GM ≦ AM Trick

"General Manager smaller than Assistant Manager"

Put 27 cubes (Length =a, Width=b, Height =c), inside a big Cube of side (a+b+c), can't fill the big Cube completely.
$latex 27abc leq (a+b+c)^{3} $
$latex sqrt [3] {abc} leq frac {a+b+c}{3}$
Generalized:
GM ≦AM
$latex sqrt [n] {x_1 x_2 x_3 dots x_n} leq frac {x_1+x_2+x_3...+xn}{n}$

GM = Geometric Mean
AM = Arithmetic Mean

$latex mbox {Application}$
$latex mbox {Prove:} : 1003^{2005} > 2005!$
Solution: use 'Promote' technique to general case:
$latex sqrt[n] {1.23...n} < frac {1+2...+n}{n}$
Let n=2005
$latex sqrt[2005] {1.23...2005} < frac {1+2...+2005}{2005}$
$latex sqrt[2005] {2005!} < frac {(2005)(2006)/2}{2005}$
$latex sqrt[2005] {2005!} <1003$
$latex 1003^{2005} > 2005! $ [QED]