2(x²-xy+y²)-2(x+y-1)
=(x²-2x+1)+(y²-2y+1)+(x²-2xy+y²)
=(x-1)²+(y-1)²+(x-y)²
平方大于等于0
所以(x-1)²+(y-1)²+(x-y)²>=0
所以2(x²-xy+y²)-2(x+y-1)>=0
所以2(x²-xy+y²)>=2(x+y-1)
所以x²-xy+y²>=x+y-1
2(x²-xy+y²)-2(x+y-1)
=(x²-2x+1)+(y²-2y+1)+(x²-2xy+y²)
=(x-1)²+(y-1)²+(x-y)²
平方大于等于0
所以(x-1)²+(y-1)²+(x-y)²>=0
所以2(x²-xy+y²)-2(x+y-1)>=0
所以2(x²-xy+y²)>=2(x+y-1)
所以x²-xy+y²>=x+y-1