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?

**Answer**

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

**Attribution***Source : Link , Question Author : math110 , Answer Author : Hanno*