Prove that the function has a maximum and a minimum value

Question

A function $$f\colon D \to \mathbb R$$ where $$(x,y,z) \in D \iff (x^2+y^2+4z^2-6 =0)$$ is given by this formula: $$f(x, y, z) = xyz$$ Prove - without finding them - that this function has a maximum and a minimum value.

I have thought about this problem and I came up with a possible:

We can either show that $D$ is a compact set, then $f$ will be uniformly continuous, and so $f$ will be continuous, and so - by the theorem of continuous functions - will have a minimum and a maximum value on $D$. However, it is not easy to prove that $D$ is compact.

Is there a better alternative to solving this problem, something a trifle less tedious?

Answer 1

It isn't too much work to show that $D$ is closed (it is, for instance, the inverse image of the closed set $\{0\}\subseteq \Bbb R$ for some suitable continuous function $\Bbb R^3\to\Bbb R$), and that it is bounded (none of the coordinates can exceed $3$ in absolute value). And for subsets of $\Bbb R^n$, closed and bounded means compact.

Answer 2

Alt. hint: $\;\sqrt[3]{|x| \cdot |y| \cdot 2 |z|} \le \sqrt{\dfrac{x^2+y^2+4z^2}{3}}=\sqrt{2}\;$ by the root-mean square vs. GM inequality, and the equality holds iff $\,|x|=|y|=2|z|\,$.