Among parallelepipeds of equal volume, which one has the smallest sum of the squares of the three edges meeting at one vertex?
Among the parallelepipeds of a given volume $V$, it is sufficient to consider only the rectangular parallelepipeds (cuboids) in which any of the three edges meeting at a vertex is perpendicular to the plane defined by the other two. If, for example, in a parallelepiped $A B C D E F G H$, the edge $A E$ is not perpendicular to the plane defined by the other two edges $A B, A D$ meeting at $A$, then by transitioning to the parallelepiped $A B C D E^{\prime} F^{\prime} G^{\prime} H^{\prime}$, where the face $E^{\prime} F^{\prime} G^{\prime} H^{\prime}$ lies in the plane of $E F G H$ and $A E^{\prime}$ is perpendicular to $A B C D$, we arrive at a case that is more favorable for us, with the same volume as the original but a smaller sum of the squares of the edges, since $A E^{\prime}<A E$ and thus $A B^{2}+A D^{2}+A E^{\prime 2}<A B^{2}+A D^{2}+A E^{2}$.

Now, let the edges meeting at a vertex of a cuboid of a given volume $V$ be $a, b, c$, i.e., $a b c=V$, and write the inequality between the arithmetic and geometric means of the squares of the edges:

$$
\frac{a^{2}+b^{2}+c^{2}}{3} \geq \sqrt[3]{a^{2} b^{2} c^{2}}=\sqrt[3]{V^{2}}
$$

from which

$$
a^{2}+b^{2}+c^{2} \geq 3 \sqrt[3]{V^{2}}
$$

that is, the sum of the squares of the edges in question is not less than the constant on the right-hand side, and equality holds if and only if the numbers in question are equal: $a^{2}=b^{2}=c^{2}$, i.e., in terms of segments, if $a=b=c$. According to this, the sum of the squares of the edges is the smallest for the equilateral cuboid, i.e., the cube.

Ferenc Kolonits (Bp. VIII., Piarist g. III. o. t.)