Inequality in tetrahedron You are given a tetrahedron $ABCD$. $ACB = ADB = 90^\circ$. $AC = CD = DB$. Prove that $AB < 2 * CD$.
I know that $AD = CB$ and $CBD = DCB = ADC = CAD$.
 A: Let $\theta$ be the angle of smallest measure in $\triangle ACB$; note that the angle of smallest measure in $\triangle ADB$ must have the same measure.  Assume that $AC,DB$ are the sides opposite this angle of smallest measure in each of these triangles, since otherwise the claim $AB \lt 2\cdot CD$ will be true automatically.  Since $AC=DB=CD$, the smallest possible value of $\theta$ is $\pi\over 6$, which occurs when $\triangle ABC$ is coplanar with $\triangle ABD$ (which is not a tetrahedron), and the largest possible value of $\theta$ is $\pi\over 4$, since it is one of the angles in a right triangle.
At $\theta=\frac \pi 6$, we have $AB=2\cdot CD$, and at $\theta=\frac \pi4$ we have $AB=\sqrt 2\cdot CD$.  It remains to show that the ratio between $AB$ and $CD$ is decreasing as $\theta$ goes from $\pi\over 6$ to $\pi\over 4$.
First, note that $CD$ lies in a plane $\ell$ such that the midpoint of $CD$ is the point in $\ell$ which is closest to $AB$.  This means that $AC$ has smallest possible length relative to $AB$ when $\theta=\frac\pi 6$, and $AC$ increases in length relative to $AB$ as $\theta$ increases.  But $AC=CD$, so $2\cdot CD \gt AB$ whenever $\theta\gt\frac\pi 6$.
