(x²-xy+y²)-(x+y-1)
=[x²-xy+y²+x²-xy+y²-2(x+y-1)]/2
=[(x²-2x+1)+(y²-2y+1)+(x²+y²-2xy)]/2
=[(x-1)²+(y-1)²+(x-y)²]/2≥0
所以 (x²-xy+y²)-(x+y-1)≥0
即 x²-xy+y²≥x+y-1
(x²-xy+y²)-(x+y-1)
=[x²-xy+y²+x²-xy+y²-2(x+y-1)]/2
=[(x²-2x+1)+(y²-2y+1)+(x²+y²-2xy)]/2
=[(x-1)²+(y-1)²+(x-y)²]/2≥0
所以 (x²-xy+y²)-(x+y-1)≥0
即 x²-xy+y²≥x+y-1