Let us simplify your constraints. Namely, set $x=r\sin\phi$, $y=r\cos\phi$. Then it simplifies to $u\sin^2\phi + v\cos^2\phi \geq \sin\phi \cos\phi$, for all $0\leq \phi\leq 2\pi.$ Then divide both sides by $\cos^2\phi$, and set $z:=\frac{\sin\phi}{\cos\phi}$. You'll get quadratic inequality $f(z):=uz^2-z+v\geq 0$, for all $z$.
So we have to describe the set of $u$ and $v$ such that the latter always holds. The discriminant of $f(z)$ is $1-4uv$, and so one has $1-4uv\leq 0$, otherwise $f(z)$ has two distinct real roots, and $f$ can't be nonnegative everywhere. Note that also we see, by setting $x$ or $y$ to 0, that $u\geq 0$ and $v\geq 0$.
So we simplified your constraints, getting rid of $x$ and $y$, to the following form: $uv\geq 1/4$, $u\geq 0$, $v\geq 0$. The rest looks like a standard exercise in "elementary" nonlinear optimization.
as suggested by the comment by Noah below, an easier way is to directly specify the constraint is to write down $$ux^2+vy^2-xy=(x,y)\begin{pmatrix}u &-1/2\\-1/2 & v\end{pmatrix}\begin{pmatrix} x\\y\end{pmatrix}$$ and note that it is nonnegative everywhere iff $\begin{pmatrix}u &-1/2\\-1/2 & v\end{pmatrix}$ is positive semidefinite.