Prove
Consider ordered real numbers $x_1 \le x_2 \le \dots \le x_n$ and $y_1 \le y_2 \le \dots \le y_n$.
Let $ \sigma : \{ 1,2,\dots,n\} \rightarrow \{ 1,2,\dots,n\}$ be a permutation on the integers $1,2,\dots,n$.
Show that $\sum\limits_{k=1}^n x_ky_{\sigma(k)} \le \sum\limits_{k=1}^n x_ky_k$.
My Intuition
Algebraically, to maximize a weighted sum, we want to put the heaviest weights $x_k$ on the heaviest values of $y_{\sigma(k)}$. The easiest way to do that is to order them in the same way. In other words,
$max\{ \sum\limits_{k=1}^n x_ky_{\sigma(k)} \} = \sum\limits_{k=1}^n x_ky_k$
Geometrically, the permutation of $y_{\sigma(k)}$ which satisfies this condition would be the one which is ordered in the same direction as $x_k$. To maximize the dot product, we maximize the cosine or minimize the angle between these two vectors. Since permutations would only change the direction of the vector, not its length, we pick the permutation which points in the same increasing direction as $x_k$.
I'm having trouble formalizing a proof for these observations.
Any hints on either a geometric or algebraic approach for a proof would be appreciated.