Suppose I have two undirected graphs $G_1$ and $G_2$ with the same vertex set $V$ and let $A_1$ and $A_2$ denote their respective adjacency matrices. Define the intersection of the two graphs $G_\cap$ as the graph having adjacency matrix $A_\cap$ whose elements are $$ A^{i,j}_{\cap} \equiv A^{i,j}_1 A^{i,j}_2. $$
Similarly, define the union of the two graphs $G_\cup$ as the graph having adjacency matrix $A_\cup$ with elements $$ A^{i,j}_\cup \equiv A^{i,j}_1 + A^{i,j}_2 - A^{i,j}_1 A^{i,j}_2. $$
Let $L_1$, $L_2$, $L_{\hspace{0.025cm}\cap}$ and $L_{\hspace{0.025cm}\cup}$ be the combinatorial laplacian matrices associated with the graphs denoted by their subscripts. It turns out that $L_\cup = L_1 + L_2 - L_\cap$.
Define the “Gibbs state density matrix” at “inverse temperature” $\beta \in \mathbb{R}_{\geq 0}$ associated with a graph $G$ by $$ \rho_G(\beta) \equiv \frac{\exp(-\beta L_G)}{Z_G} \cdotp $$
where the normalizing partition function is $Z_G \equiv \mathrm{Tr}[\exp(-\beta L_G)]$. One can easily show that, in the operator sense, $L_{\hspace{0.025cm}\cup} \geq L_1, L_2 \geq L_{\hspace{0.025cm}\cap}$ and that this implies that the ordered eigenvalues of these laplacian matrices satisfy the following inequalities $$ \lambda^{\hspace{0.025cm}\cup}_i \geq \lambda^{(1)}_i \geq \lambda^\cap_i, \quad \lambda^{\hspace{0.025cm}\cup}_i \geq \lambda^{(2)}_i \geq \lambda^\cap_i, \quad i \in \{1,2,\dots,|V|\}. $$
I am having trouble proving the following (numerically-observed) inequality regarding the partition functions
$$ Z_\cup \geq \frac{Z_1 Z_2}{Z_1 + Z_2} \cdotp $$
Any ideas? I've tried using the Weyl (and dual Weyl) inequalities but have not been able to used to them to prove this inequality.