$\det\left(6(A^3+B^3+C^3)+I_{2}\right)\ge 5^2\det(A^2+B^2+C^2)$ for $2 \times 2$ matrices

Motivated by this question I propose a simplification:

Question Let matrices $A,B,C\in M_{2}(\mathbb{C})$ be Hermitian and positive definite, such that:$$A+B+C=I_2$$ Show that: $$\det\left(6(A^3+B^3+C^3)+I_{2}\right)\ge 5^2\det(A^2+B^2+C^2)$$ where $I_{2}$ is the identity matrix.

The original question is of unknown origin and I am hoping for a substantial simplification in the $2 \times 2$ case where the following expansion holds:

$$\left[ \begin{array}{cc} x_1 & y + iz \\ y - iz & x_2\end{array}\right] = \left[ \begin{array}{cc} x_1 & 0 \\ 0 & x_2\end{array}\right] + y\left[ \begin{array}{cc} 0 & 1 \\ 1 & 0\end{array}\right] + z\left[ \begin{array}{rc} 0 & i \\ - i & 0\end{array}\right]$$

maybe with expansion to Pauli spin matrices this is solvable.

There is a nice article by Knutson and Tao on Honeycombs that might be of assistnce:

• http://www.ams.org/journals/jams/1999-12-04/S0894-0347-99-00299-4/

Let $L=\{(A,B,C)\in M_2(\mathbb{C})^3; A,B,C\text{ are hermitian }\geq 0,A+B+C=I_2\}$ and

$res:L\rightarrow \det(6(A^3 +B^3+C^3)+I_2)-25\det(A^2 +B^2+C^2)$.

Since we seek $\min_L res$, we may assume that $A$ is a diagonal matrix and that $res$ is defined on the set $K$ parameterized by

$A=diag(u,v),B=\begin{pmatrix}p&r+is\\r-is&q\end{pmatrix},C=\begin{pmatrix}1-u-p&-r-is\\-r+is&1-v-q\end{pmatrix}$, where

$u,v,p,q,1-u-p,1-v-q,pq-r^2-s^2,(1-u-p)(1-v-q)-r^2-s^2\geq 0$.

Note that $K$ is a compact convex subset of $\mathbb{R}^6$. The edge of $K$, $\partial K$ is the set of points of $K$ s.t. at least one of the $8$ terms above is zero and $K\setminus \partial K$ is the interior of $K$.

Of course, $\min_K res$ is reached in at least one point of $K$. We know that $\min_K res\leq 0$ and that, if $AB=BA$, then $res(A,B,C)\geq 0$.

We want to show

$\textbf{Theorem}.$ One has $\min_K res=0$ and this bound is reached only by commuting triplets of matrices.

$\bullet$ In the sequel, we assume that $AB\not= BA$.

An essential tool will be the Grobner basis theory on $\mathbb{C}$ and on $\mathbb{R}$. Since the theory over $\mathbb{R}$ is much less powerful, the idea is to reduce our problem to a polynomial system on $\mathbb{C}$ constituted by polynomials simpler than $res$ and to show that this system has no real solutions satisfying the $8$ conditions above.

$\textbf{Lemma}.$ We may assume $v>u\geq 0$ and $p,q,1-u-p,1-v-q>0$.

$\textbf{Proof}$. For example, if $pq=0$, then $r=s=0$ and $AB=BA$. $\square$

$\textbf{Proposition 1.}$ If $u>0,pq>r^2+s^2,(1-u-p)(1-v-q)>r^2+s^2$ (that is, $\det(A),\det(B),\det(C)>0$), then $(A,B,C)$ does not realize $\min_K res$.

$\textbf{Proof}$. Here $(A,B,C)\in K\setminus\partial K$; if it reaches $\min_K res$, then it must cancel the partial derivatives of $res$.

Step 1. We use the Maple library FGb (working over $\mathbb{C}$) to study the $3$ systems

$\dfrac{\partial res}{\partial u}=\dfrac{\partial res}{\partial v}=\dfrac{\partial res}{\partial p}=\dfrac{\partial res}{\partial q}=\dfrac{\partial res}{\partial r}=\dfrac{\partial res}{\partial s}=0,u-v\not= 0$, AND

$[(rs\not=0)$ OR $(r\not=0,s=0)$ OR $(r=0,s\not=0)]$,

asking (among other polynomials) for relations between $u,v$.

We obtain, in each case, a Grobner basis containing (in particular) $3$ polynomials $P,Q,R$ which only depend on $u,v$.

Step 2. We use the Maple library RAG (working over $\mathbb{R}$) to study the system

$P=Q=R=0,u,v-u,1-u,1-v>0$.

We obtain that there are no solutions in each of the $3$ cases. $\square$

$\textbf{Proposition 2}$. If $\det(A)$ or $\det(B)$ or $\det(C)=0$, then $(A,B,C)$ does not realize $\min_K res$.

$\textbf{Proof}$. We may assume that $\det(A)=0$, that is, $u=0$. If we put $a=r^2+s^2$ (with $a>0$ because $AB\not= BA$), then (Maple form)

$$res(0,v,p,q,r,s)=f(v,p,q,a)=-36*a*p^2*v^2-144*a*p*q*v^2-72*a*p*v^3-36*a*q^2*v^2-36*a*q*v^3-36*a*v^4-324*p^2*q^2*v-324*p^2*q*v^2+72*a^2*v^2+648*a*p*q*v+432*a*p*v^2+108*a*q*v^2+108*a*v^3+224*p^2*q^2+548*p^2*q*v+224*p^2*v^2+324*p*q^2*v+324*p*q*v^2-324*a^2*v-448*a*p*q-548*a*p*v-324*a*q*v-183*a*v^2-224*p^2*q-224*p^2*v-224*p*q^2-548*p*q*v-224*p*v^2-126*q^2*v-126*q*v^2+224*a^2+224*a*p+224*a*q+198*a*v+76*p^2+224*p*q+224*p*v+76*q^2+202*q*v+76*v^2-72*a-76*p-76*q-76*v+24.$$

We use the Maple library RAG (working over $\mathbb{R}$) to study the system

$f=0,v > 0, p > 0, q > 0, 1-p > 0, 1-v-q > 0, a > 0, a < 1$.

We obtain that there are no solutions.

Remark that $U=\{(v,p,q,a);v > 0, p > 0, q > 0, 1-p > 0, 1-v-q > 0, a > 0, a < 1\}$

is an open convex subset of $\mathbb{R}^4$ and that $f(0.1,0.3,0.4,0.1)\approx 5.6>0$.

We conclude that $f>0$ on $U$ and that $\min_K res$ is not reached on $U$. $\square$

$\textbf{Proof of Theorem.}$ According to Propositions 1 and 2, $\min_K res$ is reached only by triplets $(A,B,C)$ s.t. $AB=BA$; therefore $\min_k res=0$ and is reached only (up to simultaneous similarity) by the triplets

i) $diag(1/2,1/2),diag(1/2,0),diag(0,1/2)$

ii) $A=B=C=1/3.I_2$

iii) $A=B=0.5.I_2,C=0_2$

iv) $A=diag(1/3,1/2),B=diag(1/3,0),C=diag(1/3,1/2)$.