If $\prod x_k\neq0$ and $\sum\frac{x_k}{x_{k+1}}=0$, then $|\sum x_kx_{k+1}|\le (\max|x_k|- \min|x_k|)\sum x_k$

Solution 1:

Let $Q$ be the product of $\max_{1\le k\le n}|x_{k}|$ and $\min_{1\le k\le n}|x_{k}|$, that is: $$Q := \max_{1\le k\le n}|x_{k}|\cdot\min_{1\le k\le n}|x_{k}|\,.$$

We can how represent each product of two neighboring (cyclically) numbers like $$ x_1 x_2 = Q\dfrac{x_1}{x_2} + x_1(x_2 - \dfrac{Q}{x_2})\,.$$

Then we have $$ x_1 x_2 + x_2 x_3 + \cdots + x_n x_1 = x_1(x_2 - \dfrac{Q}{x_2}) + x_2(x_3 - \dfrac{Q}{x_3}) + \cdots + x_n(x_1 - \dfrac{Q}{x_1}) $$ since $n$ terms of the type $Q\dfrac{x_{k-1}}{x_k}$ sum to zero due to the equality condition on our numbers. On the other hand, for each $l = 1\ldots n$ we have: $$ | x_l - \dfrac{Q}{x_l} | \le \max_{1\le k\le n}|x_{k}| - \min_{1\le k\le n}|x_{k}|\,,$$ that implies: $$ | x_{l-1}(x_l - \dfrac{Q}{x_l}) | \le |x_{l-1}|\Bigg(\max_{1\le k\le n}|x_{k}| - \min_{1\le k\le n}|x_{k}|\Bigg)\,.$$ When we sum all $n$ such inequalities, we’ll virtually obtain the inequality that has to be demonstrated (up to applying the triangle inequality for the absolute value function at the left-hand side).