Considering an invertible matrix $A \in \mathbb{R}^{nxn}$ and the Frobenius norm, prove the following inequality:
$\lVert A^{-1}\rVert_F \ge \frac{1}{\lVert A\rVert}_F$
, knowing that $\lVert A\rVert_F = \sqrt{\sum_{i,j=1}^n \lvert a_{ij}\rvert^2} $.
This exercise isn't too difficult with other norms (the infinite one for example), as we can use the properties of a norm, and the fact that $I=AA^{-1}$, where I is the identity matrix:
$\lVert I \rVert_x = \lVert AA^{-1}\rVert_x \leq \lVert A \rVert_x \lVert A^{-1} \rVert_x$ for $x=1,2,\infty $
$(\textit{This inequality remains valid of course for the Frobenius norm.})$
With it, we can conclude that:
$\lVert A^{-1} \rVert_x \ge \frac{\lVert I \rVert_x }{\lVert A \rVert_x} $
We also know, by definition, that $\lVert I \rVert_x =1$ for $x=1,2,\infty $, and we can then directly get the upper inequality. But this isn't the same for the Frobenius norm, as $\lVert I \rVert_F = \sqrt{n}$, and from there I don't know what to do to solve the exercise.
Your help would be grateful :)