Chebyshev's Sum Inequality/Discrete/Also presented as

Chebyshev's Sum Inequality (Discrete): Also presented as

Some sources present Chebyshev's sum inequality as:

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

where:

$a_1, a_2, \ldots, a_n$ are real numbers such that:

$a_1 \ge a_2 \ge \cdots \ge a_n$

$b_1, b_2, \ldots, b_n$ are real numbers such that:

$b_1 \ge b_2 \ge \cdots \ge b_n$

Proof

We have by hypothesis that the sequences $\sequence {a_k}$ and $\sequence {b_k}$ are both decreasing.

For $j, k \in \set {1, 2, \ldots, n}$, consider:

$\paren {a_j - a_k} \paren {b_j - b_k}$

Therefore $a_j - a_k$ and $b_j - b_k$ have the same sign for all $j, k \in \set {1, 2, \ldots, n}$.

So:

$\forall j, k \in \set {1, 2, \ldots, n}: \paren {a_j - a_k} \paren {b_j - b_k} \ge 0$

Hence:

$\ds 0$

$\le$

$\ds \sum_{j \mathop = 1}^n \sum_{k \mathop = 1}^n \paren {a_j - a_k} \paren {b_j - b_k}$

$\ds $

$=$

$\ds \sum_{j \mathop = 1}^n \sum_{k \mathop = 1}^n \paren {a_j b_j - a_k b_j - a_j b_k + a_k b_k}$

$\ds $

$=$

$\ds \sum_{j \mathop = 1}^n \sum_{k \mathop = 1}^n a_j b_j - \sum_{j \mathop = 1}^n \sum_{k \mathop = 1}^n a_k b_j - \sum_{j \mathop = 1}^n \sum_{k \mathop = 1}^n a_j b_k + \sum_{j \mathop = 1}^n \sum_{k \mathop = 1}^n a_k b_k$

$\ds $

$=$

$\ds n \sum_{j \mathop = 1}^n a_j b_j - \sum_{k \mathop = 1}^n a_k \sum_{j \mathop = 1}^n b_j - \sum_{j \mathop = 1}^n a_j \sum_{k \mathop = 1}^n b_k + n \sum_{k \mathop = 1}^n a_k b_k$

General Distributivity Theorem

$\ds $

$=$

$\ds 2 n \sum_{k \mathop = 1}^n a_k b_k - 2 \sum_{k \mathop = 1}^n a_k \sum_{k \mathop = 1}^n b_k$

renaming and gathering equal terms

$\ds $

$=$

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

dividing through by $2$

Then we may manipulate it into the required form:

$\ds $

$=$

$\ds \dfrac 1 n \sum_{k \mathop = 1}^n a_k b_k - \dfrac 1 {n^2} \sum_{k \mathop = 1}^n a_k \sum_{k \mathop = 1}^n b_k$

dividing through by $n^2$

$\ds \leadsto \ \ $

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

$\ge$

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

The result follows.

$\blacksquare$

Sources

1964: Milton Abramowitz and Irene A. Stegun: Handbook of Mathematical Functions ... (previous) ... (next): $3$: Elementary Analytic Methods: $3.2$ Inequalities: Chebyshev's Inequality: $3.2.7$
1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 36$: Inequalities: Chebyshev's Inequality: $36.11$
2009: Murray R. Spiegel, Seymour Lipschutz and John Liu: Mathematical Handbook of Formulas and Tables (3rd ed.) ... (previous) ... (next): $\S 37$: Inequalities: Chebyshev's Inequality: $37.11.$