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Arithmetic and Geometric Means


Date: 11/05/2000 at 15:34:52
From: Rocio Romero
Subject: Lagrange multipliers

I have to prove that the geometric mean is equal to or less than the 
arithmetic mean for all x, y, and z using Lagrange multipliers.

     Geometric mean:   G(x,y,z) = (xyz)^(1/3)
     Arithmetic mean:  A(x,y,z) = (x+y+z)/3

where x, y and z are positive numbers.

First, I took the third power of G to make it easier to differentiate, 
and I got the following gradients:

     Gradient of G = (yz,xz,xy)
     Gradient of A = (1/3,1/3,1/3)

Then I got the following system of equations:

     yz = lambda 1/3
     xz = lambda 1/3
     xy = lambda 1/3
     (x+y+z)/3 = 0 

And this is where I am confused, because the answer to this system is 
x = y = z = 0 (the only way that the fraction will be 0). Given that 
x, y and z are positive, should I use a generic value S on the right 
side of my fourth equation? Help!


Date: 11/05/2000 at 19:34:30
From: Doctor Anthony
Subject: Re: Lagrange multipliers

If you consider the maximum value of (xyz)^(1/3), subject to the 
condition that x+y+z = N where N is a positive constant, you will find 
the solution is at the point (N/3, N/3, N/3) so that both arithmetic 
and geometric means are equal at N/3. At all other values of x, y and 
z satisfying the condition that the sum is N,

     (xyz)^(1/3) < N/3

It follows that the arithmetic mean is greater than (or equal to) the 
geometric mean.

     f(x,y,z) = (xyz)^(1/3)
     g(x,y,z) = x + y + z - N = 0

The Lagrange equation is

     phi(x,y,z) = f(x,y,z) - k*g(x,y,z)

where k is the 'multiplier'

     phi(x,y,z) = (xyz)^(1/3) - k(x + y + z - N)

Then with partial differentials we have

     d[phi]/dx = (1/3)x^(-2/3)*(yz)^(1/3) - k  = 0   .......[1]
     d[phi]/dy = (1/3)y^(-2/3)*(xz)^(1/3) - k  = 0   .......[2]
     d[phi]/dz = (1/3)z^(-2/3)*(xy)^(1/3) - k  = 0   .......[3]
     g(x,y,z)  = x + y + z - N                 = 0   .......[4]

From [1], [2] and [3] the symmetry of the expressions require

     x = y = z

and then from [4]

     3x = N
      x = N/3

and the required result (as described above) follows.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Calculus
College Number Theory

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