# Inequality of the means

\begin{question} Is is possible to pack $n^n$ rectangular $n$-dimensional boxes each of which has side lengths $a_1,a_2,\ldots,a_n$ inside an $n$-dimensional cube with side length $a_1 + a_2 + \ldots a_n$? \end{question}

Taking the arithmetic/geometric mean inequality \[ (a_1 a_2 \ldots a_n)^{1/n} \le \frac{a_1 + a_2 + \ldots a_n}{n} \] multiplying both sides by $n$ and then raising both sides to the $n^{th}$ power yields: \[ n^n \cdot a_1 a_2 \ldots a_n \le (a_1 + a_2 + \ldots a_n)^{n}.\] So, in the above question, the volume of the cube is at least the sum of the volumes of the rectangular boxes. Furthermore, a positive solution to this question would yield a strengthening of the arithmetic/geometric mean inequality.

For $n=1$ the problem is trivial, for $n=2$ it is immediate, and for $n=3$ it is tricky, but possible. It is also known that a solution for dimensions $n$ and $m$ can be combined to yield a solution for dimension $nm$. Thus, the question has a positive answer whenever $n$ has the form $2^a 3^b$. It is open for all other values.

See \href[Bar-Natan's page]{http://www.math.toronto.edu/~drorbn/projects/ArithGeom/} for more.

## Bibliography

[BCG] E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways for Your Mathematical Plays, Academic Press, New York 1983.

* indicates original appearance(s) of problem.