Why is $\text{maximize} \frac{1}{\lVert x \rVert}$ equivalent to $\text{minimize}\ \lVert x \rVert^2$? I know it is possible to solve $\text{minimize}\ \lVert x \rVert^2$ instead of
$\text{maximize} \frac{1}{\lVert x \rVert}$, since the former behaves better around zero. However, I am looking for a theorem or a rule that tells me this is allowed.
Edit: This is the objective of a constrained optimization problem. I just omit the constraints here.
 A: Consider a similar situation:

Maximising $f(x)$ is equivalent to minimising $-f(x)$.

As Michael Hardy mentioned in the comment, it is easy to see that as $f(x)$ gets bigger, $-f(x)$ gets smaller. We can say the same thing about reciprocal:

Maximising $\displaystyle\frac 1{\Vert x \Vert}$ is equivalent to minimising $\Vert x \Vert$.

Moverover, $\Vert x\Vert^2$ is an increasing function of $\Vert x\Vert$; when $\Vert x\Vert$ increases, $\Vert x\Vert^2$ increases as well; therefore

Minimising $\Vert x \Vert$ is equivalent to minimising $\Vert x \Vert^2$.

Combining the results above, we arrive at the conclusion

Maximising $\displaystyle\frac 1{\Vert x \Vert}$ is equivalent to minimising $\Vert x \Vert^2$.

You are right about the reason we choose to minimise $\Vert x \Vert^2$: it behaves better around zero, namely, it is differentiable, and the first and second derivatives are important for optimisation problems.
A: Since,
\begin{align*}
{\lVert x \rVert}^2 \leq {\lVert y \rVert}^2 \Leftrightarrow& \,{\lVert x \rVert} \leq {\lVert y \rVert}\\
\Leftrightarrow& \, \frac{1}{\lVert y \rVert} \leq \frac{1}{\lVert x \rVert}
\end{align*}
we have that the $x$ for which ${\lVert x \rVert}^2 $ is the minimum, is the $x$ for which $\frac{1}{\lVert x \rVert}$ is the maximum. (Just fix $x$ to be this specific $x$, and vary $y$ over all the possible values it can take.)
