Prove that $x^4 + y^4 - 3xy = 2$ is compact The exercise consists of showing that the function $f(x,y)=x^4 + y^4$ has a global minimum and maximum under the constraint $x^4 + y^4 - 2xy = 2$.
In the solution to the exercise, it it follows that the constraint is compact if we can show that $\lim_{x^2 + y^2 \rightarrow \infty} x^4 + y^4 - 3xy - 2 \rightarrow \infty$.Why this is the case? 
My intuition tells me that this is because the $x^4$ and $y^4$ terms dominates the other two terms when $x$ and $y$ gets large. This would then imply that $x$ and $y$ cannot get arbitrarily big without violating the constraint. Does this imply that if the limit of the constraint was $0$, that the domain would not be compact? Is my reasoning valid?
Many thanks,

Note that
$$\eqalign{x^4+y^4-2xy&={1\over2}(x^2+y^2)^2+{1\over2}(x^2-y^2)^2-2xy\geq{1\over2}(x^2+y^2)^2-(x^2+y^2)\cr
&=(x^2+y^2)^2\left({1\over2}-{1\over x^2+y^2}\right)\geq{1\over4}(x^2+y^2)^2\geq4\ ,\cr}$$
as soon as $x^2+y^2\geq4$. It follows that the constraint defines a closed and bounded, hence compact, set in the plane.
