3. Given a rectangular parallelepiped. The perimeters of each of its three mutually perpendicular faces are equal to the sides of a new rectangular parallelepiped. What can be the minimum ratio of the volume of the new parallelepiped to the volume of the original?

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# Solution

Let $x, y, z$ be the sides of the original parallelepiped. Then the volume of the new one is

$$
V_{2}=2 \cdot(x+y) \cdot 2 \cdot(y+z) \cdot 2 \cdot(z+x)
$$

The desired ratio of volumes is

$$
\begin{gathered}
\frac{V_{2}}{V_{1}}=\frac{8(x+y)(y+z)(z+x)}{x y z}=\frac{8\left(x y+y^{2}+x z+y z\right)(z+x)}{x y z}= \\
=\frac{8\left(2 x y z+y^{2} z+x z^{2}+y z^{2}+x^{2} y+x y^{2}+x^{2} z\right)}{x y z}= \\
=8\left(2+\frac{y}{x}+\frac{x}{y}+\frac{z}{y}+\frac{y}{z}+\frac{z}{x}+\frac{x}{z}\right)
\end{gathered}
$$

We use the fact that $\frac{a}{b}+\frac{b}{a} \geq 2$, thus we get

$$
\frac{V_{2}}{V_{1}} \geq 8(2+2+2+2)=64
$$

Answer: 64.