<s>[INST] 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.  
 [/INST] Another way to see this is as follows:
Suppose $x^2+xy+y^2 <0$, then $xy<-x^2-y^2< 0$, hence $2xy<xy<-x^2-y^2$ and thus $x^2+2xy+y^2<0$, which is absurd.
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