The $k^{\rm th}$ largest eigenvalue (arranged in decreasing order) of the sum of two $N \times N$ Hermitian (real symmetric) matrices $\bf{A}$ and $\bf{B}$ can be stated using the Weyl inequalities as

$L_k \leq \lambda_k({\bf A} + {\bf B}) \leq U_k$

with the lower and upper bounds given by

$L_k = {\rm max}\left[\lambda_i({\bf A}) + \lambda_j(\bf B)\right]$ with $i+j=k+N$

$U_k = {\rm min}\left[\lambda_i({\bf A}) + \lambda_j(\bf B)\right]$ with $i+j=k+1$.

The bounds on the largest eigenvalue $\lambda_1$ of the matrix sum ${\bf A}+{\bf B}$ are therefore

$L_1 = {\rm max}\left[\lambda_i({\bf A}) + \lambda_j(\bf B)\right]$ with $i+j=1+N$

$U_1 = {\rm min}\left[\lambda_i({\bf A}) + \lambda_j(\bf B)\right]$ with $i+j=2$.

Finding the upper bound is very simple; $U_1 = \lambda_1({\bf A})+\lambda_1({\bf B})$ since there is only one solution to the index equation $i+j=2$. However, finding the lower bound is much more difficult; the maximum of

$\left[\lambda_1({\bf A})+\lambda_N({\bf B}), \lambda_2({\bf A})+\lambda_{N-1}({\bf B}), \ldots, \lambda_N({\bf A})+\lambda_1({\bf B}) \right]$

must be found.

Are there any simplifications or theorems that can be applied to obtain a simpler expression for the lower bound (without restricting $\bf B$ to be a slight variation or permutation to $\bf A$) ?