n的最大值为4解法：∵1/(x-y)+1/(y-z)≥n/(x-z)(不等式两边同时乘以（x-z) 由x>y>z得x-y>0,y-z>0,x-z>0）∴(x-z)/(x-y)+(x-z)/(y-z)≥n（再通分）∴(x-z)*(x-z)/{(x-y)*(y-z)}≥n此时令x-y=a,y-z=b,则显然(a+b)*(a+b...