Does $\text{Tr}(A\rho A^\dagger \sigma)-\text{Tr}(A^\dagger A\rho\sigma) \geq 0$ hold if $\rho$ and $\sigma$ are positive semidefinite and Hermitian?

Question

Given an arbitrary $N\times N$ matrix $A$ and positive semidefinite $N\times N$ Hermitian matrices $\rho$ and $\sigma$ (i.e. $\rho, \sigma\geq 0$ and $\rho=\rho^\dagger, \sigma=\sigma^\dagger$), does $\text{Tr}(A\rho A^\dagger \sigma)-\text{Tr}(A^\dagger A\rho\sigma) \geq 0$ always hold?

Answer 1

In the comments I gave a counterexample to the OP's statement. In fact, there is a huge class of counterexamples. The Arraki-Lieb-Thirring inequality tells us that for $A,B \in \text{Mat}_{n\times n}$ with $A,B\geq 0$ and $r\in [0,1), q\in (0,\infty)$ we have

$$ \text{Tr}((B^r A^r B^r)^q) \leq \text{Tr}((BAB)^{rq}). $$

Now, for $A=A^\dagger, A\geq 0$ and $\rho=B=\sigma, B\geq 0$, $q=2, r=1/2$ Arraki-Lieb-Thirring gives us

$$\text{Tr}( \rho A \rho A)= \text{Tr}((\rho^{1/2} A \rho^{1/2})^2) = \text{Tr}((\rho^{1/2} (A^2)^{1/2} \rho^{1/2})^2) \leq \text{Tr}(\rho A^2 \rho), $$ which is the reverse inequality of the OP.