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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 $0 \le p_1 < q_1 \le 1$.

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  • $\begingroup$ This inequality is a conjecture. $\endgroup$ – jII Oct 2 '18 at 14:39
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    $\begingroup$ Can you prove it for n=2? n=3? $\endgroup$ – marty cohen Oct 2 '18 at 15:57
  • $\begingroup$ For $n=2$ yes (the only solution to the resulting system would be $\mathbf{p} = \mathbf{q}$; the inequality holds for all other sequences). $\endgroup$ – jII Oct 2 '18 at 16:17
  • $\begingroup$ @martycohen I have appended the proof for $n=2$. Perhaps this suggests a proof by induction? Not sure how that would be established here. $\endgroup$ – jII Oct 2 '18 at 16:27

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