An upper estimate for $|\det(A+B)|$

There is a paper by Chi-Kwong Li and Roy Mathias about lower and upper bounds for $|\det(A + B)|$ in terms of the singular values of $A$ and $B$. (The determinant of the sum of two matrices)

Okay, here is the proof that $C(n) = \frac{2^{n-1}}{\sqrt{n}^n}$.

Firstly, it is enough to find best constant $c(n)$ in $|\det(A)| \le c(n)||A||^n$. Indeed, we have $$|\det(A+B)| \le c(n)||A+B||^n \le c(n)(||A|| + ||B||)^n \le 2^{n-1}c(n) (||A||^n + ||B||^n),$$ and for $A = B$ we have here equality.

Now let $A$ be a matrix with $||A|| = 1$ and maximum possible $|\det A|$ (such a matrix obviously exists by compactness). Let $A = (v_1, \ldots , v_n), v_k\in \mathbb{C}^n$. We will prove that $(v_k, v_m) = 0, k\ne m$ and $(v_k, v_k) = (v_m, v_m)$.

Assume that $(v_k, v_m) \ne 0, k\ne m$. Then we can add $tv_k$ to $v_m$ without changing determinant. But if $(v_k, v_m) \ne 0$ then we can choose $t$ such that $(v_m + tv_k, v_m + tv_k) < (v_m, v_m)$. But then we will have matrix with smaller Hilbert-Schmidt norm and the same $|\det|$, which is impossible since $A$ gives us maximum.

Similarly assume that $(v_k, v_k) \ne (v_m, v_m)$. Then we can instead consider $\lambda v_k$, $\frac{1}{\lambda} v_m$ without changing $\det$. But then for $\lambda$ close enough to $1$ we can again decrease norm of $A$ which is forbidden.

Thus, all columns of $A$ are orthogonal and have same length. Then $A$ is a multiple of unitary matrix and since $\det$ of all unitary matrices is $1$ and Hilbert-Schmidt norm of all unitary matrices is $\sqrt{n}$ we are done.