令x-y=a,y-z=b,则x-z=a+b
1/a+2/b-n/(a+b)>=0
同乘 ab(a+b)
b(a+b)+2a(a+b)-nab>=0
b²+2a²+3ab-nab>=0
2(a-b/√2)²+2√2ab+3ab-nab>=0
所以只要2√2ab+3ab-nab>=0
n