If $x$ and $y$ are not both $0$ then $ x^2 +xy +y^2> 0$ Can't quite finish this proof:

Prove that if $x$ and $y$ are not both $0$ then $ x^2 +xy +y^2> 0$

$ x^2 +xy +y^2 +xy -xy> 0$
$ (x +y)^2 -xy> 0$ Without loss of generality define $x\geq y$
Firstly the case where $y<0$ and $x>0$ is trivial as both $ (x +y)^2$ and $-(xy) >0$. 
I want to say by transitivity  as each are >0 that there sum must be ${}> 0$ can I do that?
The second trivial case is $y=0$ as we simply have $ x^2 >0 $ where $x>0$.
Lastly we have define $c= x+y$  since both $x$ and $y$ must have the same sign we know $|c|>x$ since $x\geq y$ we know that $ x^2 \geq xy$ finally since  $ |c|^2 = c^2$ $ \forall c \in \mathbb{R} $
We have $ c^2 >x^2 \geq xy$ hence $c^2 -xy >0$
I feel like I have all the pieces but it doesn't feel finished how do I fix it?
Edit: I am really glad i asked this question love the variety of different good answers no idea what a discriminant is but im going to go look that up. I wanted to give the accepted answer to Edward as he actually answered my question but Anurag
s proof was just so much better then my attempt. ^^
 A: Here's another way:
If $x=y$, then obvious. Otherwise:
$$x^2+xy+y^2=\frac{x^3-y^3}{x-y}>0$$
as $f(x)=x^3$ is monotonically increasing.
A: You can simply note that
$$
x^2+xy+y^2=(x+y/2)^2+3y^2/4\geq 0.
$$
Equality is iff $x=y=0$. So if $x,y$ are not both $0$, then $x^2+xy+y^2>0$.
A: Let $y$ a fixed non zero real. The discriminant of the quadratic expression $x^2+xy+y^2$ is $\Delta=y^2-4y^2=-3y^2< 0$ hence the expression doesn't change the sign and we have
$$x^2+xy+y^2>0$$
A: Your proof is correct. The first two cases are obvious. The third case follows directly from the AM-GM inequality:
$$\frac{x+y}{2}\ge \sqrt{xy}$$
$$\Rightarrow x+y>\sqrt{xy}$$
A: We only need to consider 2 cases:-
(1) Both x and y are negative
Then, $x^2 + xy + y^2 = (x – y)^2 + 3xy = (+ve) + (+ve) > 0$
(2) Only one is negative
Then, $x^2 + xy + y^2 = (x + y)^2 - xy = (+ve) – (-ve) > 0$
A: Switching to polar coordinates ($x=r\cos \theta,y=r \sin \theta)$, we need to prove
$$1+\frac{1}{2} \sin 2 \theta>0, $$
which is obvious.
A: Consider $T=x^2+xy+y^2$. then $2T=x^2+y^2+(x+y)^2 \geq 0$. The only way $T=0$ is when $x=0,y=0$, but that has already been ruled out by the conditions given.
