x,y为任意实数求证x^2+xy+y^2>=3(x+y-1)

Question

x,y为任意实数 求证 x^2+xy+y^2>=3(x+y-1)

Answer 1

设f(x,y)= x^2+xy+y^2-3(x+y-1)
令t=x+y,因为x,y为任意实数,t也为任意实数.
f(x,y)=f(t,x)= t^2-(3+x)t+x^2+3
f(t,x)为开口向上的抛物线
由根的判别式,△=[-(3+x)]^2-4(x^2+3)=-3(x-1)^2=0,
即 x^2+xy+y^2>=3(x+y-1)