It was asked in the Putnam exam of 1969, to list all sets which can be the range of polynomials in two variables with real coefficients. Surprisingly, the set $(0,\infty )$ can be the range of such polynomials. These don't attain their global infimum although they are bounded below. But is it also possible that such polynomials with range $(0,\infty )$ also have a non zero gradient everywhere?

$(1+x+x^2y)^2+x^2$ 


Too long for a comment. I've got to wonder just how difficult this is. Anyway, one thing did work out, at least locally: We have a polynomial function $F(x,y)$ that is assumed to have a nonvanishing gradient. Then the vector field $$ \left( \frac{\partial F}{\partial x}, \; \frac{\partial F}{\partial y} \right) $$ has integral curves that foliate the plane. On the other hand, the integral curves of $$ \left( \frac{  \partial F}{\partial y}, \; \frac{\partial F}{\partial x} \right) $$ are level curves as well, and foliate the plane. We know that if one of these level curves is a simple closed curve, using the Jordan Curve theorem it has an interior. If the function is constant on this it has zero gradient within, otherwise it achieves its maximum or minimum within and again has a critical point. $$ $$ After this I'm stuck. In particular, I simply don't see what polynomial does for us. Homogeneous polynomial would be different. There is a conjecture of Thom about local behavior that he apparently settled for homogeneous polynomials only. I would like to say that the picture that is being built up resembles that for $F(x,y) = e^x$ and that is absurd for a polynomial. Well, perhaps. I've also got to wonder how much the OP knows. 

