Chebyshev's sum inequality

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Chebyshev, after which the inequality is named

In mathematics, Chebyshev's sum inequality, named after Pafnuty Chebyshev, states that if

a1≥a2≥⋯≥an and b1≥b2≥⋯≥bn,

then

1n∑k=1nakbk≥(1n∑k=1nak)(1n∑k=1nbk).

In words, if we are given two sequences that are both non-increasing or non-decreasing, then the product of their averages is less than the average of their (termwise) product.

Similarly, if

a1≤a2≤⋯≤an and b1≥b2≥⋯≥bn,

then

1n∑k=1nakbk≤(1n∑k=1nak)(1n∑k=1nbk).[1]

Proof

Consider the sum

S=∑j=1n∑k=1n(aj−ak)(bj−bk).

The two sequences are non-increasing, therefore Template:Math and Template:Math have the same sign for any Template:Math. Hence Template:Math.

Opening the brackets, we deduce:

0≤2n∑j=1najbj−2∑j=1naj∑j=1nbj,

hence

1n∑j=1najbj≥(1n∑j=1naj)(1n∑j=1nbj).

An alternative proof is simply obtained with the rearrangement inequality, writing that

∑i=0n−1ai∑j=0n−1bj=∑i=0n−1∑j=0n−1aibj=∑i=0n−1∑k=0n−1aibi+kmodn=∑k=0n−1∑i=0n−1aibi+kmodn≤∑k=0n−1∑i=0n−1aibi=n∑iaibi.

Continuous version

There is also a continuous version of Chebyshev's sum inequality:

If f and g are real-valued, integrable functions over [a, b], both non-increasing or both non-decreasing, then

1b−a∫abf(x)g(x)dx≥(1b−a∫abf(x)dx)(1b−a∫abg(x)dx)

with the inequality reversed if one is non-increasing and the other is non-decreasing.

See also

Notes

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