Does a polynomial that's bounded below have a global minimum?
No. The polynomial $f(x, y) = (1 - xy)^2 + x^2$ is bounded below by $0$, cannot actually take the value $0$, but can take the value $\epsilon$ for any $\epsilon > 0$. I learned this example from Richard Dore on MO.
@Agusti Roig. $f$ has no critical points in the interior of the disc, so you have to look for the minimum at the unit circle. One way would be to parametrize $x=\cos\theta, y=\sin\theta$, and minimize.