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pr.likelihood – How can I show Chebyshev’s sum inequality with probabilistic strategies?

I’d love to show Chebyshev’s sum inequality, which states that:

If $a_1geq a_2geq cdots geq a_n$ and $b_1geq b_2geq cdots geq b_n$, then

$$

frac{1}{n}sum_{okay=1}^n a_kb_kgeq left(frac{1}{n}sum_{okay=1}^n a_kright)left(frac{1}{n}sum_{okay=1}^n b_kright)

$$

I’m close with the non-probabilistic proof, however I necessity a probabilistic one.

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