Arithmetic/Geometric Mean Inequality Theorem

Date: 04/15/2001 at 21:54:36
From: Jeff
Subject: AM-GM Inequality Theorem

Can you give a proof of the AM-GM (arithmetic mean - geometric mean) 
inequality theorem? That is, prove that:

   (x1 + x2 + x3 + ... + xn)/n >= (x1 * x2 * x3 * ... * xn)^(1/n)

for any n nonnegative integers x1, x2, x3, ..., xn.

This theorem has come up a lot in problems I've studied recently, and 
I've searched for a proof but have been unsuccessful.

Thanks,
- Jeff

Date: 04/16/2001 at 07:02:36
From: Doctor Floor
Subject: Re: AM-GM Inequality Theorem

Hi, Jeff,

Thanks for writing.

Let functions f and g from R_{+0}^n to R (where R_{+0} represents the
nonnegative real numbers, and R the real numbers) be given by:

     f(X) = f(x1, x2, ..., xn) = x1 * x2 * ... * xn
and
     g(X) = g(x1, x2, ..., xn) = x1 + x2 + ... + xn - n

We assume g(X) = 0, and with that condition we look for a value 
A = (a1, a2, ..., an) in such a way that f(A) is a maximum. 
Of course for all j, Aj <> 0.

We can apply the Lagrange Multiplier theorem, which tells us here that 
the vectors

     ( part(dg/dx1) , ... , part(dg/dxn) ) = (1,1,...,1)
and
     ( part(df/dx1) , ... , part(df/dxn) ) = (f(A)/a1, ..., f(A)/an)

are linearly dependent, from which we see that:

     1/a1 = 1/a2 = ... = 1/an

and thus

     a1 = a2 = ... = an

Since we assumed that g(X) = 0 and thus g(A) = 0, we see that

     a1 = a2 = ... = an = 1

so that consequently

     a1 * a2 * ... * an = 1

We conclude that in general if x1 + x2 + ... + xn = n, then 
x1 * x2 * ... * xn <= 1, because we had taken A to be a maximum.

Now if:

     x1 + x2 + ... + xn = S

and thus

     (n/S)x1 + (n/S)x2 + ... + (n/S)xn = n

we conclude that

     (n/S)^n * x1 * x2 * ... * xn <= 1

               x1 * x2 * ... * xn <= (S/n)^n

       (x1 * x2 * ... * xn)^(1/n) <= S/n

as desired.

For some explanation of Lagrange Multipliers, you might visit the Dr. 
Math archives:

   Lagrange Multipliers
   http://mathforum.org/dr.math/problems/aleja1.8.98.html   

If you have more questions, write us back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/