∵1/(x-y)+1/(y-z)≥n/(x-z)两边同时乘以（x-z)：（注：x-z>0）(x-z)/(x-y)+(x-z)/(y-z)≥n通分:[(x-z)*(x-z)]/[(x-y)*(y-z)]≥n令x-y=a,y-z=b则(a+b)*(a+b)=(x-z)*(x-z)∴[(x-z)*(x-z)]/[(x-y)*(y-z)]≥n=====＞(a+...