Arithmetic and Geometric MeansDate: 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/ |
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