Let $H_n$ denote the space of all hermitian $n$ by $n$ matrices. Let $\mathbb{R}[H_n]$ denote the space of all real-valued polynomials $f$ with real coefficients in the entries of a variable matrix $A \in H_n$. Further, denote by $\mathbb{R}[H_n]^G$ the space of all such polynomials $f$, which are invariant under the adjoint action of $G = U(n)$ on $H_n$. In other words, it consists of polynomials $f \in \mathbb{R}[H_n]$ such that $f(gAg^{-1}) = f(A)$ for any unitary $n$ by $n$ matrix $g$.
Note that this is pretty much the same setting as in the Chern-Weil homomorphism (for the case where $G$ is the unitary group), except we are working mostly over $\mathbb{R}$ as the ground field.
Now let $f \in \mathbb{R}[H_n]^G$ be homogeneous of degree $k$. Denote also by $f$ its complete polarization, so that $f$ is a symmetric multilinear form depending of $k$ hermitian matrices $A_1, \ldots, A_k$.
Assume that $f(A_1,\ldots,A_k) \geq 0$ in the case where the $A_i$ are hermitian positive semidefinite and pairwise commute with each other. Does the inequality then hold for all hermitian positive semidefinite matrices $A_1, \ldots, A_k$? Or is there some counterexample to this "principle", by which I mean that there is some $f$ such as above such that the inequality $f(A_1,\ldots,A_k) \geq 0$ holds for hermitian positive semidefinite matrices $A_1, \ldots, A_k$ which pairwise commute with each other, but not for all $k$-tuples of positive semidefinite hermitian matrices?
I think it is a reasonable thing to ask.
I found a counterexample... For example, one may consider
$$\DeclareMathOperator{\tr}{tr} f(A_1,A_2,A_3) = \tr(A_1 A_2 A_3) + \tr(A_1 A_3 A_2)$$
with $n = 2$ (i.e. the $A_i$ are hermitian $2$ by $2$ matrices). Then by randomly generating $1000$ random triples of hermitian positive definite matrices, I found that for that sample, $977$ satisfy the inequality $f(A_1,A_2,A_3) \geq 0$ but the other $23$ do not... So there goes that idea...
Here is a concrete/explicit counterexample.
$$A_1 = \left( \begin{array}{cc} 10 & -4 + 5i \\ -4 - 5i & 5 \end{array} \right),$$
$$A_1 = \left( \begin{array}{cc} 4 & -1 - 3i \\ -1 + 3i & 4 \end{array} \right),$$
$$A_1 = \left( \begin{array}{cc} 2 & 3 - 0.5i \\ 3 + 0.5i & 5 \end{array} \right),$$
then $A_1$, $A_2$ and $A_3$ are positive definite hermitian matrices but $f(A_1,A_2,A_3) = -71 < 0$.