# Lagrange Multipliers & AM-GM

As you’ve possibly realized in my last article (check it out if you don’t know the theorem), I’m a fan of Lagrange Multipliers! I remember when my professor announced that he’d now teach us Lagrange Multipliers. That made me immensely happy.

The goal in this text is to prove the AM-GM inequality.

Theorem (AM-GM). Let $a_1, a_2, \dots, a_n \geq 0$ be real numbers, then

$\displaystyle \frac{a_1+a_2+\dots+a_n}{n} \geq \sqrt[n]{a_1 a_2 \dots a_n}.$

The left and right sides are called arithmetic and geometric mean respectively.

I’d dare to say that this is the most well-known inequality (besides perhaps the trivial inequality) in the world of math olympiads. AM-GM and Cauchy – any mathlete should know these.

The case n = 2 has a nice geometric interpretation.

Usually, AM-GM is proven by Cauchy Induction. Fancy people also like to apply Jensen. We want to be even fancier and use Lagrange Multipliers.

The key idea is to notice that the inequality is homogenous. Hence, we may de-homogenize it and WLOG let $a_1 + a_2 + \dots + a_n = n.$ So our inequality becomes

$\displaystyle 1 \geq a_1 a_2 \dots a_n \text{ with constraint } a_1 + a_2 + \dots + a_n = n.$

Our task has become an optimization problem. We want to maximize $a_1 \dots a_n$ on the compact set $S = \{ (a_1, \dots, a_n) \in \mathbb{R}^n : a_1 + \dots + a_n = n \}.$ This process is quite straightforward. Functions in question:

$f, g : \mathbb{R}^n \to \mathbb{R}, \ f(a_1, \dots, a_n) = a_1 a_2 \dots a_n, \quad g(a_1, \dots, a_n) = a_1 + \dots + a_n - n.$

We can compute

$\displaystyle \nabla f(a_1, \dots, a_n) = \left(\frac{a_1 \dots a_n}{a_1}, \dots, \frac{a_1 \dots a_n}{a_n} \right)$  and  $\nabla g(a_1, \dots, a_n) = (1, 1, \dots, 1).$

As the constraint set $S$ is compact, a global maximum exists, so a Lagrange Multiplier exists. Evidently, $a_1 \dots a_n = 0$ if any of $a_1, \dots, a_n$ is 0. That’s clearly a minimum case. So let all of the numbers be non-zero. Hence

$a_1 a_2 \dots a_n = \lambda a_1, \ a_1 a_2 \dots a_n = \lambda a_2, \dots, a_1 a_2 \dots a_n = \lambda a_n$

at a maximum.

That implies $a_1 = \dots = a_n$ and therefore $a_1 = \dots = a_n = 1$ due to the constraint $a_1 + \dots + a_n = n.$ Clearly, the inequality $1 \geq a_1 \dots a_n$ is true for $(a_1, \dots, a_n) = (1, \dots, 1)$ and thus for all choices of $(a_1, \dots, a_n).$ We’re done – simple as that!

## One thought on “Lagrange Multipliers & AM-GM”

1. cd says:

S is not compact. You need to restrict to the positive orthant and check the boundaries. Works otherwise.

Like