Let $M:=\{(x,y,z) \in \mathbb R^{3} : x^2+y^2+z^2\geq 36\}$ and $f: \mathbb R^{3} \to \mathbb R$, $f(x,y,z)=\frac{x^2}{2}+\frac{y^2}{4}+\frac{z^2}{6}$
Justify why $f|_{M}$ takes on a minimum
The solution of my professor just confuses me:
$M$ is closed, $f$ is continuous and $\lim_{\vert (x,y,z)\vert \to \infty}f(x,y,z)\leq \lim_{\vert (x,y,z)\vert \to \infty}(x^2+y^2+z^2)$
And therefore, $f$ takes on a minimum on $M$ but does not take on a maximum on $M$.
How does this one-liner prove that $f|_{M}$ takes on a minimum??? Is there a theorem that I am forgetting?
The statement $\lim_{|(x,y,z)|\to\infty}f(x,y,z)\le\lim_{|(x,y,z)|\to\infty}(x^2+y^2+z^2)$ is a little weird, as both limits are infinite. However, we can use the fact that $\lim_{|(x,y,z)|\to\infty}f(x,y,z)=\infty$ to prove the claim. The basic idea is that since $f(x,y,z)$ gets large for large $|(x,y,z)|$, we don't need to worry about what happens very far away, so we can restrict to a closed, bounded set, and use the extreme value theorem.
More formally: Pick any $n$ such that $n=f(x,y,z)$ for some $(x,y,z)\in M$. Because $f(x,y,z)\to\infty$ as $|(x,y,z)|\to\infty$, we can find some $R$ such that $f(x,y,z)>n$ whenever $|(x,y,z)|>R$. Now $f$ has a minimum on $X:=\{(x,y,z)\in M:|(x,y,z)|\le R\}$ by the extreme value theorem, because this set is closed and bounded. Call this minimum $m$. Since $f$ achieves the value $n$ somewhere on this set, we must have $m\le n$. Now, outside of $X$, we have $|(x,y,z)|>R$ and therefore (by choice of $R$) $f(x,y,z)>n\ge m$. Hence $m$ is the minimum of $f$ on all of $M$.