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 HilbertSchmidt norm of a matrix. I am looking for stronger/better/more elegant estimates (with references included) that would imply the above inequality.

5$\begingroup$ Are you interested in some generalizations of this inequality or best possible $C(n)$? I have reasonably short proof that $C(n) = \frac{2^{n1}}{\sqrt{n}^n}$ (that is, $A = B = I$ is optimal), but if you want something else could you please specify it somehow. $\endgroup$ – Aleksei Kulikov Dec 24 '18 at 1:11

1$\begingroup$ The bound $\det(A+B)=\detA+B \le \prod_i (a_i+b_{ni+1})$, where $a_i$ and $b_i$ are the singular values of $A, B$, respectively (sorted in the usual decreasing order) easily holds. Moreover, it follows from the following even more elegant inequality: $\det(A+B)^2 \le \det(A+B)\det(A^*+B^*)$, where $A$ denotes the matrix absolute value ($A=(A^*A)^{1/2}$). $\endgroup$ – Suvrit Dec 24 '18 at 12:51
Okay, here is the proof that $C(n) = \frac{2^{n1}}{\sqrt{n}^n}$.
Firstly, it is enough to find best constant $c(n)$ in $\det(A) \le c(n)A^n$. Indeed, we have $$\det(A+B) \le c(n)A+B^n \le c(n)(A + B)^n \le 2^{n1}c(n) (A^n + B^n),$$ 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 HilbertSchmidt 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 HilbertSchmidt norm of all unitary matrices is $\sqrt{n}$ we are done.

3$\begingroup$ we could use Hadamard inequality $\det A\leqslant \prod_{i=1}^n \Ae_i\\leqslant (\frac1n \sum \Ae_i\^2)^{n/2}=n^{n/2} \A\_{HS}^n$ $\endgroup$ – Fedor Petrov Dec 24 '18 at 9:23
There is a paper by ChiKwong 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)