I have two positive definite $N\times N$ Hermitian matrices $A$ and $A$ and am interested in bounding the eigenvalues of $A+B$ in terms of the eigenvalues of $A$ and $B$. Let $\lambda_k(\cdot)$ be the $k$th largest eigenvalue of the argument.
I have studied the (dual) Lidskii and Weyl inequalities, but feel that I can come up with a better bound for a special case. My question is whether there exist a general version of the bound (I assume that the answer is yes...)
I can (easily) show that $$\lambda_1(A+B)+\lambda_2(A+B)\geq \lambda_1(A)+\lambda_1(B)+\lambda_N(A)+\lambda_N(B).$$ Equality occurs if $$\mathrm{Tr}(Q_1Q_2^\ast)=0,$$ where $$A=Q_1\Lambda_1Q_1^\ast,\quad \quad B=Q_2\Lambda_2Q_2^\ast.$$
What is the name of this? What is the general form of this inequality?