If $A$ and $B$ are $n\times n$ matrices, then it easily follow from the definition of the determinant by sum over permutations, and from the Young inequality that
$$
|\det (A+B)|\leq C(n)(\Vert A\Vert^n+\Vert B\Vert^n),
$$
where $\Vert \cdot\Vert$ stands for the Hilbert-Schmidt norm of a matrix. I am looking for stronger/better/more elegant estimates (with references included) that would imply the above inequality.



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


