Prove that $x^2 + xy + y^2 \ge 0$ by contradiction Using the fact that $x^2 + 2xy + y^2 = (x + y)^2 \ge 0$, show that the assumption $x^2 + xy + y^2 < 0$ leads to a contradiction...
So do I start off with...
"Assume that $x^2 + xy + y^2 <0$, then blah blah blah"?
It seems true...because then I go $(x^2 + 2xy + y^2) - (x^2 + xy + y^2) \ge 0$.
It becomes $2xy - xy \ge 0$, then $xy \ge 0$.
How is this a contradiction? 
I think I'm missing some key point.  

Suppose that $x^2+xy+y^2<0$; then $x^2+2xy+y^2<xy$, so $(x+y)^2<xy$. Subtracting $3xy$ from both sides of the original inequality, we see that $x^2-2xy+y^2<-3xy$, so $(x-y)^2<-3xy$. Squares are non-negative, so on the one hand $xy>0$, and on the other hand $-3xy>0$ and therefore $xy<0$.
