6
$\begingroup$

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:

Example of 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 $a+b+c$.

$\endgroup$

1 Answer 1

3
$\begingroup$

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 $x_1 = \dots = x_n$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .