Is anyone aware of any general relationship (inequality for instance) between
$${\rm tr}\,(A^{-1}B)$$ and $${\rm tr}\,(B)/{\rm tr}\,(A)$$ where both matrices are positive semidefinite?
For instance, they are discussed as approximations in the paper: Trace Ratio vs. Ratio Trace for Dimensionality Reduction
But I could not find any proof or explanation of how they are related.
There is no relationship between the two, even in the simplest case where the $n\times n$ matrices $A$ and $B$ are simultaneously diagonalizable. In that case, we can write the expressions in terms of the eigenvalues $\lambda^{(A)}_{j}$ and $\lambda^{(B)}_{j}$, of which there are $n$ each (counting with multiplicity). $${\rm tr}\left(A^{-1}B\right)=\sum_{j=1}^{n}\frac{\lambda^{(B)}_{j}}{\lambda^{(A)}_{j}}.$$ In contrast, the ratio of the traces is $$\frac{{\rm tr}\,B}{{\rm tr}\,A}=\frac{\sum_{j=1}^{n}\lambda^{(B)}_{j}}{\sum_{j=1}^{n}\lambda^{(A)}_{j}}.$$ Note that by taking one of the $\lambda_{j}^{(A)}$ to $0$, the magnitude of ${\rm tr}\left(A^{-1}B\right)$ can be made arbitrarily large (and furthermore, the sign may be either positive or negative if the eigenvalues are not restricted by the question's demand that they be nonnegative), while the other expression is minimally affected. Furthermore, interchanging the eigenvalues associated with two eigenvector leaves the ratio of traces unchanged but can have a drastic effect on ${\rm tr}\left(A^{-1}B\right)$.
