因?yàn)?x^2+y^2+1)-(x+y+xy)
=x^2+y^2+1-x-y-xy=1/2*(2x^2+2y^2+2-2x-2y-2xy)
=1/2*[(x^2-2xy+y^2)+(x^2-2x+1)+(y^2-2y+1)]
=1/2*[(x-y)^2+(x-1)^2+(y-1)^2]
又因?yàn)?x-y)^2≥0且(x-1)^2≥0且(y-1)^2≥0,
所以(x-y)^2+(x-1)^2+(y-1)^2≥0,
所以1/2*[(x-y)^2+(x-1)^2+(y-1)^2]≥0,
即(x^2+y^2+1)-(x+y+xy)≥0,
所以x^2+y^2+1≥x+y+xy.