# Cauchy’s proof of the Arithmetic/Geometric mean inequality

March 31, 2010 Leave a comment

The Arithmetic/Geometry mean inequality, often called the AM-GM inequality , states that:

There are quite a few proof of this, but there is a really interesting one due to Cauchy. It uses induction, but induction that goes backwards as well as forwards.

Define *P(n)* to mean that the AM-GM inequality holds for arbitrary . In this notation, the stages of the proof are:

- Show
*P(2)*holds - Show that if
*P(n)*holds, then*P(2n)*holds. - Show that if
*P(n)*holds, then*P(n-1)*holds.

Of course, this will show that *P(n)* holds for all *n*. It is an interesting way of going about it though. And the great thing is, proving each of these sub-cases is quite easy!

*P(2)* is simple, as the true inequality can be expanded, *4xy* added to each side and factorised to give , which is what we want.

Suppose that *P(n)* is true. Then *P(2n)* easily follows by just splitting it into two products of *n* numbers:

Between the first and second line *P(n)* has been applied, and between the third and forth lines *P(2)* is applied.

Finally, we must show that if *P(n)* holds, then *P(n-1)* also holds. The obvious way to go about this is to take the *n-1* numbers and add a new number to the end of it to make *n* numbers, and hope we can choose a number which doesn’t change anything. It’s not too obvious what this number should be, but let’s call it *A*:

We want to somehow be able to factor the *A* back out so that we have:

A value of *A* which does this nicely is:

As then we have:

Hence *P(n-1)* and so we are done.