Valid AM-GM inequality proof?

Question

We have to prove that $$\frac{(x_1+x_2+x_3+...+x_n)}{n} \geq (x_1\cdot x_2\cdot x_3\cdots x_n)^{1/n}$$

Attempt: Raising both sides to the nth power gives $\left(\frac{x_1+x_2+x_3+...x_n}{n}\right)^{n} \geq x_1x_2x_3...x_n$ This is equivalent to proving that if a sum of random numbers is a fixed number, then their maximum product occurs when all the number are equal. So we have to find which combination of numbers gives the maximum product if their sum is fixed.

Obviously $$\left({x_1+x_2+x_3+...x_n}\right)^{n} > x_1x_2x_3...x_n$$

then, a maximizer must exist.

Now suppose that we have this combination of products: $x_1x_2x_3...x_n$ If i take the arithmetic mean of any 2 numbers and replace them by their arithmetic mean, I will get a bigger product (By the simple 2 case AM-GM inequality which is easy to proof), obviously the arthimetic mean of 2 numbers will not change the sum of the numbers, we can write that:

$$\frac{x_1+x_2}{2}\cdot\frac{x_1+x_2}{2}\cdot x_3 \cdots x_n\ \geq x_1x_2x_3\cdots x_n$$

Which means if we have any combination of product of numbers, I could increase that product by making any 2 numbers equal (arthimetic mean) without changing the sum of the numbers.

Now supposing that the maximum product does not have all of the numbers equal, then I can increase its product by having 2 numbers equal, so this mean that it's not the maximum product, then, by contradiction, the maximum product must occur when all the numbers are equal (Sum divided by the number of numbers)

Which means that $$\left(\frac{x_1+x_2+x_3+...x_n}{n}\right)^{n} \geq x_1x_2x_3...x_n$$

Is this a valid proof? It's my first time here and I don't know how to do the math notations thingy on this website so forgive me, thanks.

Answer 1

Hint: Your idea is correct. Here is a north to let things rigorous.

Consider $f, \psi : U \to \mathbb R$ defined as $$ f(x) = x_1 \cdot x_2 \cdots x_n \,\,\, \text{and} \,\,\, \psi (x) = x_1 + x_2 + \ldots + x_n$$

for all $x = (x_1, x_2, \ldots , x_n) \in U$. Fix $s > 0$ and search for the critical points of $f|_M$ where $M = \psi^{-1}(s)$.

Observe that $\mathrm{grad} \, \psi (x) = (1,1,\ldots, 1)$ for any $x \in U$ and $\mathrm {grad}\, f (x) = (\alpha_1 , \ldots , \alpha_n)$, with $a_i = \prod_{j\neq i} x_j$. Show that the maximum is necessarily in $M$ and is the only critical point of $f|_M$.

Answer 2

That's pretty good.

One aspect I'd want to fiddle with is the existence of the maximizer. You argued that it's bounded (which as a commenter noted could be made more explicit) and so a maximizer exists; this requires some kind of compactness argument, and the continuity of $\mathbb R^n\to\mathbb R$, $(x_1,\dotsc,x_n)\to\prod_{k=1}^n x_k$. That's all fine, but it would be better to make it explicit.

If you don't want to use topological concepts like that, then the argument can be adjusted not to assume the existence of the maximizer, but to prove it. The idea is to repeatedly replace pairs of $x_i$ until they're all the same, and argue (as you did) that we can do this keeping the AM constant but increasing the GM. The only tricky bit is that, if you replace pairs of numbers with their average, this process might not ever stop. (Example: $(1,2,2)\to (\frac32,\frac32,2)\to (\frac32,\frac74,\frac74) \to\dotsm$.) So the replacement step has to be tweaked. The resulting proof can be found in a note by Dijkstra, EWD1140.