Chebyshev's Sum Inequality/Continuous

Theorem

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:

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

Proof

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Sources

• 1 February 2015: Adam Besenyei: Picard's weighty proof of Chebyshev's sum inequality (MAA Mathematics Magazine Vol. 88, no. 1: pp. 1 – 6)