# Determinant inequality involving Hermitian, positive definite matrices

Let $$A,B,C\in M_{n}(\mathbb C)$$ be Hermitian and positive-definite matrices such that $$A+B+C=I_{n}$$.
Show that $$\det\left(6(A^3+B^3+C^3)+I_{n}\right)\ge 5^n\det(A^2+B^2+C^2)$$

This question has been asked on MSE without receiving any answers. Just some comment by achille hui and myself.
I think this result is true because for $$n=1$$ we have $$6(A^3+B^3+C^3)+1\ge 5(A^2+B^2+C^2)$$
but unfortunately I didn't succeed to generalize for $$n\ge2$$.

N.B.: My apologize if this question is not suitable on math overflow.

• Why do your expect this to be true? Have you compiled extensive evidence or is this posed somewhere as a problem? Sep 7, 2014 at 16:03
• @MichaelRenardy You can look at the link on MSE, it's a problem but achille hui was thinking that perhaps this problem is false, but as I said for $n=1$ is true because we can prove that $6(A^3+B^3+C^3)+I_n\ge 5(A^2+B^2+C^2)$ with proviso that $A+B+C=I_n$. Sep 7, 2014 at 16:09
• @MichaelRenardy Sigh. That particular OP posts question after question from contests, mostly in China but not always. I continue to think that the best approach is to get the OP to get the answers from the contest organizers, which are typically published eventually. Sep 7, 2014 at 20:13
• @ChristianRemling, sure. I find many contest problems fascinating. I do think people should be aware that this is a contest problem, someone knows the answer very well, and the person who originally asked on MSE does not care very much about it, there seems to be a bit of compulsion involved. Meanwhile, if I remain the only vote to close, there will be plenty of time for answers to appear here; also answers can be posted at MSE any time regardless. It is probably fair to say that I dislike the machine-gun rapid-fire postings more than I dislike contest problems. Sep 7, 2014 at 21:12
• @WillJagy But what if that OP doesn't get access to the solutions?
– C.S.
May 30, 2016 at 16:08