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 be real numbers, then

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 So our inequality becomes

Our task has become an optimization problem. We want to maximize on the compact set This process is quite straightforward. Functions in question:

We can compute

and

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

at a maximum.

That implies and therefore due to the constraint Clearly, the inequality is true for and thus for all choices of We’re done – simple as that!

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

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