Chebyshev's Sum Inequality/Discrete

Theorem
Let $a_1, a_2, \ldots, a_n$ be real numbers such that:
 * $a_1 \ge a_2 \ge \cdots \ge a_n$

Let $b_1, b_2, \ldots, b_n$ be real numbers such that:
 * $b_1 \ge b_2 \ge \cdots \ge b_n$

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

Proof
We have 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:

The result follows.