# 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.

• What are $x$ and $y$? – preferred_anon Sep 30 '13 at 23:05
• It's amazing how all of the answers seemed to appear at once :) – Clayton Sep 30 '13 at 23:12
• I think it's just all real numbers. – Kat Sep 30 '13 at 23:14

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 $xy<0$.

• how do you get x^2 + 2xy + y^2 < xy ? – Kat Sep 30 '13 at 23:32
• @user96110: By adding $xy$ to both sides of $x^2+xy+y^2<0$. – Brian M. Scott Sep 30 '13 at 23:55
• thanks, and also, how did (x-y)^2 < -3xy become 0 < -3xy ? – Kat Oct 1 '13 at 0:04
• @user96110: $0\le(x-y)^2<-3xy$, since $(x-y)^2$ is a square. – Brian M. Scott Oct 1 '13 at 0:04
• How did you for from (x+y)^2 to (x-y)^2? Is it a typo? – Kevin Oct 1 '13 at 0:13

By completing square,

$$x^2+xy+y^2 = x^2+2x\frac{y}{2}+\frac{y^2}{4} + \frac{3y^2}{4} = \left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}\ge 0$$

• Sorry if this is kind of dumb, but i don't really get this. how would i complete the square? – Kat Sep 30 '13 at 23:22
• By completing square, you wish to find a form $x^2+xy+y^2 = (x+p)^2+q = x^2+2px+p^2+q$. Therefore, you should actually take $p=\frac{y}{2}$ to match the $xy$ term, then split the $y^2$ term to $\left(\frac{y}{2}\right)^2$ as $p^2$ and the remaining portion as $q$. – peterwhy Sep 30 '13 at 23:32

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.

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.

• i think i kind of get this. because even if x, y <0 ,the x^2 and y^2 would make the positive. Right? So then it would have to be >= 0. – Kat Sep 30 '13 at 23:13
• You had already arrived at the conclusion $xy\ge 0$. – Ted Shifrin Sep 30 '13 at 23:17

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.

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 $x^2+xy+y^2\lt0$.