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.  
 A: Assume $x^2+xy+y^2<0$. Adding and subtracting $xy$ on the left-hand side gives $x^2+2xy+y^2-xy=(x+y)^2-xy<0$, and therefore $0\leq(x+y)^2<xy$. Conversely, $x^2+xy+y^2<0$ implies $xy<-(x^2+y^2)\leq0$. Combining these, we have $$xy<0<xy,$$ a clear contradiction.
A: Assuming everything in the picture is a real number, you've arrived at $xy\ge 0$, and so $x^2+xy+y^2\ge 0$. This contradicts our assumption.
A: 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.
A: 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 $xy<0$.
A: By completing square,
$$x^2+xy+y^2 = x^2+2x\frac{y}{2}+\frac{y^2}{4} + \frac{3y^2}{4} = \left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}\ge 0$$
A: 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 $x^2+xy+y^2\lt0$.
