Hlawka inequality for determinants of positive definite matrices

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.

• It also seems to hold for 4 p.s.d. matrices $$\det(A_1+\cdots+A_4) + \sum_{i<j}\det(A_i+A_j)\ge \sum_{i<j<k}\det(A_i+A_j+A_k)+\sum_i \det(A_i)$$ and so I'd conjecture the general one for $n$ such matrices ($det$ of sums of lengths $n,n-2,...$ on LHS, $det$ of sums of lengths $n-1,n-3,...$ on RHS). – Wolfgang Sep 30 '14 at 20:08
• This $n$ being unrelated to the previous $n$ (size of the matrices), I guess. – Robert Israel Sep 30 '14 at 20:49
• Ok, I think I proved the general case by proving a stronger result (your conjecture about the Hlawka for immanants is true). The typesetting is cumbersome, so I'll post a link. – Suvrit Sep 30 '14 at 23:03

My previous answer mis-attributed a claim to the paper of Paksoy, Turmen, and Zhang (also cited in the OP). Their claim is indeed strictly weaker than the inequality conjectured by Wolfgang. The details are slightly messy to type up, but they follow from the following more general operator Hlawka inequality

Theorem. Let $A, B, C$ be positive definite operators. Then, for each integer $k \ge 1$, we have \begin{equation*} \otimes^k(A+B+C) + \otimes^k A + \otimes^k B + \otimes^k C \ge \otimes^k(A+B) + \otimes^k(A+C) + \otimes^k(B+C). \end{equation*}

From this operator inequality, using "standard" arguments, we obtain the following result.

Corollary. Let Let $G$ be a subgroup of the symmetric group $S_n$, and $\chi$ is an irreducible character for $G$. Define the $G$-immanant of an arbitrary complex matrix \begin{equation*} d_{\chi}(X) := \sum_{\sigma \in G}\chi(\sigma)\prod_{i=1}^n x_{i,\sigma(i)}. \end{equation*} Then, for psd matrices $A, B, C$ it holds that \begin{equation*} d_\chi(A+B+C) + d_\chi(A)+d_\chi(B) + d_\chi(C) \ge d_\chi(A+B) + d_\chi(A+C) + d_\chi(B+C). \end{equation*}

We can use induction to generalize this claim to $n$ matrices. The conjecture mentioned by Wolfgang in the comments should also hold, though I haven't had time to think about it yet.

The details of the above results can be found here: PDF of the proof

• Unless I'm mistaken, the most general version of the conjecture (stated in comments to the original post) also follows from the theorem above. I'll expand out the details in the linked PDF as soon as I get a chance. – Suvrit Oct 1 '14 at 6:40
• Wow. (BTW, I realize something is wrong with my implementation of ´matperm´.) – Wolfgang Oct 1 '14 at 7:27
• Yes I think your proof by induction generalizes to $n>3$. Supposedly very technical to write up, though. Is there a better place to store your current pdf? With Dropbox, I'm afraid it will disappear after some weeks. – Wolfgang Oct 9 '14 at 8:25