Let $A$ be a symmetric positive semidefinite $n \times n$ matrix. How can I show that the sum of the largest $n-k+1$ eigenvalues of $A - k\cdot \textrm{diag}(A)$ is nonpositive, for any $k \in \{1, \dots, n\}$?
For example, equality holds for $k = 1$ and for $k = n$ when $\textrm{rank}(A) = 1$. I have ad-hoc proofs for $k \in \{2,n-1\}$ and have checked it with random matrices on numpy for other values of $k$.
We can reduce it to the case when $A$ has rank 1 by convexity, and in that case we can write down the characteristic polynomial using the matrix determinant lemma. It doesn't seem to help.
There also seems to be a more general version: if the eigenvalues are $\lambda_1 \ge \cdots \ge \lambda_n$, then for any $m \in \{1, \dots, n\}$ we have $\lambda_m + \cdots + \lambda_{n - m(k-1)} \le 0$. The problem above is the special case $m = 1$. When $m = n/k$ this means $\lambda_{n/k} \le 0$, which is not hard to show by looking at the eigenvalues of $D^{-1/2} A D^{-1/2}$ where $D = \textrm{diag}(A)$. I don't know how to solve it for any other $m$.
