Inequality AM/HM <= (AM/GM)^n with packing problem interpretation I have stumbled upon the following inequality while exploring "Hoffman's packing problem", which I'm pretty convinced is true, but unable to prove. Let $n \geq 2$ be a natural number and let $x_1, \dots, x_n$ be positive real numbers. How do I prove that:
$$
\frac{AM}{HM} \leq \left(\frac{AM}{GM}\right)^n,
$$
where the harmonic mean $HM$, geometric mean $GM$ and arithmetic mean $AM$ are defined as:
$$
HM = \frac{n}{\sum_{i=1}^{n}{\frac{1}{x_i}}}, \quad
GM = \sqrt[n]{\prod_{i=1}^{n}{x_i}} \quad
\text{and} \quad
AM = \frac{1}{n}\sum_{i=1}^{n}{x_i}.
$$
What I know so far
We have equality if $n=2$ or if $x_1 = \dots = x_n$.
I haven't been able to prove the inequality in general, but for $n=3$ (let's use $x_1,x_2,x_3 = a,b,c$ for clarity) we obtain:
$$
\frac{\frac{a+b+c}{3}}{\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}}
\leq
\left(\frac{\frac{a+b+c}{3}}{\sqrt[n]{abc}}\right)^3
\Leftrightarrow\\
\frac{(a+b+c) \cdot (bc + ac + ab)}{9abc} \leq \frac{(a+b+c)^3}{27abc}
\Leftrightarrow\\
3(bc + ac + ab) \leq (a+b+c)^2 \quad (*)
\Leftrightarrow\\
0 \leq a^2 + b^2 + c^2 - bc - ac - ab
\Leftrightarrow\\
0 \leq 2a^2 + 2b^2 + 2c^2 - 2bc - 2ac - 2ab
\Leftrightarrow\\
0 \leq (a - b)^2 + (b - c)^2 + (c - a)^2,
$$
which is clearly true.
Interpretation as packing problem
The inequality can be interpreted as an $n-1$ dimensional packing problem. For instance if $n=3$ the inequality can be rearranged as $3(bc + ac + ab) \leq (a+b+c)^2$, see (*) above. We can look at this like fitting three $a \times b$, three $b \times c$ and three $c \times a$ rectangles inside a square with side-length $a+b+c$. Here is an example of a packing:

Caption: Suppose $a < b < c$. This is a packing of 3 red $a \times b$, 3 blue $b \times c$ and 3 green $c \times a$ rectangles inside a square with side-length $a+b+c$.
 A: For $n=4$ the inequality expands to
$$ 
16 (abc+abd+acd+bcd) \le (a+b+c+d)^3
$$
which is solved in https://artofproblemsolving.com/community/c6h494463
using Maclaurin's inequality.

The same approach works for arbitrary $n \ge 2$:
$$ \tag{*}
\frac{AM}{HM} \leq \left(\frac{AM}{GM}\right)^n
$$
is equivalent to
$$
 \left(\frac 1n\sum_{i=1}^{n}{\frac{1}{x_i}}\right) \left(\prod_{i=1}^{n}{x_i} \right)\le \left(\frac{\sum_{i=1}^{n}{x_i}}{n}\right)^{n-1}
$$
which can be written as
$$
 \frac{\sum\limits_{1 \le i_1  < i_2 < \cdots < i_{n-1}\le n}x_{i_1}x_{i_2}\cdots x_{i_{n-1}}}{n} \le \left( \frac{\sum_{i=1}^{n}{x_i}}{n}\right)^{n-1} \, .
$$
With the "averages" $S_k$ defined as
$$
 S_k = \frac{\sum\limits_{1 \le i_1  < i_2 < \cdots < i_{k}\le n}x_{i_1}x_{i_2}\cdots x_{i_k}}{\binom{n}{k}} 
\quad (1 \le k \le n)
$$
the inequality becomes
$$  \tag{**}
 S_{n-1} \le S_1^{n-1}
$$
Maclaurin's inequality
states that
$$
S_1 \ge S_2^{1/2} \ge \dots \ge S_{n-1}^{1/(n-1)}  \ge S_n^{1/n}
$$
with equality if and only if all $x_i$ are equal.
It follows that $(**)$ – and therefore the given inequality $(*)$ –
is true, and equality holds if and only if $n=2$ or $x_1 = \dots = x_n$.
