If $\operatorname{Tr}(H)\ge 0$, is $\operatorname{Tr}(PH)\ge 0?\ $ $H$ is Hermitian, $P$ is positive definite. Old: Given $\operatorname{Tr}(DH)\ge 0$ for positive real diagonal $D$ and Hermitian $H$, is $\operatorname{Tr}(H)\ge 0$?
Let $D$ be positive real diagonal and $H$ be Hermitian such that $H$ may have negative eigenvalues. Both matrices are of course of the same size $n\times n$. 
Given that $\operatorname{Tr}(DH)\ge 0. \ $ Is $\operatorname{Tr}(H)\ge 0$?

Post-edit:
Thanks @Aweygan for providing a counter-example to the initial question above.
Let's look at it the other way round by letting $\operatorname{Tr}(H)\ge 0$. $H$ is $n\times n$ Hermitian, which may have negative eigenvalue(s).


*

*$D$ be positive real diagonal. Is $\operatorname{Tr}(DH)\ge 0$?

*Let us generalize the question for when $n\times n$ complex matrix $P$ is positive definite. Is $\operatorname{Tr}(PH)\ge 0$?


I have edited the title to accommodate the changes made.

Not necessarily.  Consider the following example:
$$D=\begin{pmatrix}2&0\\0&1\end{pmatrix},\qquad H=\begin{pmatrix}2&0\\0&-3\end{pmatrix}.$$
