## Task 4 - 191234

Investigate whether among all tetrahedra $ABCD$ with a given volume $V$ and right angles $\angle BDC, \angle CDA, \angle ADB$, there exists one with the smallest possible sum $AB + AC + AD + BC + BD + CD$.

If this is the case, determine (in dependence on $V$) this smallest possible sum.
Due to the right angles, according to the Pythagorean theorem, we have \( |AB| = \sqrt{|AD|^2 + |BD|^2}, |AC| = \sqrt{|AD|^2 + |CD|^2} \) and \( |BC| = \sqrt{|BD|^2 + |CD|^2} \).

For any positive real numbers \( x \) and \( y \), it holds that \( \frac{x^2 + y^2 - 2xy}{2} = \frac{(x - y)^2}{2} \geq 0 \), thus

\[
x^2 + y^2 \geq \frac{x^2 + y^2 + 2xy}{2} = \frac{(x + y)^2}{2} \quad \text{and therefore} \quad \sqrt{x^2 + y^2} \geq \frac{\sqrt{2}}{2} \cdot (x + y)
\]

where equality holds only for \( x = y \).

Substituting this, we obtain

\[
\begin{gathered}
|AB| + |AC| + |AD| + |BC| + |BD| + |CD| \geq \\
\geq \frac{\sqrt{2}}{2} \cdot (|AD| + |BD|) + \frac{\sqrt{2}}{2} \cdot (|AD| + |CD|) + |AD| + \frac{\sqrt{2}}{2} \cdot (|BD| + |CD|) + |BD| + |CD| = \\
= (|AD| + |BD| + |CD|) \cdot \left(1 + \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right) = (|AD| + |BD| + |CD|) \cdot (1 + \sqrt{2})
\end{gathered}
\]

where equality holds only for \( |AD| = |BD| = |CD| \).

The volume \( V \) of the tetrahedron \( ABCD \) is determined by the right angles at \( D \) as \( V = \frac{1}{6} \cdot |AD| \cdot |BD| \cdot |CD| \), so \( \sqrt[3]{|AD| \cdot |BD| \cdot |CD|} = \sqrt[3]{6V} \) follows.

For any positive real numbers \( x, y \) and \( z \), the inequality between the geometric and arithmetic mean holds: \( \sqrt[3]{xyz} \leq \frac{x + y + z}{3} \), where equality holds only for \( x = y = z \).

Thus,

\[
|AD| + |BD| + |CD| = 3 \cdot \frac{|AD| + |BD| + |CD|}{3} \geq 3 \cdot \sqrt[3]{6V}
\]

and therefore

\[
|AB| + |AC| + |AD| + |BC| + |BD| + |CD| \geq 3 \cdot (1 + \sqrt{2}) \cdot \sqrt[3]{6V}
\]

where equality holds only for \( |AD| = |BD| = |CD| \). Thus, the term on the right side is exactly the smallest sum sought.