Let $\mathbf{p} := (p_1,\dots,p_n)$ and $\mathbf{q} := (q_1,\dots,q_n)$ be two finite sequences of real numbers such that:
$$0 \le p_i \le 1, \sum_{i=1}^{n}p_i = 1,$$ $$0 \le q_i \le 1, \sum_{i=1}^{n}q_i = 1.$$
Assume that $q_1 > p_1$ and for all $k$ $(1 < k < n)$, we have $q_k - p_k \ge q_{k+1} - p_{k+1}$.
Using these facts and assuming without loss of generality that $|q_1 - p_1| > |p_n - q_n|$, how would you establish that:
$$\sum_{i=1}^n\sum_{j=1}^{i-1}q_jp_i - \sum_{i=1}^{n}\sum_{j=1}^{i-1}p_jq_i > (q_1-p_1)^2?$$
For the case that $n=2$, subtracting the right-hand-side from the left-hand-side gives:
$$ \begin{align} q_1p_2 - p_1q_2 -(q_1-p_1)^2 &= q_1(1-p_1)-p_1(1-q_1)-(q_1-p_1)^2\\ &= q_1-q_1p_1-p_1+p_1q_1-(q_1-p_1)^2\\ &= (q_1-p_1)-(q_1-p_1)^2\\ &> 0 \end{align} $$ where the final inequality follows from $0 \le p_1 < q_1 \le 1$.