Given two positive definite matrices $A,B$, I seek convenient ways to analytically compute or estimate $\frac{\operatorname{Tr}(AB)}{\operatorname{Tr}(A)}$, where $\operatorname{Tr}$ denotes the trace.

I know essentially two things (beyond the representation in terms of the entries). First I know trivial bounds: the ratio is between $\lambda_{min}(B)$ and $\lambda_{max}(B)$. Second, I know that these are not at all tight. The following example quickly illustrates the point:

$$A=\begin{bmatrix} \epsilon & 0 \\ 0 & 1 \end{bmatrix} \\ B=\begin{bmatrix} 1 & 0 \\ 0 & \epsilon \end{bmatrix} \\ AB=\begin{bmatrix} \epsilon & 0 \\ 0 & \epsilon \end{bmatrix}$$

Thus $\frac{\operatorname{Tr}(AB)}{\operatorname{Tr}(A)}=\frac{2\epsilon}{1+\epsilon} \to 0$ as $\epsilon \to 0$. Yet $\lambda_{max}(B)=1$ for all $\epsilon$.

What happened in this example always happens if $A,B$ are simultaneously diagonalizable. Namely, if they share eigenvectors $v_1,\dots,v_n$, the corresponding eigenvalues of $A$ are $\lambda_1,\dots,\lambda_n$, and the corresponding eigenvalues of $B$ are $\eta_1,\dots,\eta_n$, then the ratio is $\frac{\sum_{i=1}^n \lambda_i \eta_i}{\sum_{i=1}^n \lambda_i}$. So the ratio in the simultaneously diagonalizable case is an average of the eigenvalues of $B$, but with completely arbitrary weights (determined by the eigenvalues of $A$).

Is there any similar way to think about the case when $A,B$ are not simultaneously diagonalizable? In particular, the "trivial bounds" above imply that the ratio is a weighted average of the eigenvalues of $B$. Is there any way to estimate the weights?

If it helps, you may assume that the Cholesky factorizations $A=C^T C,B=D^T D$ are given.

Some relevant literature: http://ieeexplore.ieee.org/document/1695991/ http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.636.8021&rep=rep1&type=pdf etc.; search "trace of product of symmetric matrices" for more such papers. Some of these papers have relevant results but I have found them difficult to apply in my own context.