Chebyshev's Sum Inequality

Theorem

Discrete Version

Let $a_1, a_2, \ldots, a_n$ be real numbers such that:

$a_1 \ge a_2 \ge \cdots \ge a_n$

Let $b_1, b_2, \ldots, b_n$ be real numbers such that:

$b_1 \ge b_2 \ge \cdots \ge b_n$

Then:

$\ds \dfrac 1 n \sum_{k \mathop = 1}^n a_k b_k \ge \paren {\dfrac 1 n \sum_{k \mathop = 1}^n a_k} \paren {\dfrac 1 n \sum_{k \mathop = 1}^n b_k}$

Continuous Version

Let $u, v: \closedint 0 1 \to \R$ be integrable functions.

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Let $u$ and $v$ both be either increasing or decreasing.

Then:

$\ds \paren {\int_0^1 u \rd x} \cdot \paren {\int_0^1 v \rd x} \le \int_0^1 u v\rd x$

Also known as

Chebyshev's Sum Inequality is also known as Chebyshev's Inequality.

However, some sources use this name to mean the Bienaymé-Chebyshev Inequality, which is a completely different result.

Hence, to avoid confusion, the name Chebyshev's Inequality is not used on $\mathsf{Pr} \infty \mathsf{fWiki}$.

Source of Name

This entry was named for Pafnuty Lvovich Chebyshev.