Assume $ CD>1$ and let $ AB=x$. Let $ P,Q,R$ be the feet of perpendicular from $ C$ to $ AB$ and $ \triangle ABD$ and from $ D$ to $ AB$, respectively. Suppose $ BP>PA$. We have that $ CP=\sqrt{CB^2-BT^2}\le\sqrt{1-\frac{x^2}4}$, $ CQ\le CP\le\sqrt{1-\frac{x^2}4}$. We also have $ DQ^2\le\sqrt{1-\frac{x^2}4}$. So the volume of the tetrahedron is $ \frac13\left(\frac12\cdot AB\cdot DR\right)CQ\le\frac{x}6\left(1-\frac{x^2}4\right)$. We want to prove that this value is at most $ \frac18$, which is equivalent to $ (1-x)(3-x-x^2)\ge0$. This is true because $ 0<x\le 1$.