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.

AM-GM_am_Halbkreis.svgUsually, 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!

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