(x2+y2)-(x+y+xy-1)
=(x^2-xy+y^2)-(x+y-1)
=[(x^2-2xy+y^2)+(x^2-2x+1)+(y^2-2y+1)]/2
=[(x-y)^2+(x-1)^2+(y-1)^2]/2≥0,
当x-y=x-1=y-1,
即x=y=1时取等号.
所以(x2+y2)≥(x+y+xy-1)
比较x2+y2与x+y+xy-1的大小?
(x2+y2)-(x+y+xy-1)
=(x^2-xy+y^2)-(x+y-1)
=[(x^2-2xy+y^2)+(x^2-2x+1)+(y^2-2y+1)]/2
=[(x-y)^2+(x-1)^2+(y-1)^2]/2≥0,
当x-y=x-1=y-1,
即x=y=1时取等号.
所以(x2+y2)≥(x+y+xy-1)