Chebyshev's Sum Inequality/Discrete/Proof

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.