Arithmetic/Geometric Mean Inequality TheoremDate: 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/ |
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