(x²-xy+y²)-(x+y-1)
=[(x²-2xy+y²)+(x²-2x+1)+(y²-2y+1)]/2
=[(x-y)²+(x-1)²+(y-1)²]/2
≥0
∴x²-xy+y²≥x+y-1
当且仅当x=y=1时取等号
已知x,y∈R,求证x2-xy+y2>=x+y-1
(x²-xy+y²)-(x+y-1)
=[(x²-2xy+y²)+(x²-2x+1)+(y²-2y+1)]/2
=[(x-y)²+(x-1)²+(y-1)²]/2
≥0
∴x²-xy+y²≥x+y-1
当且仅当x=y=1时取等号