a-b)(b-c)+2(a-c)/((a-b)(b-c)(c-a)=2/(a-b)(b-c)+-2/(a-b)(b-c)=0
所以[1/(a-b)]^2+[1/(b-c)]^2+[1/(c-a)]^2=1-0=1