Let $A$ and $B$ be positive definite Hermitian matrices. They need not necessarily commute. Let $a_m$ be the minimum eigenvalue of $A$.
Is it true that $$ \text{Tr} (AB) \geq a_m \text{Tr} (B) $$
My attempt: let $a_i$ and $b_i$ be the eigenvalues (which are all real and positive) of $A$ and $B$. If $A$ and $B$ are simultaneously diagonalizable by a similarity transformation then $ \text{Tr} (AB) = \sum_i a_i b_i \geq a_m \sum_i b_i = a_m \text{Tr} (B) $. But is this generalizable to arbitrary $A$ and $B$ as laid out above?
Note that this is equivalent to proving that $\text{tr}((A - a_m I)B) \ge 0$. In fact the following more general claim is true:
To see this we use the fact that there exist matrices $C, D$ such that $A = C^{\dagger} C, B = D^{\dagger} D$ where $^{\dagger}$ denotes the adjoint. Then we compute that
$$\text{tr}(AB) = \text{tr}(C^{\dagger} CD^{\dagger} D) = \text{tr}(DC^{\dagger} CD^{\dagger}) = \text{tr}(E^{\dagger} E) \ge 0$$
where $E = CD^{\dagger}$.