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.
 A: 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 $P(z)=z^2+z+1$.
A: Don't give up so soon! Your idea also works when they have opposite sign:
$$  x^2+xy+y^2=(x+y)^2-xy>0$$
A: 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 !!
A: $x^2+xy+y^2 = \frac{3}{4}(x+y)^2+\frac{1}{4}(x-y)^2 >0$, when $x \ne 0$ and $y \ne 0$.
A: $x^2 + xy + y^2$ is a quadratic form and can be written
$$x^2 + xy + y^2 = \begin{bmatrix} x & y\end{bmatrix}\begin{bmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & 1 \end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}$$
A matrix $\mathbf{A}$ is positive-definite if $\mathbf{z'}\mathbf{A}\mathbf{z}>0 $ for all $\mathbf{z}\neq 0$
Show that $\begin{bmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & 1 \end{bmatrix}$ is a positive-definite matrix and you will have a very nice solution to your problem
A: Some of the answers here are rigorous but overly complicated. Consider that $x^2$ is positive and $y^2$ is positive. The only term that could be negative is $xy$.
Suppose that $xy$ is negative. If abs($x$)>abs($y$), then $x²>-xy,$ so the whole expression is positive. The same logic applies for abs($y$)>abs($x$). The $x=y$ case is clearly positive.
Therefore the whole expression is positive.
A: Why not simply $x^2+xy+y^2=(x+\frac{y}{2})^2+\frac{3}{4}y^2>0$?
A: As you say, if $x,y$ have the same sign then: $x^2+xy+y^2=(x-y)^2+3xy>0$. If $x,y$ have opposite signs then $x^2+xy+y^2=(x+y)^2-xy>0$.
A: Writing the given expression in the quadratic matrix form:
$$
x^2 + xy + y^2 = \begin{bmatrix} x \\ y \end{bmatrix}^T
\underbrace{\begin{bmatrix} 1 & a \\ (1-a) & 1 \end{bmatrix}}_{A}
\begin{bmatrix} x \\ y \end{bmatrix} \quad\quad\quad (a \in [0,1])
$$
According to the Gershgorin Circle Theorem, both the eigenvalues of $A$ may lie in $[0,2]$, and two of them cannot be zero at the same time; so A is at least positive semi-definite.
But, if we manually check for extreme cases (i.e.; $a=0$ and $a=1$) for a more strict analysis, we see that both the eigenvalues are $1$ when $a=1$ and $a=0$.
Therefore, $A$ is positive definite. The given expression is always positive under the given conditions.
