A Lower Bound of the Number of Threshold Functions in Terms of Combinatorial Flags on the Boolean Cube

Let $ E={ (1, b_1, \ldots , b_n)\in R^{n+1} \mid ; b_i= \pm 1 ,; i=1, \ldots, n }$,$E^{\times n}{\ne 0} := { W=(w{i_1}, \ldots , w_{i_n}) \mid w_{i_k}\in E, , k=1, \ldots, n, , dim , span(w_{i_1}, \ldots , w_{i_n}) = n },$and$q^W_l := |span(w_{i_{n-l+1}}, \ldots , w_{i_n}) \cap E|.$ Then for any weights $p=(p_1,\dots ,p_{2^n})$, $p_i\in R$, ${\sum }_{i=1}^{2^n}{p}_i=1$ we have for the number of threshold functions $P(2,n)$ the following lower bound $$P(2,n)\ge 2\sum _{W\in E_{\mathrm{\ne }0}^{\times n}}\frac{1-p_{i_1}-p_{i_2}-\cdots -p_{i_{q_n^W}}}{q_n^W\cdot q_{n-1}^W\cdots q_1^W},$$ and the right side of the inequality doesn’t depend on the choice of $p$. Here the indices used in the numerator correspond to vectors from $span(w_{i_1}, \ldots , w_{i_n})\cap E = \left{w_{i_1}, \ldots, w_{i_n}, \ldots w_{i_{q_n^W}}\right}$.