Cofactor an geometrical mean in $\mathit{SPD}_3$: a Gårding-like inequality The cofactor map $A\mapsto\widehat A$ is polynomial homogeneous of degree $n-1$ over $\mathbf M_n({\mathbb R})$. It can be polarized into an $(n-1)$-linear symmetric map. When $n=3$, this provides a bilinear map $(A,B)\mapsto\widehat{A,B}$, defined by
$$\widehat{A,B}=\frac12\left(\widehat{A+B}-\widehat A-\widehat B\right).$$
Let me focus on the subspace of symmetric matrices. If $A$ is symmetric (resp. positive definite), then so is $\widehat A$. If $A,B\in\mathbf{Sym}_3$, then obviously $\widehat{A,B}$ is symmetric. More interesting and a little less obvious is the fact that if $A,B\in\mathbf{SPD}_3$, then $\widehat{A,B}$ is still positive definite. This is a consequence of the fact that the determinant is a hyperbolic polynomial over $\mathbf{Sym}_3$, with future cone $\mathbf{SPD}_3$; actually the result extends to positive symmetric matrices of arbitrary size.
Now let me recall the geometric mean of positive definite matrices $A$, $B$:
$$X\mathbin\sharp Y=Y^{\frac12}\left(Y^{-\frac12}XY^{-\frac12}\right)^{\frac12}Y^{\frac12}.$$
My question is about comparing two symmetric positive definite matrices:

Is it true that whenever $A,B\in\mathbf{SPD}_3$, we have
$$\widehat A\mathbin\sharp\widehat B\prec\widehat{A,B},\tag{$\dagger$}\label{dagger}$$
in the sense of the order between quadratic forms?

Remark that both sides are homogeneous of degree $1$ with respect to either argument. I have a positive answer in the following subcases:

*

*$A=I_3$, then it reduces to the arithmetic-geometric inequality,

*$A\vec e_1=0$ (which is a limit case when $A$ is semi-definite), the calculation being more involved.

*$B=A$, trivial because both sides equal $\widehat A$.

The inequality \eqref{dagger} can be seen as a variant of Gårding's Inequality for hyperbolic polynomials. If $P:{\mathbb R}^N\to{\mathbb R}$ is homogeneous of degree $d$, hyperbolic with forward cone $\Gamma$, then the associated $d$-linear form $\phi$ satisfies
$$\phi(a_1,\dotsc,a_d)\ge P(a_1)^{\frac1d}\dotsb P(a_d)^{\frac1d},\qquad\forall a_1,\dotsc,a_d\in\Gamma.$$
Here $d=2$, ${\mathbb R}^N\sim\mathbf{Sym}_3$, the cofactor map stands for $P$, and the inequality between numbers is replaced by Loewner's order between symmetric matrices.
 A: Because of the lack of answers, I continued my investigations, and eventually got it !
Notation : because some expressions are too long for the command widehat, I'll sometimes denote ${\rm Cof}A$ for $\widehat A$.
Recall that $\widehat A=(\det A)A^{-1}$ (the matrices are positive definite). This implies ${\rm Cof}(A^\alpha)=({\rm Cof}A)^\alpha$ for every $\alpha\in{\mathbb R}$. Using also the identity $\widehat M\widehat N=\widehat{NM}$ and the explicit formula for the mean, we infer
$$\widehat A\sharp\widehat B={\rm Cof}(A\sharp B).$$
Comparing this with $\widehat{A,B}$ amounts to comparing $x^T{\rm Cof}(A\sharp B)x$ with $x^T\widehat{A,B}x$ for unit vectors $x$. Because of orthogonal equivariance, it is enough to make the comparison when $x=e_1$ is the first vector of the canonical basis, hence to compare the upper-left entries of both matrices.
For this, let us denote $A',B',S'$ the lower-right $2\times2$ blocks of $A,B$ and $A\sharp B$ respectively. On the one hand
$${\rm Cof}(A\sharp B)_{11}=\det S'.$$
Because
$$\begin{pmatrix} A & S \\ S & B \end{pmatrix}$$
is positive semi-definite, the principal submatrix
$$\begin{pmatrix} A' & S' \\ S' & B' \end{pmatrix}$$
is so. Thus $S'\prec A'\sharp B'$, by definition of the mean. In particular
$$\det S'\le\det A'\sharp B'=\sqrt{\det A'\det B'}.$$
We now use the fact that ${\rm Det}$ is a hyperbolic polynomial over ${\bf Sym}_2({\mathbb R})$, the forward cone being ${\bf SPD}_2$. By Garding's inequality, we have
$$\sqrt{\det A'\det B'}\le\phi(A',B')$$
where $\phi$ is the bilinearization of ${\rm Det}$. Therefore
$$\det S'\le\frac12(a_{22}b_{33}+a_{33}b_{22}-2a_{23}b_{23}).$$
Finally the rhs is the upper-left entry of $\widehat{A,B}$. We have thus proved
$${\rm Cof}(A\sharp B)\prec\widehat{A,B}$$
for every $A,B\in{\bf SPD}_3$.
