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Importance: High ✭✭✭

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Keywords:	arithmetic mean
	geometric mean
	Inequality
	packing

Recomm. for undergrads: no

Posted	by:	mdevos
	on:	April 6th, 2009

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 $a_1 + a_2 + \ldots a_n$ ?

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 Bar-Natan's page 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.

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