Let $f : \mathbb{R}^n \to \mathbb{R}^p$. I'm wondering if it is true that \begin{equation*} \arg\min_{x\in \mathbb{R}^n} \,\lVert f(x)\rVert = \arg\max_{x\in \mathbb{R}^n} \, \frac{1}{\lVert f(x)\rVert} \end{equation*} for each continuous $f$. If yes, which it seems to be at a first glance, how can be proved such equality? Maybe one can use some trick based on the logarithms of the costs.
Note: for me is perfectly fine to have an infinite cost, the important is that the solutions of the two optimization problems are always the same.