AMGM(n) help

1. Nov 17, 2007

ehrenfest

1. The problem statement, all variables and given/known data
http://math.stanford.edu/~vakil/putnam07/07putnam1.pdf

Can someone help me with problem 8 part b?

I plugged that in to AMGM(n) and the LHS easily transforms into the LHS of AMGM(n-1) but I am not sure how the RHS transforms to the RHS of AMGM(n-1). Specifically, what happens to the exponent?

2. Relevant equations

3. The attempt at a solution

2. Nov 17, 2007

Kummer

This is the famous Cauchy proof of Am-Gm. Post what you dd so far.

3. Nov 17, 2007

ehrenfest

Again, I plugged in the hint to AMGM(n) and the LHS easily transforms into the LHS of AMGM(n-1).

So I have the LHS of AMGN(n-1) is greater than or equal to $$(\frac{a_1...a_{n-1} (a_1+...+a{n-1})}{(n-1)})^{1/n}$$

and I do not know where to go from there.

4. Nov 18, 2007

morphism

Work with
$$\left( \frac{\sum a_i}{n} \right)^n \geq \prod a_i$$

Keep an eye on the possibility of canceling something convenient off both sides after you apply the hint.

5. Nov 18, 2007

ehrenfest

After I apply the hint, it looks like this:

$$\left( \frac{\sum_{i=1}^{n-1} a_i}{n-1} \right)^n \geq 1/(n-1)\sum_{i=1}^{n-1} a_i \prod_{i=1}^{n-1} a_i$$

which is the same as

$$\left( \frac{\sum_{i=1}^{n-1} a_i}{n-1} \right)^{n-1} \geq \prod_{i=1}^{n-1} a_i$$

because everything is nonnegative.

On to part c).

Last edited: Nov 18, 2007
6. Nov 18, 2007

ehrenfest

And of course I got stuck on part c) as well. I tried doing something similar like letting a_i = a_i + a_{n+i} or a_i = a_i *a_{i+n} but that only works for one side at a time. I tried to do the case k=2 "by hand" but there is the (a_1 +a_2+a_3+a_4)^4 part that is prohibitive. And I tried other stuff that was just as bad. So I am out of ideas.

7. Nov 18, 2007

morphism

Here, let's assume k=2 holds and show k=4.

$$(a_1 a_2 b_1 b_2)^{1/4} = \left( (a_1 a_2)^{1/2} (b_1 b_2)^{1/2} \right)^{1/2} \leq \frac{(a_1 a_2)^{1/2} + (b_1 b_2)^{1/2}}{2}$$

Can you take it from here?