The operators in this question are in Hilbert space, but I'm going to word it as a discrete spectrum and a finite dimensional (linear algebra) answer is fine by me.
Operators $A$ and $\Delta$ are bounded, self-adjoint and non-commutative, and $B\equiv e^{- A} \Delta e^{ A}$. Some of the eigenvalues of $A$ or $\Delta$ may be negative. I want to prove that the $B \Delta$ is positive semi-definite and that the eigenvalue $0$ has the same multiplicity in $B\Delta$ as it does in $\Delta$ (I'm not sure that this second part is true).
$B$ is a similarity transformation, so the eigenvalues, $\Delta_i$, of $\Delta$ are the same as those of $B$, but the eigenvectors are different, so the eigenvalues of $B\Delta$ are not $\Delta_i^2$. The $e^A$ form (positive eigenvalues) is important, since the product of general similar operators doesn't have a non-negative spectrum in general.
I'm tempted to expand in bases of the eigenvectors $|x\rangle = \sum_i b_i |B_i\rangle = \sum_j d_j |\Delta_j\rangle$, $|B_i\rangle = \sum_j c_{i,j} |\Delta_j\rangle \implies d_j = \sum_i b_ic_{i,j}$, to find
$ \begin{aligned} \langle x | B\Delta | x \rangle & = \sum_{i,j} b_i^* d_j \langle B_i | B\Delta | \Delta_j \rangle = \sum_{i,j} b_i^* d_j \Delta_i \Delta_j \langle B_i| \Delta_j \rangle =\sum_{i,j,k} b_i^* d_j c_{i,k}^* \Delta_i\Delta_j \delta_{kj} \\ & = \sum_{i,j} b_i^* d_j c_{i,j}^* \Delta_i\Delta_j \end{aligned} $
which doesn't help much, because I haven't used that $e^A$ is positive. If I expand in the basis of eigenvectors of $A$ and $\Delta$, I get a similar but slightly more complicated expression, and I think there must be a simpler approach.