设：x=a+b,y=b+c,m=c+a
则a-b=m-y,b-c=x-m,c-a=y-x
(a-b)/(a+b)=(m-y)/x (b-c)/(b+c)=(x-m)/y (c-a)/(c+a)=(y-x)/m
得：原式= (m-y)/x +(x-m)/y+ (y-x)/m+(m-y)(x-m)(y-x)/(xym)
= (my-y²+x²-mx)/(xy) + (y-x)[xy+(m-y)(x-m)]/(xym)
= (x-y)(x+y-m)/(xy) - (x-y)[xy+(mx-m²-xy+my)]/(xym)
= (x-y)(mx+ym-m²)/(xym) - (x-y)(mx-m²+ym)/(xym)
= 0
如果你不想用这种方法,你就要通分消去分母计算,等式两边同时乘以(a+b)(b+c)(c+a),很麻烦.