A relation between the 2-norm condition number of a matrix $A$ and its eigenvalues is stated as $$ \kappa_2(A) \geq {\max{|\lambda_i(A)|} \over \min {|\lambda_i(A)|} } \tag{1} $$ where $A$ is not necessarily symmetric.
If we take for instance $$ A = \left[ \matrix{ 1 & 2 \cr 3 & 4 \cr} \right], $$ then $$ \kappa_2(A) = 14.9330, \ \max |\lambda_i(A)| = 5.3723, \min |\lambda_i(A)| = 0.3723 $$ and (1) can be verified as $$ {\max |\lambda_i(A)| \over \min |\lambda_i(A)| } = 14.4300 $$
If $A$ is symmetric, then we can use its special properties and in fact, the result (1) holds as an equality.
$$ \kappa_2(A) = {\sigma_\max(A) \over \sigma_\min(A)} = {|\lambda_\max(A)| \over |\lambda_\min(A)|} $$ (for symmetric matrices)
For the general case, how to establish the result (1)?
I attempted a proof using the property that $$ \rho(A) \leq \Vert A \Vert_2 $$ where $\rho(A)$ is the spectral radius of $A$.
This shows that $$ \max|\lambda_i(A)| \leq \Vert A \Vert_2 \tag{2} $$ and next we need to show that $$ {1 \over \min|\lambda_i(A)|} \leq \Vert A^{-1} \Vert_2 $$
We also note that $$ \rho\left( A^{-1} \right) \leq \Vert A^{-1} \Vert_2 $$
As noted in the comments below, the eigenvalues of $A^{-1}$ are the reciprocals of the eigenvalues of $A$.
Hence, it is immediate that $$ {1 \over \min{ | \lambda_i(A) |}} = \rho\left( A^{-1} \right) \leq \Vert A^{-1} \Vert_2 \tag{3} $$
Combining (2) and (3), the result (1) follows, viz.
$$ {\max |\lambda_i(A)| \over \min |\lambda_i(A)|} \leq \kappa_2(A) $$
In fact, this result holds true for the condition number of $A$ ($\kappa(A)$) with respect to any operator norm of $A$, since $\rho(A) \leq \Vert A \Vert$ for any operator norm.
