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.

I am posting 2 ways of solving this:
$(1)$ If $x,y$ belong to positive real numbers only:
We know that $x^2 + y^2 \geq 2xy$.
Hence we can say that $x^2 + y^2 > xy$
Hence even if $xy$ is negative $x^2 + y^2$,which is positive, is always greater than $xy$ making the sum $x^2 + y^2 + xy$ always positive.
$(2)$ If $x,y$ belong to real numbers:
$$x^2 + y^2 + xy$$
$$ = x^2 + 2(x)(\dfrac{y}{2}) + \dfrac{y^2}{4} + \dfrac{3y^2}{4}$$
$$={(x+\dfrac{y}{2})}^2 + \dfrac{3y^2}{4}$$which is always positive.
Hope the answer is clear now !!
