TITLE: How does one prove the determinant inequality $\det\left(6(A^3+B^3+C^3)+I_{n}\right)\ge 5^n\det(A^2+B^2+C^2)$?
QUESTION [276 upvotes]: Let $\,A,B,C\in M_{n}(\mathbb C)\,$ be Hermitian and positive definite matrices such that $A+B+C=I_{n}$, where $I_{n}$ is the identity matrix. Show that $$\det\left(6(A^3+B^3+C^3)+I_{n}\right)\ge 5^n \det \left(A^2+B^2+C^2\right)$$

This problem is a  test question from China (xixi). It is said one can use the equation
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$$
but I can't use this to prove it. Can you help me?

REPLY [7 votes]: Here is a partial and positive result, valid around the "triple point"
$A=B=C= \frac13\mathbb 1$.
Let $A,B,C\in M_n(\mathbb C)$ be Hermitian satisfying $A+B+C=\mathbb 1$, and additionally assume that
$$\|A-\tfrac13\mathbb 1\|\,,\,\|B-\tfrac13\mathbb 1\|\,,\,
\|C-\tfrac13\mathbb 1\|\:\leqslant\:\tfrac16\tag{1}$$
in the spectral or operator norm. (In particular, $A,B,C$ are positive-definite.)
Then we have
$$6\left(A^3+B^3+C^3\right)+\mathbb 1\:\geqslant\: 5\left(A^2+B^2+C^2\right)\,.\tag{2}$$
Proof:
Let $A_0=A-\frac13\mathbb 1$ a.s.o., then $A_0+B_0+C_0=0$, or
$\,\sum_\text{cyc}A_0 =0\,$ in notational short form.
Consider the

Sum of squares
$$\sum_\text{cyc}\big(A_0 + \tfrac13\mathbb 1\big)^2 
\:=\: \sum_\text{cyc}\big(A_0^2 + \tfrac23 A_0+ \tfrac19\mathbb 1\big)
 \:=\: \sum_\text{cyc}A_0^2 \:+\: \tfrac13\mathbb 1$$
Sum of cubes
$$\sum_\text{cyc}\big(A_0 + \tfrac13\mathbb 1\big)^3 
\:=\: \sum_\text{cyc}\big(A_0^3 + 3A_0^2\cdot\tfrac13 
+ 3A_0\cdot\tfrac1{3^2} + \tfrac1{3^3}\mathbb 1\big) \\
\;=\: \sum_\text{cyc}A_0^3 \:+\: \sum_\text{cyc}A_0^2 \:+\: \tfrac19\mathbb 1$$
to obtain
$$6\sum_\text{cyc}\big(A_0 + \tfrac13\mathbb 1\big)^3+\mathbb 1
\;-\; 5\sum_\text{cyc}\big(A_0 + \tfrac13\mathbb 1\big)^2
\:=\: \sum_\text{cyc}A_0^2\,(\mathbb 1 + 6A_0) \:\geqslant\: 0$$
where positivity is due to each summand being a product of commuting positive-semidefinite matrices.
$\quad\blacktriangle$

Two years later observation:
In order to conclude $(2)$ the additional assumptions $(1)$ may be weakened a fair way off to
$$\tfrac16\mathbb 1\:\leqslant\: A,B,C\tag{3}$$
or equivalently, assuming the smallest eigenvalue of each matrix $A,B,C\,$ to be at least $\tfrac16$.
Proof:
Consider the very last summand in the preceding proof.
Revert notation from $A_0$ to $A$ and use the same argument, this time based on $(3)$, to obtain
$$\sum_\text{cyc}\big(A-\tfrac13\mathbb 1\big)^2\,(6A -\mathbb 1)\:\geqslant\: 0\,.\qquad\qquad\blacktriangle$$