Trace of product of positive matrices Let $A$, $B$ symmetric matrices over $\mathbb{R}$ with the same dimension. If $A$ has only positive eigenvalues and $B$ has only nonnegative eigenvalues, is $\text{trace}(AB)\ge 0?$
If yes, prove it. If no, counterexample it.

Yes.  In fact all eigenvalues of $A B$ are nonnegative.  This is because $A^{1/2} B A^{1/2}$ is positive semidefinite, where 
$A^{1/2}$ is the positive definite square root of $A$, and
$A B = A^{1/2} (A^{1/2} B)$ and $(A^{1/2} B) A^{1/2}$ have the same
eigenvalues (the products of two matrices in either order always have the same nonzero eigenvalues).
