Chebyshev Sum Inequality

If

$$a_1 \geq a_2 \geq \ldots \geq a_n$$

$$b_1 \geq b_2 \geq \ldots \geq b_n,$$

then

$$n \sum_{k=1}^n a_kb_k \geq \left({\,\sum_{k=1}^n a_k}\right)\left({\,\sum_{k=1}^n b_k}\right).$$

This is true for any distribution.

See also Cauchy Inequality, Hölder Sum Inequality

References

Gradshteyn, I. S. and Ryzhik, I. M. Tables of Integrals, Series, and Products, 5th ed. San Diego, CA: Academic Press, p. 1092, 1979.

Hardy, G. H.; Littlewood, J. E.; and Pólya, G. Inequalities, 2nd ed. Cambridge, England: Cambridge University Press, pp. 43-44, 1988.