# 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)\det\left(6(A^3+B^3+C^3)+I_{n}\right)\ge 5^n\det(A^2+B^2+C^2)?

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

This problem is a test question from China (xixi). It is said one can use the equation

$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$$

but I can’t use this to prove it. Can you help me?

Here is a partial and positive result, valid around the “triple point”
$$A=B=C= \frac13\mathbb 1A=B=C= \frac13\mathbb 1$$.

Let $$A,B,C\in M_n(\mathbb C)A,B,C\in M_n(\mathbb C)$$ be Hermitian satisfying $$A+B+C=\mathbb 1A+B+C=\mathbb 1$$, and additionally assume that
$$\|A-\tfrac13\mathbb 1\|\,,\,\|B-\tfrac13\mathbb 1\|\,,\, \|C-\tfrac13\mathbb 1\|\:\leqslant\:\tfrac16\tag{1}\|A-\tfrac13\mathbb 1\|\,,\,\|B-\tfrac13\mathbb 1\|\,,\, \|C-\tfrac13\mathbb 1\|\:\leqslant\:\tfrac16\tag{1}$$
in the spectral or operator norm. (In particular, $$A,B,CA,B,C$$ are positive-definite.)
Then we have
$$6\left(A^3+B^3+C^3\right)+\mathbb 1\:\geqslant\: 5\left(A^2+B^2+C^2\right)\,.\tag{2}6\left(A^3+B^3+C^3\right)+\mathbb 1\:\geqslant\: 5\left(A^2+B^2+C^2\right)\,.\tag{2}$$

Proof:
Let $$A_0=A-\frac13\mathbb 1A_0=A-\frac13\mathbb 1$$ a.s.o., then $$A_0+B_0+C_0=0A_0+B_0+C_0=0$$, or
$$\,\sum_\text{cyc}A_0 =0\,\,\sum_\text{cyc}A_0 =0\,$$ in notational short form.
Consider the

• Sum of squares
$$\sum_\text{cyc}\big(A_0 + \tfrac13\mathbb 1\big)^2 \:=\: \sum_\text{cyc}\big(A_0^2 + \tfrac23 A_0+ \tfrac19\mathbb 1\big) \:=\: \sum_\text{cyc}A_0^2 \:+\: \tfrac13\mathbb 1\sum_\text{cyc}\big(A_0 + \tfrac13\mathbb 1\big)^2 \:=\: \sum_\text{cyc}\big(A_0^2 + \tfrac23 A_0+ \tfrac19\mathbb 1\big) \:=\: \sum_\text{cyc}A_0^2 \:+\: \tfrac13\mathbb 1$$
• Sum of cubes
$$\sum_\text{cyc}\big(A_0 + \tfrac13\mathbb 1\big)^3 \:=\: \sum_\text{cyc}\big(A_0^3 + 3A_0^2\cdot\tfrac13 + 3A_0\cdot\tfrac1{3^2} + \tfrac1{3^3}\mathbb 1\big) \\ \;=\: \sum_\text{cyc}A_0^3 \:+\: \sum_\text{cyc}A_0^2 \:+\: \tfrac19\mathbb 1\sum_\text{cyc}\big(A_0 + \tfrac13\mathbb 1\big)^3 \:=\: \sum_\text{cyc}\big(A_0^3 + 3A_0^2\cdot\tfrac13 + 3A_0\cdot\tfrac1{3^2} + \tfrac1{3^3}\mathbb 1\big) \\ \;=\: \sum_\text{cyc}A_0^3 \:+\: \sum_\text{cyc}A_0^2 \:+\: \tfrac19\mathbb 1$$
to obtain
$$6\sum_\text{cyc}\big(A_0 + \tfrac13\mathbb 1\big)^3+\mathbb 1 \;-\; 5\sum_\text{cyc}\big(A_0 + \tfrac13\mathbb 1\big)^2 \:=\: \sum_\text{cyc}A_0^2\,(\mathbb 1 + 6A_0) \:\geqslant\: 06\sum_\text{cyc}\big(A_0 + \tfrac13\mathbb 1\big)^3+\mathbb 1 \;-\; 5\sum_\text{cyc}\big(A_0 + \tfrac13\mathbb 1\big)^2 \:=\: \sum_\text{cyc}A_0^2\,(\mathbb 1 + 6A_0) \:\geqslant\: 0$$
where positivity is due to each summand being a product of commuting positive-semidefinite matrices.
$$\quad\blacktriangle\quad\blacktriangle$$

Two years later observation:
In order to conclude $$(2)(2)$$ the additional assumptions $$(1)(1)$$ may be weakened a fair way off to
$$\tfrac16\mathbb 1\:\leqslant\: A,B,C\tag{3}\tfrac16\mathbb 1\:\leqslant\: A,B,C\tag{3}$$
or equivalently, assuming the smallest eigenvalue of each matrix $$A,B,C\,A,B,C\,$$ to be at least $$\tfrac16\tfrac16$$.

Proof:
Consider the very last summand in the preceding proof.
Revert notation from $$A_0A_0$$ to $$AA$$ and use the same argument, this time based on $$(3)(3)$$, to obtain
$$\sum_\text{cyc}\big(A-\tfrac13\mathbb 1\big)^2\,(6A -\mathbb 1)\:\geqslant\: 0\,.\qquad\qquad\blacktriangle\sum_\text{cyc}\big(A-\tfrac13\mathbb 1\big)^2\,(6A -\mathbb 1)\:\geqslant\: 0\,.\qquad\qquad\blacktriangle$$