Decide whether there is a polynomial of two variables bounded below with its set of values being an open subset of $\mathbb{R}$. My teacher gave me this problem:
Decide whether there is a polynomial of two variables bounded below with its set of values being an open subset of $\mathbb{R}$.
That means I can think of this function:
$$f(x,y)=Ax^2+Bxy+Cy^2+D$$
The fact that it is bounded below means that both $A$ and $B$ must be positive, right? But at this point I am stuck and I have no idea how to proceed. My guess is that it is unfeasible, because I can't think of a function that would satisfy all the requirements. 

Consider the polynomial $f : \mathbb{R}^2 \to \mathbb{R}$ given as $f(x,y) = (1-xy)^2 + x^2$.
$\DeclareMathOperator{\Ima}{Im}$The image of $f$ is $\langle 0, +\infty \rangle$, an open set which is bounded from below.
Indeed, we have $f(x,y) \ge 0$. Assume $0 = f(x,y) = (1-xy)^2 + x^2$. This gives $x = 0$ and $xy = 1$, a contradiction. Thus, $\Ima f \subseteq \langle 0, +\infty \rangle$.
Furthermore, for every $\varepsilon > 0$, we have:
$$f\left(\sqrt{\varepsilon}, \frac1{\sqrt{\varepsilon}}\right) = \left(1 -
 \sqrt{\varepsilon}\cdot \frac1{\sqrt{\varepsilon}}\right)^2 + \varepsilon = \varepsilon$$
So $\Ima f =\langle 0, +\infty \rangle$.
The example is from this answer.
