Let $A\in\mathbb{R}^{n\times n}$ be an invertible matrix and $\mathbf{x}\in\mathbb{R}^n$. I am trying to prove that
$$\sup_{||\mathbf{x}||=1}||A^{-1}\mathbf{x}||\inf_{||\mathbf{x}||=1}||A\mathbf{x}||=1,$$
but I failed to find relation between $||A^{-1}\mathbf{x}||$ and $||A\mathbf{x}||$. Some numerical results suggest that if $||.||$ is Euclidean norm, one can generalize the above equation to
$$\sup_{||\mathbf{x}||=r}||A^{-1}\mathbf{x}||\inf_{||\mathbf{x}||=r}||A\mathbf{x}||=r^2,$$
for any $r>0$. Any help would be appreciated. Thank you.
Note that
$$\dfrac{1}{\inf_{\|{\bf x}\|=1} \|A{\bf x}\|} = \sup_{\|{\bf x}\|=1} \dfrac{1}{\|A{\bf x}\|} = \sup_{\|{\bf x}\|=1} \dfrac{\|A^{-1}(A{\bf x})\|}{\|A{\bf x}\|} = \sup_{{\bf x} \in \mathbb{R}^n \setminus \{{\bf 0}\}} \dfrac{\|A^{-1}(A{\bf x})\|}{\|A{\bf x}\|}$$
Since $A$ is nonsingular, we have $A(\mathbb{R}^n \setminus \{{\bf 0}\}) = \mathbb{R}^n \setminus \{{\bf 0}\}$ and therefore $$\sup_{{\bf x} \in \mathbb{R}^n \setminus \{{\bf 0}\}} \dfrac{\|A^{-1}(A{\bf x})\|}{\|A{\bf x}\|} = \sup_{{\bf y} \in \mathbb{R}^n \setminus \{{\bf 0}\}} \dfrac{\|A^{-1}{\bf y}\|}{\|{\bf y}\|} = \sup_{\|{\bf y}\| = 1} \|A^{-1}{\bf y}\|$$