Consider a weighted connected graph $G=(V,E,W)$ with n vertices such that all edge weights are some positive numbers. The minimum edge weight in $G$ is some $\epsilon>0$. Let $L$ be its laplacian matrix, which has n orthonormal eigenvectors $v_1, v_2, \cdots, v_n$ with eigenvalues $0=\mu_1 \leq \mu_2\leq \cdots \leq \mu_n$. Define $$L^{-1}=\sum_{i=2}^{n} \frac{v_i v_i^T}{\mu_i}.$$ Note $\mu_2>0$ since $G$ is connected.
For any vertex u and v, we define unit vectors $e_u=(0,0,\cdots,1,0\cdots) \in R^{V}$ and $e_v=(0,0,\cdots,1,0\cdots) \in R^{V}$. I want to prove the quadratic form inequality $$(e_u-e_v)^TL^{-}(e_u-e_v)\leq \frac{n}{\epsilon}$$
To prove this, I expand the formula of $L^-$ given above, and obtain $$ \begin{split} (e_u-e_v)^T L^{-}(e_u-e_v) &= (e_u-e_v)^T \left(\sum_{i=2}^{n} \frac{1}{\mu_i} v_i v_i^T \right) (e_u-e_v) \\ &= \sum_{i=2}^{n} \frac{1}{\mu_i}\left<e_u-e_v,v_i\right>^{2} \\ &\leq \frac{2}{\mu_2}, \end{split} $$ since $\sum_i \left<e_u-e_v,v_i\right>^2=2$. To finish the proof, I just need to show $$\mu_2 \geq \frac{2\epsilon}{n}$$ I know how people can use test vector to get an upper bound for $\mu_2$, but I'm not sure how to prove this lower bound. Any thoughts are welcomed!\
New idea: In this problem we have $$tr(L)=\sum_{i=2} \mu_i \geq n\epsilon$$ This might help us relate $\epsilon$ to $\mu_2$.