Chebyshev's Sum Inequality/Discrete/Equality

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 equality in Chebyshev's Sum Inequality (Discrete), that is:
 * $\ds \dfrac 1 n \sum_{k \mathop = 1}^n a_k b_k = \paren {\dfrac 1 n \sum_{k \mathop = 1}^n a_k} \paren {\dfrac 1 n \sum_{k \mathop = 1}^n b_k}$

holds :
 * $\forall j, k \in \set {1, 2, \ldots, n}: a_j = a_k, b_j = b_k$