1. Which is Greater, inequality?

(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

2. Originally Posted by Ideasman
(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
If x_i>=0 then the first one is greater.

That is the AM-GM inequality.

3. 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

4. hopefully that makes sense even tho latex is bust. if not ill pdf it for you

5. Originally Posted by chogo
hopefully that makes sense even tho latex is bust. if not ill pdf it for you
That no makes no sense to me.

It seems you made this inequality look too easy.
The standard method is by max/minimizing the multivarible function. Or by Lagrange multipliers.

However! The cleanest proof uses the best inequality (the Cauchy-Swartzh Inequality).

6. i thought that was the standard induction proof. Yeah i liked the cauchy proof (the one u did in the previous thread for n=2). Though i would try this tho

7. If you have the proof in PDF (since LateX is down) I'd appreciate it if you could post it. I just need a simple explanation.

Thanks.

8. Originally Posted by ThePerfectHacker
If x_i>=0 then the first one is greater.

That is the AM-GM inequality.
x_i?

9. If you have access to a reasonable mathematics library, you may find AN INTRODUCTION TO INEQUALITIES by Beckenbach & Bellman very useful. Thy have a rather complete discussion of this topic in that book.

10. 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?