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.

Hint: $\displaystyle x^2+xy+y^2=y^2 \left( \left( \frac{x}{y} \right)^2+ \frac{x}{y} +1 \right)$; so you just have to study the polynomial $P(z)=z^2+z+1$.
