# Arithmetic-geometric mean inequality

1. Sep 20, 2004

### Ed Quanta

The arithmetic-geometric mean inequality is

a1...an<=[(a1+...+an)/n]^n where all of the a terms (a1,a2,etc) are non-negative real numbers. How do I go about proving this is true for 2^n terms? Thanks.

2. Sep 21, 2004

### TenaliRaman

There are many proofs of AM-GM inequality,
the smallest one i know comes from rearrangement inequality.
(Rearrangement inequality itself is easily proved through induction).

if u wish u can try to prove AM - GM yourself using rearrangement inequality (its pretty easy) else post back and i will sketch a small proof of it.

Another way to prove is using chebyshev's inequality which again can be proved from rearrangement inequality.

Another simpler way of proving it is using jensen inequality but proving jensen inequality requires some handy work.

Ofcourse u can prove AM-GM ground up using induction but really this is the most tiring proofs of all.

-- AI

3. Sep 21, 2004

### matt grime

since it is specifically for the case of 2^r terms (you have n meaning two different things in the same sentence), then some kind of induction ought to work since you can split the sum of 2^r terms into two smaller sums of 2^{r-1} terms and use the am-gm formula by induction on everything in sight.

4. Sep 21, 2004

### Gokul43201

Staff Emeritus
I spent a little time trying this last night...I tired rapidly and quit.

5. Sep 28, 2004