Question

设实数x,y满足x+y=4,则根号x^2+y^2-2x+2y+2（这几个一起开根号）的最小值

Answer 1

x+y=4

y=4-x

代入x^2+y^2-2x+2y+2

=x^2+x^2-8x+16-2x-2x+8+2

=2x^2-12x+26

=2(x-3)^2+8

所以当x=3,y=1时

x^2+y^2-2x+2y+2最小=8

所以原式最小值=2*根号2