已知x>y>0,xy=1,求x^2+y^2/x-y的最小值.
问题描述：
已知x>y>0,xy=1,求x^2+y^2/x-y的最小值.
答
(x^2+y^2+2-2)/x-y=(x^2+y^2-2xy+2)=[(x-y)^2+2]/x-y=(x-y)+2/x-y>=2^(1/2)
答
(x^2+y^2)/(x-y)=[(x-y)^2+2xy]/(x-y)∵xy=1,∴(x^2+y^2)/(x-y)=[(x-y)^2+2]/(x-y)=(x-y)+2/(x-y)∵x>y>0∴x-y>0∴根据基本不等式:(x-y)+2/(x-y)>=2√[(x-y)*2/(x-y)]=2√2当且仅当x-y=2/(x-y),即:x-y=√2时等号成立...