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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, $$trace(ABC)\leq 0$$

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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 $(u^Tv)(v^Tw)(w^Tu)=\left(-\frac12\right)^3<0$.

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