Example 1 In the tetrahedron $A B C D$, $A D=D B=A C$ $=C B=1$, then its maximum volume is
Analysis: As shown in Figure 2, let $E$ and $F$ be the midpoints of $CD$ and $AB$ respectively, then it is easy to know that $AB \perp CD$, and $EF$ is the common perpendicular of $AB$ and $CD$. Let $AB = 2x, CD = 2y$. From the fact that $\triangle ACD$, $\triangle BCD$, and $\triangle EAB$ are all isosceles triangles, it is easy to get $EF = \sqrt{1 - x^2 - y^2} (x > 0, y > 0, x^2 + y^2 < 1)$. Therefore, by the basic conclusion 3, the volume of this tetrahedron is $V = \frac{1}{6} \cdot 2x \cdot 2y \cdot \sqrt{1 - x^2 - y^2}$. Using the arithmetic-geometric mean inequality in three variables, we can find that the maximum volume is $\frac{2 \sqrt{3}}{27}$.