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)\, 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


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 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}

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.

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.

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

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

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