If $x \neq 0,y \neq 0,$ then $x^2+xy+y^2$ is ..... I came across the following problem that says:   

If $x \neq 0,y \neq 0,$ then $x^2+xy+y^2$ is
  1.Always positive
  2.Always negative
  3.zero
  4.Sometimes  positive and sometimes negative.  

I have to determine which of the aforementioned options is right.   
Now since $x \neq 0,y \neq 0$, so $ x^2+xy+y^2=(x-y)^2+3xy > 0$,if $x,y$ are of  same sign. But if $x,y$ are of different sign,I am not sure about the conclusion.
Can  someone point me in the right direction?  Thanks in advance for your time.

Don't give up so soon! Your idea also works when they have opposite sign:
$$  x^2+xy+y^2=(x+y)^2-xy>0$$
