Let $A$, $B$ be a square, complex $4 \times 4$ matrices such that $Tr(A) = Tr(B) = 0$ Let $s_i(X)$ be i-th biggest singular value of matrix X. Let $\lambda_i(X)$ be i-th biggest eigenvalue (by magnitude).
From the fact that $A$ is traceless we have $\sum_{i=1}^4\lambda_i(A) = 0$, we have $\lvert \lambda_1(A) \rvert \leq \sum_{i=2}^4 \lvert \lambda_i(A) \rvert$. With that
$$ \frac{\lvert\lambda_1(A)\rvert}{\lvert\lambda_1(A)\rvert + \lvert\lambda_2(A)\rvert + \lvert\lambda_3(A)\rvert + \lvert\lambda_4(A)\rvert} \leq \frac{\lvert\lambda_2(A)\rvert + \lvert\lambda_3(A)\rvert + \lvert\lambda_4(A)\rvert}{\lvert\lambda_2(A)\rvert + \lvert\lambda_3(A)\rvert + \lvert\lambda_4(A)\rvert + \lvert\lambda_2(A)\rvert + \lvert\lambda_3(A)\rvert + \lvert\lambda_4(A)\rvert} =\frac{1}{2} $$
I wonder if we from that we can deduce something about analogous ratio for singular values $\frac{s_1(A)}{\sum_{i=1}^4s_i(A)}$ or squares of singular values $\frac{s_1^2(A)}{\sum_{i=1}^4s_i^2(A)}$ (eigeinvalues of $A^HA$). With $\lvert\lvert A\rvert\rvert^2_F + \lvert\lvert B\rvert\rvert^2_F = \frac{1}{4}$, where $\lvert\lvert X\rvert\rvert_F$ is a Frobenius norm of $X$, I saw that ratio $\frac{s_1^2(A)}{\sum_{i=1}^4s_i^2(A)}$ $$ \frac{s_1^2(A)}{\sum_{i=1}^4s_i^2(A)} \leq \frac{1}{2} $$