TITLE: An eigenvalue problem for positive semidefinite matrix
QUESTION [6 upvotes]: Let  $  \begin{bmatrix}
A& B   \\ B^*  &C
\end{bmatrix}$ be positive semidefinite, $A,C$ are of size $n\times n$. Is it true that $$\quad \sum\limits_{i=1}^n\lambda_i\begin{bmatrix}
A& B   \\ B^*  &C
\end{bmatrix}\le \sum\limits_{i=1}^n\left(\lambda_i(A)+\lambda_i(C)\right)\quad? $$ 
Here, $\lambda_i(\cdot)$ means the $i$th largest eigenvalue of $\cdot\quad$

REPLY [4 votes]: Note that, using your notation, $\quad \sum\limits_{i=1}^{2n}\lambda_i(\begin{bmatrix}
A& B   \\ B^*  &C
\end{bmatrix}) = \text{tr}( \begin{bmatrix}
A& B   \\ B^*  &C
\end{bmatrix}) = \text{tr} (A) + \text{tr} (C) = \sum\limits_{i=1}^n\left(\lambda_i(A)+\lambda_i(C)\right),$
so your inequality can actually be made stronger. But your inequality holds, since all the eigenvalues of a positive semidefinite matrix satisfy $\lambda_i \ge 0$, so if the sum on the LHS only goes up to $n$ it will of course be lessened.