已知x,y为实数,且x^2+y^2/2+4小于等于xy+2y

题目

已知x,y为实数,且x^2+y^2/2+4小于等于xy+2y
求x,y的值
答案是x=2,y=4

答案

x^2+y^2/2+4≤xy+2y
x^2-xy+y^2/2-2y+4≤0
x^2-xy+y^2/4+y^2/4-2y+4≤0
(x-y/2)^2+(y/2-2)^2≤0
(x-y/2)^2和(y/2-2)^2非负
x-y/2=0,y/2-2=0
x=2,y=4