Suppose that we have two real and positive definite $n \times n$ matrices $B_1$ and $B_2$ and that $A$ is an arbitrary real $n \times n$ matrix. Running some numerical tests by generating random matrices it seems that we have the inequality $$\text{tr}(B_1^{-1} B_2) \geq \text{tr}((A^\text{T} B_1 A)^{+} A^\text{T} B_2 A)$$ with equality if $A$ is invertible. Here, tr denotes the trace, T matrix transposition, and $+$ the Moore-Penrose pseudo-inverse.
The equality part is straightforward using the inverse of a product and the cyclic invariance of the trace, but I don't know how to show the inequality part. Intuitively, I think it should hold, since we are essentially projecting $B_1$ and $B_2$ onto the image of $A$. Any pointers would be greatly appreciated!
Proof.
Let $Z := A^\top B_1 A$. Note that $Z^2$ is PSD.
First, we prove that, for all $t > 0$, $$\mathrm{tr}(B_1^{-1}B_2) > \mathrm{tr}\Big((Z^2 + tI)^{-1}Z \cdot A^\top B_2 A\Big). \tag{1}$$ (The proof is given at the end.)
Second, we use the property of the Moore-Penrose inverse $$Y^{+} = \lim_{t\to 0^{+}} (Y^\mathsf{T} Y + tI)^{-1} Y^\mathsf{T}.$$ Taking limit of (1), we have $$\mathrm{tr}(B_1^{-1}B_2) \ge \mathrm{tr}\Big((A^\top B_1 A)^{+} A^\top B_2 A\Big). $$
We are done.
$\phantom{2}$
Proof of (1).
(1) is written as $$\mathrm{tr}(B_1^{-1}B_2) > \mathrm{tr}\Big((Z^2 + tI)^{-1}Z A^\top B_1 \cdot B_1^{-1}B_2A\Big) $$ or $$\mathrm{tr}(B_1^{-1}B_2) > \mathrm{tr}\Big((XA + tI)^{-1}X \cdot B_1^{-1}B_2A\Big) \tag{A1} $$ where $X := A^\top B_1 AA^\top B_1$.
From $(XA + tI) X = XAX + tX = X(AX + tI)$, we have $(XA + tI)^{-1}X = X(AX + tI)^{-1}$. (A1) is written as $$\mathrm{tr}(B_1^{-1}B_2) > \mathrm{tr}\Big(X(AX + tI)^{-1} \cdot B_1^{-1}B_2A\Big) $$ or $$\mathrm{tr}(B_1^{-1}B_2) > \mathrm{tr}\Big(AX(AX + tI)^{-1} \cdot B_1^{-1}B_2\Big) $$ or $$\mathrm{tr}(B_1^{-1}B_2) > \mathrm{tr}\Big(B_1^{-1}B_2\Big) - \mathrm{tr}\Big(t(AX + tI)^{-1}\cdot B_1^{-1}B_2\Big) $$ or $$\mathrm{tr}(B_1^{-1}B_2) > \mathrm{tr}\Big(B_1^{-1}B_2\Big) - \mathrm{tr}\Big(t(AA^\top B_1 AA^\top B_1 + tI)^{-1}\cdot B_1^{-1}B_2\Big) $$ or $$\mathrm{tr}\Big(t(AA^\top B_1 AA^\top B_1 + tI)^{-1}\cdot B_1^{-1}B_2\Big) > 0$$ or $$\mathrm{tr}\Big(t(B_1AA^\top B_1 AA^\top B_1 + tB_1)^{-1}B_2\Big) > 0$$ which is true, using the fact that $\mathrm{tr}(YZ) > 0$ for two PD matrices $Y, Z$.
We are done.