Prove that $k(A) \geq \rho(A)/\min |\lambda|$ and that $k(A) \geq \rho(A) \rho(A^{-1})$ I'm trying to solve this question here. Thank you in advance for your help.
Prove that $k(A) \geq  \rho(A)/\min |\lambda|$ and that $k(A) \geq \rho(A) \rho(A^{-1}) $
We assume the matrices $A$ and $A^{-1}$ exist. Any size $A$ matrix.
I know $k(A) = ||A||\cdot ||A^{-1}||$ and $\rho(A) = \max_{\lambda\in\sigma(A)} |\lambda|=\lim_{n\to\infty}\|A^n\|^{1/n}$ with $\rho$ being the spectral radius and $k(A)$ being the condition number of the matrix. 
I'm not sure what exactly I need to set it up in order to get to both needed results.
 A: We will prove $k(A) \geq \rho(A)\rho(A^{-1}) $
We know, by definition, let $ \lambda $ be the eigenvalue for which $ \rho(A) = \max_{\lambda\in\sigma(A)} |\lambda| $. We let $x$ be an eigenvector, $ ||x||_{v} = 1 $.
$\Rightarrow \rho(A) = \max_{\lambda\in\sigma(A)} |\lambda| = ||\lambda x||_{v}$
$\Rightarrow  ||\lambda x||_{v} \leq ||Ax||_{v} \leq ||A||\cdot||x||_{v} = ||A|| $
$\Rightarrow \rho(A) \leq ||A||$
We apply the same with $A^{-1}$:
$\Rightarrow \rho(A^{-1}) = \max_{\lambda\in\sigma(A^{-1})} |\lambda| = ||\lambda x||_{v}$
$\Rightarrow  ||\lambda x||_{v} \leq ||A^{-1}x||_{v} \leq ||A^{-1}||\cdot||x||_{v} = ||A^{-1}|| $
$\Rightarrow \rho(A^{-1}) \leq ||A^{-1}||$
We combine both inequalities and get $k(A) \geq \rho(A)\rho(A^{-1})$.
To get $ k(A) \geq \rho(A)/\min_{\lambda\in\sigma(A)} |\lambda| $, we note $ k(A) \geq \rho(A)\rho(A^{-1})$. We know $\rho(A) = \max_{\lambda\in\sigma(A)} |\lambda|$ and the same with $\rho(A^{-1})$. We note that the eigenvalues of $A^{-1}$ are the reciprocals of those of $A$.
$\Rightarrow k(A)\geq \rho(A)\rho(A^{-1})$
$\Rightarrow k(A) \geq \rho(A)/\min_{\lambda\in\sigma(A)}|\lambda|$.
