Is my solution for the proof of $x^2+xy+y^2 > 0$ correct? The problem requires you to prove: $x^2 + xy + y^2 > 0$ assuming that $x$ and $y$ are not both zero
I made use of a property proven earlier ($x^3 - y^3) = (x-y)(x^2 + xy + y^2)$) and rewrote $x^2 + xy + y^2$ as $\dfrac{x^3 - y^3}{x-y}$. Then I simply tested for the three cases $x = y, x > y, x < y$
For the last two, it's easy to show that the condition holds. But when I consider $x = y$ then the statement that I got evaluates to zero, which is clearly not greater than zero, even though the statement $x^2 + xy + y^2$ should not be zero.
Edit: One stupid mistake. $x = y$ is undefined as it requires division by zero.
What do I do from here?
 A: If $x+y$, then your equation simply says that $0 = 0\cdot (x^2 + xy + y^2)$ which is correct, even if the expression $x^2+xy+y^2$ is not equal to zero.
The equation in this case is correct, but also useless. However, if you know that $x=y$, then the expression simplifies to $$x^2+x\cdot x + x^2$$ which is greater than zero because it is equal to $3x^2$.
A: If $x \neq 0 \Rightarrow x^2+xy+y^2 = \left(y+\dfrac{x}{2}\right)^2+\dfrac{3x^2}{4} \geq \dfrac{3x^2}{4}>0$. Similarly you can do the case $y \neq 0$.
A: Everyone has given a possible solution but I think they may not have answered your specific question. How to deal with the case $x=y$?. 
For $x=y$ you simple look at the expression given $x^2+xy+y^2=3x^2 >0$ when $x \neq 0$. Hence your proof will be complete. 
For a simple argument please look at my comment where I have given a link to the same question and the solution I had submitted for it.
A: Indeed, for $x\ne y$,
$$x^2+xy+y^2=\frac{x^3-y^3}{x-y}>0$$
as the numerator and denominator have the same sign.
For $x=y$, the remarkable identity cannot be used as it leads to an indeterminate form, and instead
$$x^2+xx+x^2>0,$$
unless $x=y=0$, which is false by hypothesis.
A: Seems obvious: we know $(x+y)^2>0$, and that just expands to $x^2+2xy+y^2>0$. So either $xy>0$ to begin with, or else, if $xy<0$, then positive $x^2+y^2$ dominates negative $2xy$. And if it dominates $2xy$, then it certainly dominates $xy$.
A: Hint: $$(x+y)^2-xy=x^2+xy+y^2=(x-y)^2+3xy.$$
