(x_1 + x_2 + ... + x_n)/n *** (x_1*x_2*...*x_n)^(1/n)
Fill in *** with >, <, >=, <= and please explain how you came to that conclusion.
Thanks
the ph has already explained the proof for this in a previous thread. Ill re explain it for you.
Arithmetic mean: $\displaystyle AM(a_1,....,a_n) = \frac{a_1 + .... + a_n}{n} $
Geometric mean: [tex] GM(a_1,.....a_n) = \sqrt[n]{a_1,....,a_n}
Proof by induction:
without loss of generality we may resclae $\displaystyle a_i$ so that $\displaystyle a_1.....a_n = 1$ . If all $\displaystyle a_i= 1$ the proof is trivial.
Thus assume atleast one $\displaystyle a_i > 1$ and one $\displaystyle a_i < 1$. We assume $\displaystyle a_1 > 1, a_2 < 1$. By inductive assumption we have
$\displaystyle \frac{a_1a_2 + a_3 + ......+ a_n}{n-1} \geq \sqrt[n-1]{(a_1 a_2) a3.....a_n = 1 $
thus we have
$\displaystyle a_1a_2+a_3+....a_n \geq n-1$
we need to show
$\displaystyle a_1 + a_2+a_3+....a_n \geq n$
this would follow if $\displaystyle a_1 + a_2 - (a_1a_2 + 1) \geq 0$ BUT
$\displaystyle a_1 + a_2 - (a_1a_2 + 1) = (a_1 - 1)(1-a_2) \geq 1$,
prooving what you have writen
http://www.math.princeton.edu/mathla...eanGeoMean.pdf
thats the one. PH what do you think? is that wrong? have you got a link for the maximisation proof?