A weighted Chebyshev’s order inequality

Wildon's Weblog 2018-08-02

Let $w_i \in \mathbb{R}^{\ge 0}$ be non-negative weights and let $a_1, \ldots, a_n \in \mathbb{R}^{\ge 0}$ , $b_1, \ldots, b_n \in \mathbb{R}^{\ge 0}$ be non-negative numbers such that

$\begin{aligned} &a_1 \ge a_2 \ge \ldots \ge a_n \\ & b_1 \le b_2 \le \ldots \le b_n \end{aligned}.$

Then

$\begin{aligned} \bigl( \sum_{i} w_i \bigr) \bigl(\sum_{j}w_ja_jb_j \bigr) &= \frac{1}{2} \sum_{ij} w_iw_j (a_jb_j + a_ib_i) \\ &\le \frac{1}{2} \sum_{ij} w_iw_j (a_jb_i + a_ib_j) \\ &= \bigl( \sum_i w_ia_i\bigr) \bigl( \sum_j w_ja_j \bigr)\end{aligned}$

where the middle step follows from the inequality $(a_jb_i+a_ib_j) - (a_jb_j + a_ib_i) = (a_j-a_i)(b_i-b_j) \ge 0$ . In particular, if $\sum_i w_i = 1$ then we obtain

$\sum_k w_i a_i b_i \le \bigl( \sum_i w_i a_i \bigr)\bigl(\sum_i w_i b_i\bigr).$

This is a weighted version of Chebyshev’s order inequality. The weighted Chebychev’s order inequality appears (with a different proof) in an interesting article in MAA Mathematics Magazine, where it is used to deduce Jensen’s inequality.