# Prove the A-G-M Inequality using Lagrange multipliers.

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?

• 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? – farawaybeach Mar 13 '13 at 17:59

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 $c^{1/n}$.