\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ge \frac{1}{x} \sum_{p\in{\cal P}}\left(\sum_{i=1}^x |E_i^{p}| - 1\right)^2 \hspace{-2mm}= \frac{n(n-1-x)^2}{x}  \ \ \ \ \ \ \ \ \ \ \ \ (1) 