Chebyshev's Sum Inequality/Discrete/Equality

Theorem

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 equality in Chebyshev's Sum Inequality (Discrete), that is:

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

holds if and only if:

$\forall j, k \in \set {1, 2, \ldots, n}: a_j = a_k, b_j = b_k$

Proof

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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.

Sources

• 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): Chebyshev's inequality: 2.