The Fan-Hoffman inequality for singular values is discussed in Bhatia's "Matrix Analysis" on page 73, stating that for any square matrix $A$:$$\lambda_j^\downarrow(\text{Re(A)}) \leq s_j^\downarrow(A)$$ where $\lambda_j^\downarrow(\text{Re(A)})$ are the eigenvalues of $\text{Re}(A)=(A+A^\dagger)/2$ in decreasing order and $s_j^\downarrow(A)$ are the singular values of $A$ in decreasing order. From this inequality, it directly follows that $$\sum_j|\lambda _j(\text{Re(A)})| \leq \sum_js_j(A)$$
Suppose now that $A$ is Hermitian and traceless, then $\text{Re}(A)=A$ and the inequality becomes trivial. Rather, we can write $A$ in terms of it's symmetric and skew-symmetric parts $A=A_s+iA_a$ where $A_s^T=A_s$ and $A_a^T=-A_a^T$. I'm trying to apply a Fan-Hoffman like inequality to show that $$\sum_j|\lambda_j(A_s)| \leq \sum_js_j(A).$$ This would follow directly from the relation $$\lambda_j^\downarrow(A_s) \leq s_j^\downarrow(A)$$ Is there an easy way to show this relation?