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I’m trying to prove the Arithmetic-Geometric-Mean Inequality (A-G-M) using Lagrange multipliers. For positive real numbers $ x_{1},x_{2},\ldots,x_{n} $, we want to show that $$ (x_{1} x_{2} \cdots x_{n})^{1/n} \leq \frac{x_{1} + x_{2} + \cdots + x_{n}}{n}. $$ Consider the function $ f(x_{1},x_{2},\ldots,x_{n}) = x_{1} + x_{2} + \cdots + x_{n} $ subject to the constraint $$ x_{1} x_{2} \cdots x_{n} = c, $$ where $ c $ is a constant.

So I’m using Lagrange multipliers to solve this. I get $ \dfrac{1}{n} = \dfrac{\lambda c}{x_{i}} $ for all $ 1 \leq i \leq n $. Then $ \dfrac{x_{i}}{n} = \lambda c $. Then the sum of all the $ \dfrac{x_{i}}{n} $-terms yields $ x_{i} = \lambda cn $. I’m not sure where to go from here, or if I’ve made a mistake somewhere. Any tips?

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  • $\begingroup$ So that proves the AGM the case of equality, yes? I'm not clear on where the inequality comes from. When xi does not equal xj? $\endgroup$ – farawaybeach Mar 13 '13 at 17:59
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Your result shows that all the $x_i$ are equal to $n \lambda c$, so their product is $(n \lambda c)^n$.

If this equals $c$, $c = (n \lambda c)^n$ so $\lambda = c^{-1+1/n}/n$ and the sum of the $x_i$ is $n^2 \lambda c =n^2 c (c^{-1+1/n}/n) =n c^{1/n} $ and each $x_i$ is $c^{1/n} $.

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