设x>y>z,且1/(x-y)+1/(y-z)≥k/(x-z)恒成立,则实数k的最大值是
问题描述：
设x>y>z,且1/(x-y)+1/(y-z)≥k/(x-z)恒成立,则实数k的最大值是
答
最大为4
令x-y=a,y-z=b,则x-z=a+b
则原式=(a+b)^2/(ab)=(a^2+b^2+2ab)/(ab)>=4
答
根据调和平均数≤算术平均数,即2/(1/a+1/b)≤(a+b)/2 （a,b>0）2同时除以两边,可得1/a+1/b≥4/(a+b)因为x>y>z,所以x-y,y-z>0所以运用1/a+1/b≥4/(a+b)可得1/(x-y)+1/(y-z)≥4/(x-z)又因为1/(x-y)+1/(y-z)≥k/(x-z)恒...