x^2+y^2-(xy+x+y-1)
=(x^2-2xy+y2+x^2-2x+1+y^2-2y+1)/2
=[(x-y)^2+(x-1)^2+(y-1)^2]/2
≥0
所以x2＋y2＞等于xy＋y＋x－1
求证x2＋y2＞等于xy＋y＋x－1
x^2+y^2-(xy+x+y-1)
=(x^2-2xy+y2+x^2-2x+1+y^2-2y+1)/2
=[(x-y)^2+(x-1)^2+(y-1)^2]/2
≥0
所以x2＋y2＞等于xy＋y＋x－1