Denote the eigenvalues of an $n\times n$ matrix $\mathbf{X}$ by $\lambda_i(\mathbf{X})$ and its singular values by $\sigma_i(\mathbf{X})$, $i=1,\ldots,n$. When $\mathbf{X}$ is Hermitian, we know that $\sigma_i(\mathbf{X})=|\lambda_i(\mathbf{X})|$
Now, suppose $\mathbf{A}$ is Hermitian, $\mathbf{B}$ is complex-valued but not Hermitian, and $b\neq 0$ is a complex scalar. We know that, $\mathbf{B}+\mathbf{B}^*$ is a Hermitian matrix, as is $b\mathbf{B}+ b^*\mathbf{B}^*$. Additionally, I am given that $\operatorname{Tr}[\mathbf{B}+\mathbf{B}^*]=\sum_{i=1}^n\lambda_i(\mathbf{B}+\mathbf{B}^*)=0$ and $\operatorname{Tr}[\mathbf{A}]=\sum_{i=1}^n\lambda_i(\mathbf{A})=0$ (though not sure that that fact helps).
I am wondering if there is a lower bound the sum of singular values
$$\sum_{i=1}^n\sigma_i(\mathbf{A}+b\mathbf{B}+b^*\mathbf{B}^*)$$
in terms of the scalar $b$, the eigenvalues $\lambda_i(\mathbf{A})$, $\lambda_i(\mathbf{B}+\mathbf{B}^*)$, and/or the singular values $\sigma_i(\mathbf{A})$, $\sigma_i(\mathbf{B}+\mathbf{B}^*)$.
My usual approach to these kinds of problems involves application of Weyl's inequality, but that doesn't seem to work here (though perhaps I am not seeing something obvious). Any ideas?
