x²+2xy+y²-4=0
(x+y)²=4
x²+y²=(x+y)²-2xy
=4-2xy x²+y²≥2xy -2xy≥-(x²+y²)
≥4-(x²+y²)
2(x²+y²)≥4
x²+y²≥2
所以当x=±1,y=±1时x²+y²有最小值2