Let $A, B, C$ be Hermitian matrix and $C \geq 0, A \leq B$. Then, $$tr(CA) \leq tr(CB)$$
My attempt:
As we have that $C \geq 0$ and $B - A \geq 0$ and trace is a linear operator, then we just need to show that $tr(CB - CA) \geq 0$. We also know that $tr(M)$ is the sum of its eigenvalues, so let $\lambda_1, ..., \lambda_n$ be the eigenvalues of $CB - CA$. Then, our problem reduces to show that $\sum_{i=1}^n \lambda_i \geq 0$.
Can someone help me? Thank you!
$\text{tr}(C(B-A)) = \text{tr}(C^{1/2} (B-A) C^{1/2}) \ge 0$, where $C^{1/2}$ is the positive semidefinite square root of $C$.