x,y为实数,证明x*2-xy+y*2>=x+y-1
问题描述：
x,y为实数,证明x*2-xy+y*2>=x+y-1
答
x*2-xy+y*2-(x+y-1)
=x^2-(y+1)x+(y^2-y+1)
判别式△=(y+1)^2-4(y^2-y+1)
=-3y^2+6y-3
=-3(y-1)^2
≤0
所以，x^2-(y+1)x+(y^2-y+1)≥0
x*2-xy+y*2>=x+y-1
答
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
答
〔（x^2-xy+y^2）-（x+y-1）〕
=(x^2-2x+1）+(y^2-2y+1)+(x^2-2xy+y^2)
=(x-1)^2+(y-1)^2+(x-y)^2>=0