Let $G(V,E)$ be a connected simple graph, where $V$ and $E$ denote respectively its vertex and the edge set respectively. Let $f: V\to \{-1,1\}$ a function mapping each vertex to a value in $\{-1,1\}$. 

Let $T(E_T,V)$ be a uniformly generated random spanning tree of $G$, where $E_T\subseteq E$ denotes its edge set. 
Given any vertex $v\in V$, we denote by $D_T(v)$ its degree in $T$ and by $S_T(v)$ the sum of $f(v')$ over all vertices $v'$ adjacent to $v$ in $T$. 

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**Question:** How can we prove (or disprove) that, for all connected simple graphs $G(V, E)$, all vertices $v\in V$, and all functions $f:V\to\{1,-1\}$, we have 
$$\mathbb{E}[S_T(v)]\cdot\mathbb{E}\left[\frac{S_T(v)}{D_T(v)}\right]\ge 0\,,$$
where the expectation is taken over the generation of the random spanning tree $T$ of $G$?

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***Observations:*** I think this concentration result might be helpful. Let $X_1, X_2, \ldots, X_{|E|}$ be the Bernoulli random variables corresponding to the edges of $G$, where for each $i\in\{1,2,\ldots, |E|\}$ we have $X_i=1$ iff the $i$-th edge of $G$ is included in $T$. Note that $\{X_i\}_{i=1}^{|E|}$ are negatively correlated as well as the random variables $\{1-X_i\}_{i=1}^{|E|}$. Let $p:=\frac{1}{|E|}\sum_{i=1}^{|E|}p_i$. Using a Chernoff's bound for negatively correlated random variables we have 
$$\mathbb{P}\left[\sum_{i=1}^{|E|} X_i<p|E|-\lambda\right]\le\exp\left(-\frac{\lambda^2}{2p|E|}\right)\,,$$
and therefore we also have
$$\mathbb{P}\left[\sum_{i=1}^{|E|} X_i>p|E|+\lambda\right]\le\exp\left(-\frac{\lambda^2}{2p|E|}\right)\,.$$