Please refer to this beautiful paper on trace inequalities for matrix products. Theorem $3$ of the article (rephrased) states:
For any real $n\times n$ matrix $A$ and any real symmetric $B$ of the same size, let $\bar{A}=(A+A^T)/2$. Then \begin{align} \lambda_n(\bar{A})\operatorname{Tr}(B) &- \lambda_n(B)\Big(n \lambda_n(\bar{A}) - \operatorname{Tr}(A) \Big) \\ & \le \ \operatorname{Tr}(AB) \ \le \\ \lambda_1(\bar{A})\operatorname{Tr}(B) &- \lambda_n(B)\Big(n \lambda_1(\bar{A}) - \operatorname{Tr}(A) \Big) \end{align} where $\lambda_1$ and $\lambda_n$ denote the maximum and minimum eigenvalues respectively.
- Is this valid over the complex field $\mathbb{C}^{n\times n}$ when $A$ is positive semidefinite and $B$ is Hermitian?
- If yes, will having $A$ and $B$ as Hermitian be sufficient to ensure its validity?
The conditions mean that complex matrices $A$ and $B$ have real eigenvalues (trace also) and that $A = \bar{A}$.
While I'm of the thought that this is valid, I need its affirmation or correction. Thanks!