Question

已知x,y∈R,求证:x�0�5-xy+y�0�5≥x+y-1

Answer 1

设f（x）=x�0�5-xy+y�0�5-x-y+1看成X的函数，y看成常数 判别式△=（y+1)^2-4（y�0�5-y+1）=-3y�0�5+6y-3=-3（y-1）的平方<=o 与x轴最多只有一个交点 又因为F(X)开口向上 所以F（X）恒为非负 就得到 x�0�5-xy+y�0�5-x-y+1≥0 也就是x�0�5-xy+y�0�5≥x+y-1

Answer 2

证明：X^2-XY+Y^2>=X+Y-12X^2-2XY+2Y^2>=2Y+2X-2 两边同时乘以2(X-Y)^2+X^2-2X+1+Y^2-2Y+1>=0 移项凑完全平方项 (X-Y)^2+(X-1)^2+(Y-1)^2>=0