Elegant Proof of a simple inequality I'm looking for an elegant proof of the following identity: for $w_1,w_2,z_1,z_2\ge 0$, 
$w_1w_2+z_1z_2\le \max\{z_1,w_1\}\max\{z_2,w_2\}+\min\{z_1,w_1\}\min\{z_2,w_2\}$
The proof I currently have involves checking by cases, which is what I'd like to avoid. If there's some linear algebraic, combinatorial, or geometric way to explain this inequality intuitively, I'd love to hear it.
Thanks in advance for your time. 
 A: Here's another approach:
You have a bag full of two type of coins; one type has value $z_1$ and the other $w_1$. You are allowed to pick $w_2$ coins of one type, and $z_2$ coins of another. If you want to get the maximum possible amount, how should you go about picking the coins? (Greedy algorithm)
A: 
Consider the min-max case (blue) compared to the "mixed" case (red). We can match out the red to blue as shown and we are left with the excess shown for the min-max case.
A: Surely you can just attack the algebra head on? Using the identities $max(a,b) = \frac{1}{2}(a+b+|a-b|)$ and $min(a,b) = \frac{1}{2}(a+b-|a-b|)$ eliminates the need to check cases.
A: There are a couple of nice non-algebraic approaches; here's an algebraic approach that isn't too bad. Without loss of generality $w_1\ge z_1$; the question is whether we can increase the value of the expression $w_1w_2+z_1z_2$ by interchanging $w_2$ and $z_2$. We have
$$\begin{align*}
(w_1z_2+w_2z_1)-(w_1w_2+z_1z_2)&=w_1(z_2-w_2)-z_1(z_2-w_2)\\
&=(w_1-z_1)(z_2-w_2)\;,
\end{align*}$$
so switching $w_2$ and $z_2$ increases the value if and only if $w_1>z_1$ and $z_2>w_2$. (Recall the assumption that $w_1\ge z_1$.) And this is exactly what we wanted to show.
(It's essentially a proof that the greedy algorithm works.)
A: The sum of the sizes of two sets is the size of their union plus the size of their intersection.  Let one set be the rectangle with diagonally opposite vertices $(0,0)$ and $(w_1,w_2).$  Let the other be the rectangle with diagonally opposite vertices $(0,0)$ and $(z_1,z_2).$  Let size be area.  The product $\max\{w_1,z_1\}\max\{w_2,z_2\}$ is the area of the bounding rectangle, and therefore greater than or equal to the area of the union.  The product $\min\{w_1,z_1\}\min\{w_2,z_2\}$ is the area of the intersection.
A: This is a special case of what's sometimes called the "rearrangement inequality."  A simple, intuitive inductive proof can be found on the Art of Problem Solving wiki.
