Question 126
The diagonals of the faces of the rectangular parallelepiped $ABCD'A'B'C'D'$ satisfy the relation:
$$
CA^2 = \frac{2 \cdot B'A^2 \cdot D'A^2}{B'A^2 + D'A^2}.
$$
Prove that $AC' \le \sqrt{3} \cdot AA'$.
$$
CA^2 = \frac{2 \cdot B'A^2 \cdot D'A^2}{B'A^2 + D'A^2}.
$$
Prove that $AC' \le \sqrt{3} \cdot AA'$.