Footnotes

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I had hoped the intuitive argument had worked, but I guess not. Let

be the subset of numbers smaller than the average, and

be the subset larger. Let

be the subset of all numbers equal to the average.

and

are mutually exclusive (a number cannot be both greater, less than, or equal to the average). Call the average

, and let

be an element in

(similarly for

and

). The facts are

$\sum{(S_i - a_x)} < 0$ , true because each by definition
$\sum{(L_i - a_x)} > 0$ , true for similar reasons
$\sum{(E_i - a_x)} = 0$ , true because each by definition
$\sum{a_i} = E + L + S$ , since the sets are mutually exclusive and each number must fall into exactly one category
$\sum{a_i - a_x} = 0$ , shown earlier

Therefore,

$\displaystyle \sum{a_i} - \sum{a_x}$	$\displaystyle =$	$\displaystyle E + L + S - \sum{a_x}$
0	$\displaystyle =$	$\displaystyle \sum{(S_i - a_x)} + \sum{(L_i - a_x)} + \sum{(E_i - a_x)}$
0	$\displaystyle =$	$\displaystyle \sum{(S_i - a_x)} + \sum{(L_i - a_x)}$
$\displaystyle -\sum{(S_i - a_x)}$	$\displaystyle =$	$\displaystyle \sum{(L_i - a_x)}$

which is the result we sought to prove. Phew! Sorry for the roundabout proof, but you asked for it!

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