# Proving a challenging inequality on finite summations

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

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