Question

已知X>Y>0,且XY=1,求(X^2+Y^2)/(X-Y)的最小值及相对应的X,Y的值.

已知X>Y>0,且XY=1,求(X^2+Y^2)/(X-Y)的最小值及相对应的X,Y的值.

Answer 1

因为(X^2+Y^2)/(X-Y)=[(X-Y)^2+2XY]/(X-Y),
因为XY=1,，
所以(X^2+Y^2)/(X-Y)
=[(X-Y)^2+2]/(X-Y)
=(X-Y)+2/(X-Y)，
因为X>Y>0,
所以X-Y>0,
所以(X^2+Y^2)/(X-Y)
=(X-Y)+2/(X-Y)
>=2*根号[(X-Y)*2/(X-Y)]
=2倍根号2.
当且仅当(X-Y)=2/(X-Y),即X-Y=根号2时等号成立，
因为X-Y=根号2，XY=1,
所以X=[(根号6)+(根号2)]/2,Y=[(根号6)-(根号2)]/2,
所以(X^2+Y^2)/(X-Y)的最小值为2倍根号2，
此时X=[(根号6)+(根号2)]/2,Y=[(根号6)-(根号2)]/2.