<s>[INST] In tetrahedron ABCD, prove that $r<\frac{AB\cdot CD}{2AB+2CD}$ 
Given tetrahedron $ABCD$, let $r$ be the radius of the sphere inscribed tetrahedron. Prove that $$r<\dfrac{AB\cdot CD}{2\,AB+2\,CD}$$

This is difficult question, my friends and I could not find any hint to solve it.
In the case that $ABCD$ is a regular tetrahedron, the inequality is easy to prove.  Let $\ell$ be the side length of $ABCD$.  Then, if $M$ is the centroid of $\triangle ABC$ and $I$ is the center of the insphere of the tetrahedron, then
we have
$$MA=\frac{1}{\sqrt{3}}\,\ell\,,$$
and
$$DM=\sqrt{\frac{2}{3}}\,\ell\,.$$
Since $r=DI$, we have
$$AI=DI=DM-DI=\sqrt{\frac{2}{3}}\,\ell-r\,.$$
Hence, by the Pythagorean Theorem,
$$r^2=IM^2=AI^2-MA^2=\left(\sqrt{\frac{2}{3}}\,\ell-r\right)^2-\frac{1}{3}\,\ell^2\,.$$
So
$$r=\frac{1}{2\sqrt{6}}\,\ell<\frac{1}{4}\,\ell=\frac{AB\cdot CD}{2\,AB+2\,CD}\,.$$
 [/INST] Let $MN$ be the common perpendicular line of $AB$ and $CD$ with $M$ a point in $AB$ and $N$ in $CD$. The volume $V$ of the tetrahedron $ABCD$ can be calculated in two ways:
$$\begin{align}
 V &= \frac16 \times AB \times CD \times MN \times \sin(\alpha) \\
   &= \frac13 \times r \times (S_{\triangle ABC} + S_{\triangle BCD} + S_{\triangle CDA} + S_{\triangle DAB})
\end{align}$$
where $\alpha$ is the angle between $AB$ and $CD$.
Notice that 
$$ S_{\triangle ABC} = \frac12 \times AB \times MC > \frac12 \times AB \times MN, $$
and this is true for all other $3$ triangles. Thus 
$$\begin{align}
 V &= \frac13 \times r \times (S_{\triangle ABC} + S_{\triangle BCD} + S_{\triangle CDA} + S_{\triangle DAB}) \\
   &> \frac16 \times r \times (AB \times MN + CD \times MN + CD \times MN + AB \times MN) \\
&= \frac13 \times r \times MN \times (AB  + CD)
\end{align}$$
It follows that
$$\begin{align}
 r \times MN \times (AB  + CD) &< 3V \\
   &= \frac12 \times AB \times CD \times MN \times \sin(\alpha) \\
&\leqslant \frac12 \times AB \times CD \times MN
\end{align}$$
So finally 
$$ r < \frac{AB \times CD}{2(AB  + CD)}.$$
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