Generalised Matrix $(p, k)$-norm[1] generalizes Schatten and Ky-Fan norms as it is defined as follows: $$ N_{p, k}(A) = (\sum_{i=1}^{k}\sigma_i^{p}(A))^{\frac{1}{p}} $$ where $\sigma_i$ is i-th largest singular value of matrix $A$.
Let $A$, $B$ be $4\times 4$ complex matrices, such that: $$ \text{tr}(A^\dagger A) + \text{tr}(B^\dagger B) = \frac{1}{4} $$ where $A^{\dagger}$ is complex conjugate of $A$.
I want to show that $[N_{2,2}(A+B)]^2 = [\sigma_1(A+B)]^2 + [\sigma_2(A+B)]^2 \leq \frac{1}{2}$.
For any complex, $4 \times 4$ complex matrix $X$ $$ tr(X^{\dagger}X) = \sum_{i=1}^4(\sigma_i(X))^2 = [N_{2,4}(X)]^2 \\ N_{2,4}(X) \geq N_{2,2}(X) \\ [N_{2,4}(X)]^2 \geq [N_{2,2}(X)]^2 \\ $$
Because $N_{p,k}(A)$ is a norm, we have: $$ N_{2,2}(A+B) \leq N_{2,2}(A) + N_{2,2}(B) \text{ (from triangle inequality)} \\ [N_{2,2}(A+B)]^2 \leq [N_{2,2}(A) + N_{2,2}(B)]^2 \\ [N_{2,2}(A+B)]^2 \leq [N_{2,2}(A)]^2 + [N_{2,2}(B)]^2 + 2 N_{2,2}(A)N_{2,2}(B) \\ [\sigma_1(A+B)]^2 + [\sigma_2(A+B)]^2 \leq [N_{2,2}(A)]^2 + [N_{2,2}(B)]^2 + 2 N_{2,2}(A)N_{2,2}(B) $$
Now, we know that $[N_{2,4}(A)]^2 + [N_{2,4}(B)]^2 = \frac{1}{4}$, from that we have $[N_{2,2}(A)]^2 + [N_{2,2}(B)]^2 \leq \frac{1}{4}$. We can plug it above to obtain
$$ [\sigma_1(A+B)]^2 + [\sigma_2(A+B)]^2 \leq \frac{1}{4} + 2 N_{2,2}(A)N_{2,2}(B) $$
Now, if we assign $a = N_{2,2}(A), b = N_{2,2}(B)$, then we want to find maximum value of $2ab$, given that $a^2+b^2 \leq \frac{1}{4}$ and $a, b \geq 0$. I double checked my calculations using wolframalpha (source) and found out that maximum value of $2ab$ under above conditions is $\frac{1}{4}$. From that we have $$ [\sigma_1(A+B)]^2 + [\sigma_2(A+B)]^2 \leq \frac{1}{4} + \frac{1}{4} = \frac{1}{2} $$
I am sure that there is an error somewhere along the lines, but I cannot find it Otherwise it would be a trivial solution to a problem described here.
[1] List of papers that shows that $N_{p,k}$ is a norm - link
This boils down to computing two different squared Frobenius norms to bound your truncated squared Schatten 2 (Frobenius) norm.
$0\leq \big\Vert A-B\big\Vert_F^2$
$\implies \text{trace}\big(A^*B\big)+\text{trace}\big(B^*A\big)\leq \text{trace}\big(A^*A\big)+\text{trace}\big(B^*B\big)= \frac{1}{4}$
and
$[N_{2,2}(A+B)]^2$
$\leq [N_{2,4}(A+B)]^2$
$= \big\Vert A+B\big\Vert_F^2$
$ = \text{trace}\big(A^*A\big)+\text{trace}\big(B^*B\big) + \Big(\text{trace}\big(A^*B\big)+\text{trace}\big(B^*A\big)\Big) $
$\leq \frac{1}{4}+\frac{1}{4}$
$=\frac{1}{2}$