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.
Best Answer
The eigenvalues of $A^{-1}$ are the reciprocals of the eigenvalues of $A$. So $\frac{1}{\min |\lambda_i(A)|}=\max |\lambda_i(A^{-1})|$; thus your property finishes the proof.