, 2017

Solution:
Let the tangent lengths be $a, b, c, d$ so that
$$\begin{aligned} & a+b=2 \\ & b+c=3 \\ & c+d=5 \\ & d+a=4 \end{aligned}$$
Then $b=2-a$ and $c=1+a$ and $d=4-a$. The radius of the inscribed circle of quadrilateral $ABCD$ is given by
$$\sqrt{\frac{abc+abd+acd+bcd}{a+b+c+d}} = \sqrt{\frac{-7a^2+16a+8}{7}}$$
This is clearly maximized when $a=\frac{8}{7}$ which leads to a radius of $\sqrt{\frac{120}{49}}=\frac{2\sqrt{30}}{7}$.