It is mentioned here that if $A, B, C\in M_{n}(\mathbb C)$ are positive semidefinite, then $$\det (A+B+C)+\det C\ge \det (A+C)+\det (B+C)$$ (quoted from this article) and the special case ($C=\bf 0$) $$\det (A+B)\ge \det (A)+\det (B).$$
The latter one has many proofs. The proof of the former one given in the paper uses tensor products which are decomposed into parts that mostly vanish due to orthogonality. It is valid not only for determinants, but for all *generalized matrix functions* (a.k.a. immanents).

Now: I have numerical evidence that a sharper Hlawka inequality $$\det (A+B+C)+\det (A)+\det (B)+\det (C)\ge \det (A+B)+\det (A+C)+\det (B+C)$$ also holds. ~~But this one becomes wrong if determinants are replaced by permanents (and supposedly for the other immanents, too). This makes me think that the decomposition of the corresponding tensor products can very probably ~~*not* be used to prove this one.

Note the formal similarity with Popoviciu's inequality, though I don't think that is of any help.

Is there a way to prove the Hlawka inequality for determinants?

EDIT: (to include my comment from below) **More generally, if $A_1,...,A_r\in M_{n}(\mathbb C)$ are psd matrices and $\chi\vdash n$ (or even: $\chi$ is an irreducible character of a subgroup of $S_n$), define for $k=1,...,r$ $$s_k:=\sum_{i_1<\cdots<i_k}Imm_\chi(A_{i_1}+\cdots+A_{i_k}).$$ Then I conjecture** $$s_r+s_{r-2}+\cdots\ge s_{r-1}+s_{r-3}+\cdots.$$

*FINAL EDIT: Since then, Suvrit and myself have proved an even more general result. It is published in Linear Algebra and its Applications.*