Clean Explanation/Reasoning as to why the set takes on a minimum This is rather trivial but I am trying to wrap my head around good reasoning as to why a set that has a lower bound, yet has no upper bound would still take on a minimum. All of this should be taken in the context of constrained optimization problems:
Example:
Let $f: \mathbb R^{3} \to \mathbb R$, $f(x,y,z)=\frac{x^2}{2}+\frac{y^2}{4}+\frac{z^2}{4}$ and  $M:=\{(x,y,z)\in \mathbb R^{3}|x^2+y^2+z^2\geq 9\}$
Show that $f|_{M}$ takes on a minimum
My Approach:
Define $h:\mathbb R^3\to\mathbb R, h(x,y,z)=x^2+y^2+z^2$
i) $M = h^{-1}([9,\infty[)$ and $h$ is continuous, so thus $M$ is a closed interval. It is clear that $M$ has no upper bound. And it is clear that $M$ has a lower bound, $\inf M$. 
Define $(x,y,z) = \inf M$ and we know there exists a sequence $(a_{n})_{n}$ such that $a_{n}\to \inf M$, as $ n \to \infty$
And since $M$ is closed, we know we know that $(a_{n})_{n}$ converges in $M$ thus $\inf M \in M$ and so $M$ possesses a minimum.
But is this enough to know that $f|_{M}$ takes on a minimum? Perhaps yes, because the function are so similar, however, but what if my functions were different, for example define $f(x,y,z)=x^3-7y+z^2$.
Bottom line: There must be something obvious I am missing in order to prove that $f|_{M}$ takes on a minimum
 A: Your argument is flawed: $M$ is a subset of $\mathbb{R}^3$, so it doesn't make sense in this context for $M$ to be an interval, or for $M$ itself to have a lower bound. Even for functions $\mathbb{R} \to \mathbb{R}$, the preimage of an interval under a continuous function need not be an interval, for instance if $g(x)=x^2$ then $g^{-1}(\left[9,\infty\right[) = \left] -\infty, -3 \right] \cup \left[ 3, \infty \right[$.
What you need to prove is that the image of $f |_M$ has a minimum element; that is, what you seek is a minimum element of the set
$$\left\{ \frac{x^2}{2} + \frac{y^2}{4} + \frac{z^2}{4} ~ \middle| ~ (x,y,z) \in M \right\}$$
Here's a direct approach. Note that when $(x,y,z) \in M$ we have
$$f(x,y,z) = \frac{1}{4}(x^2+y^2+z^2) + \frac{1}{4}x^2 \ge \frac{9}{4} + \frac{1}{4}x^2 \ge \frac{9}{4}$$
so it suffices to prove that $f(x,y,z) = \frac{9}{4}$ for some $(x,y,z) \in M$.
But $(0,0,3) \in M$ and $f(0,0,3) = \frac{9}{4}$. Hence $f |_M$ attains a minimum value.

Added: To answer your more general question, there are functions whose domains are closed subsets of $\mathbb{R}^3$ which are bounded below and unbounded above but do not attain a minimum value.
For example take
$$M = \{ (x,y,z) \in \mathbb{R}^3 \mid x \ge 0, y \ge 0, z \ge 0 \}$$
and let $f(x,y,z) = e^x + e^y + e^{-z}$. Then $f|_M$ is bounded below by $2$ and takes all values $>2$, but does not attain the value $2$.
