已知1已知1
已知1记住公式a²+b²≥2ab(当且仅当a=b时成立)1<=x^2+y^2<=2等价于1<=2xy<=2所以1/2<=xy<=1故1/2<=x^2-x*y+y^2<=3
用三角换元,设X=r*cos@,Y=r*sin@(@是任意角,1<=r^2<=根号2)然后把它带入x^2-x*y+y^2,化简后就差不多了