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
$$r=\frac{1}{2\sqrt{6}}\,\ell<\frac{1}{4}\,\ell=\frac{AB\cdot CD}{2\,AB+2\,CD}\,.$$
 A: It is clear when $\vec{AB} \parallel \vec{CD}$, the tetrahedron $ABCD$ is degenerate and $r = 0$, the inequality will be trivially true. 
Let us consider the case $\vec{AB} \not\parallel \vec{CD}$.  Under this assumption, one can choose a coordinate system such that


*

*the incenter is centered at origin.

*$AB$ is lying in the plane $z = u > 0$.

*$CD$ is lying in the plane $z = -v < 0$.


Let $p = |AB|$, $q = |CD|$ and $\displaystyle\;\tau = \frac{v}{u+v} \in (0,1)$. 
Let $E, F, G, H$ be the intersection of the edge $AC$, $BC$, $AD$, $BD$ with the plane $z = 0$. The intersection of the tetrahedron $ABCD$ with the plane $z = 0$ 
will be a quadrilateral with these 4 points as vertices.
It is easy to see


*

*$\vec{EF} \parallel \vec{GH} \parallel \vec{AB}$ and $|EF| = |GH| = p\tau$.

*$\vec{EG} \parallel \vec{FH} \parallel \vec{CD}$ and $|EG| = |FH| = q(1-\tau)$.


The quadrilateral is actually a parallelogram with sides $p\tau$ and $q(1-\tau)$. 
The intersection of the insphere with the same plane will be a circle of radius $r$. It is clear this circle lies inside above parallelogram. 
Since the distance between the line $EF$ and $GH$ is at most $q(1-\tau)$ and the distance between the line $EG$ and $FH$ is at most $p\tau$, we have
$$2r \le \min( p\tau, q(1-\tau) )$$
Treat the RHS as a function of $\tau \in [0,1]$, it is maximized when 
$$p\tau = q(1-\tau) \iff \tau = \frac{q}{p+q}$$ 
This give us
$$2r \le \max_{t\in[0,1]}\min(p t, q(1-t)) = \frac{pq}{p+q} \quad\implies\quad 2r \le \frac{|AB||CD|}{|AB|+|CD|}\tag{*1}$$
This is pretty close to what we want to prove. To show above inequality is actually strict, we need two observations.


*

*The insphere touches the surface of the tetrahedron at 4 points, one at each face. Furthermore, the corresponding face is perpendicular to the radial direction there.

*In order for the equality to $(*1)$ to hold, we need
$p\tau = q(1-\tau) = 2r$. On the plane $z = 0$, the parallelogram is actually a square and the circle of radius $r$ need to touch the 4 sides of it.


Combine these two observations, we find in order for the equality to hold, the face holding $EF$ need to parallel to that holding $GH$ and the face holding $EG$ need to parallel to that holding $FH$. Their normal vectors are all orthogonal to the $z$-direction. These four faces are now bounding an infinite long cylinder instead of a tetrahedron. This is absurd and we can conclude
$$2r < \frac{|AB||CD|}{|AB|+|CD|}$$
Update
At the end is a picture which hope to illustrate the configuration. The vertices of the tetrahedron is located at
$$A,B,C,D = (2,0,2), (-2,0,2), (2,-2,-2), (-2,2,-2)$$
Both $\vec{AB}$ and $\vec{CD}$ are lying in some planes perpendicular to $z$-axis. The insphere is centered at $\left(0,0,2\left(\frac{3-\sqrt{5}}{3+\sqrt{5}}\right)\right)$ with radius $r = \frac{4}{3-\sqrt{5}}$. If one cut the tetrahedron with the plane $z = 2\left(\frac{3-\sqrt{5}}{3+\sqrt{5}}\right)$ which passes through the incenter, one obtain a parallelogram $EGHF$ ( $F$ is behind the insphere and not visible from this viewpoint ). The intersection of the insphere with that plane is a circle of radius $r$ which is contained within the parallelogram $EGHF$.
$\hspace 0.75in$ 
A: 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 
$$ r < \frac{AB \times CD}{2(AB  + CD)}.$$
