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.  

Averaging the inequalities $(x+y)^2\ge0$ and $x^2+y^2\ge0$ we get $\frac12(x+y)^2+\frac12(x^2+y^2)\ge0$, that is $x^2+xy+y^2\ge0$, contradicting the assumption that $x^2+xy+y^2\lt0$.
