Q:

Prove the arithmetic-geometric mean inequality by maximizing the product for a fixed sum

I have seen proofs for the AM-GM inequality. However, I am asked to prove
\begin{equation*}
(x_{1}x_{2}...x_{n})^\frac{1}{n}\leq\frac{x_{1}+x_{2}+...+x_{n}}{n}
\end{equation*}
And $x_i > 0$ by "maximizing $x_{1}x_{2}...x_{n}$ on a compact subset of $\left\{ x = (x_1, x_2, ... x_n) : x_i > 0 \text{ and } \sum_{i = 1}^{n}x_i = k \right\}$"
On a high level, I guess I need to show for any $k > 0$, I only need to consider some compact subset, and on this subset the product/geometric mean reaches a max, which is less than $\frac{k}{n}$. But I have no idea how to pick this subset

Edit: using Lagrange multipliers,
$$f(x_1, ... x_n) = x_{1}...x_n$$
$$g(x_1, ... x_n) = x_1 + ... x_n - k = 0$$
Each partial must equal the constant $\lambda$:
$$\frac{\partial f}{\partial x_i} = x_{1}...x_{i - 1}x_{i + 1}...x_n = \frac{\partial g}{\partial x_i} = \lambda$$
Edit: bad math Taking the quotient of any two partials, using $x_1$ and $x_2$ for example, I get $\frac{x_1}{x_2} = \frac{\lambda}{x_{3}...x_n} = \frac{x_2}{x_1}$. So all the $x_i$ are equal, so they're $\frac{k}{n}$. Is there a better way to argue that the $x_i$ are equal?
Should be: $\lambda=\frac{1}{x_i}\prod x_k$ for every $i$ so setting the products equal you can cancel to get $x_i = x_j$ for every $i$ and $j$ (courtesy of Stella Biderman)

A:

You’re slightly misinterpreting the question. It’s saying to approach the problem like this:

Fix $k\in\mathbb{R}^+$ and $n\in\mathbb{N}$. Let $S_k\leq\mathbb{R}^n$ be the subspace that satisfies $\sum x_i=k$. Define $f(x)=(x_1x_2\cdots x_n)^{1/n}$. Prove that $f(x)\leq k/n$ for all $x\in S_k$ and, in particular,
$$\sup_{x\in S_k}f(x)=\frac{k}{n}$$

To prove this, you’re best bet is to prove that the value that maximizes $f$ is $x_i=k/n$ for all $i\in[n]$. This can be done with Lagrange multipliers, but it can also be done by a symmetry argument and some basic properties of the gradient function.

