# sign of trace of product of 3 Hermitian matrices

Suppose $A$ and $B$ and $C$ are three Hermitian matrices, $A$ and $B$ are positive-definite, and all eigenvalues of $C$ are negative. $A$ and $B$ and $C$ don't commute. Can we say, $$trace(ABC)\leq 0$$

No. $$\pmatrix{2&1\\ 1&1}\pmatrix{1&-3\\ -3&10}\pmatrix{-8\\ &-1}=\pmatrix{8&-4\\ 16&-7}.$$
Edit. Here is another way to think about this problem. Suppose the hypothesis is always true. Then by a limiting argument, the hypothesis is also true when $$A,B$$ and $$-C$$ are rank-one positive semidefinite matrices. It follows that $$(u^\ast v)(v^\ast w)(w^\ast u) =-\operatorname{trace}\left(uu^\ast vv^\ast (-ww^\ast)\right)\ge0\tag{1}$$ for all nonzero vectors $$u,v$$ and $$w$$. However, when the vectors are complex, the actually isn't any reason why the LHS of $$(1)$$ is real in the first place. Even if the vectors are real, it is easy to construct a counterexample to $$(1)$$: let $$u,v,w$$ be three unit vectors lying on $$\mathbb R^2\times0$$ and angularly separated from each other by $$120^\circ$$, then $$(u^Tv)(v^Tw)(w^Tu)=\left(-\frac12\right)^3<0$$.