Proving that $\max(x_1, x_2, x_3) = x_1 + x_2 + x_3 - \min(x_1, x_2) - \min(x_1, x_3) - \min(x_2, x_3) + \min(x_1, x_2, x_3)$ $$\max(x_1, x_2, x_3) = x_1 + x_2 + x_3 - \min(x_1, x_2) - \min(x_1, x_3) - \min(x_2, x_3) + \min(x_1, x_2, x_3)$$
Is there a more elegant proof to this than just trying out all the possibilities and showing that it works?  For example, assuming that the numbers are distinct, this statement is true if
$x_1 < x_2 < x_3$ since the equation would yield $x_1 + x_2 + x_3 - x_1 - x_1 - x_2 + x_1 = x_3$, etc.
It also seems to work if the numbers are not distinct, but then there are even more combinations to try.  Also, interestingly, it looks like inclusion-exclusion but that might be a red herring.  Also not sure how to tag this question...
 A: It is essentially inclusion-exclusion.  $x_1+x_2+x_3$ counts each number once.  $-\min(x_1, x_2)-\min(x_1,x_3)-\min(x_2,x_3)$ removes the smallest element twice and the second smallest number once.  This has now under-counted the smallest number (and removed the second smallest).  Then $+\min(x_1, x_2, x_3)$ puts back the under-counted smallest number.
A: It really is inclusion/exclusion.  For each number $x_i$, define $X_i$ to be the interval $[0,x_i]$.  Then for any set $I$ of indices, $|\cup_{i\in I} X_i|=\max_{i\in I} x_i$ and $|\cap_{i\in I}X_i|=\min_{i\in I}x_i$. So inclusion/exclusion translates directly to the formula in the problem.
A: For $x_1,\ldots,x_n$ distinct,
$$
\max\{x_1,\ldots,x_n\}=\sum\limits_{i=1}^{n}x_i\prod\limits_{j\ne i}(1-{\bf 1}_{x_j>x_i})
$$

For completeness, more generally,
$$
\max\{x_1,\ldots,x_n\}=\sum\limits_{i=1}^{n}x_i{\bf 1}_{x_i\geqslant x_{j\ne i}}-\sum\limits_{i<j}x_i{\bf 1}_{x_i\geqslant x_{k\ne i,j}, x_i=x_j}+\ldots +(-)^{n+1}x_1{\bf 1}_{x_1=\ldots=x_n}
$$
A: A formula that might be of use to you is the following:
$$
\max(a,b) = \frac{a + b + |b-a|}{2}
$$
If you don't want to do case by case, you can apply this formula repeatedly, although it is true that the algebra does get voluminous. 
