The product AI*BI*CI

Consider a triangle $\triangle{ABC}$ and its Incenter, $I$. Denote $R$ and $r$ the circumradius and inradius, respectively. Also let $AI=k$; $BI=l$; $CI=m$. Then, the following identity holds

$$klm=4Rr^2$$

Proof. We make use of the semiperimeter-half-angle formula, 

$$\cos^2{\frac{\gamma}{2}}= \frac{s(s-c)}{ab}\tag{1}$$  

where $s$ is the semiperimeter and $\gamma$ denotes the angle $\angle{ACB}$. The proof for this formula can be found here.

Notice that $\cos{\frac{\gamma}{2}}=\frac{(s-c)}{m}$. Also, because of $(1)$ we have $\cos{\frac{\gamma}{2}}=\sqrt{\frac{s(s-a)}{ab}}$. Equating both expressions and solving for $m^2$, 

$$m^2=\frac{ab(s-c)}{s}$$

Similarly you get $k^2=\frac{bc(s-a)}{s}$ and $l^2=\frac{ac(s-b)}{s}$. Hence, 

$$(klm)^2=\frac{a^2b^2c^2(s-a)(s-b)(s-c)}{s^3}=\frac{a^2b^2c^2s(s-a)(s-b)(s-c)}{s^4}$$

Substituting from Heron's formula,

$$(klm)^2=\frac{a^2b^2c^2\Delta^2}{s^4}$$

Simplifying and using the well-known formulas $abc=4R\Delta$ and $\Delta=rs$ you get the desired result. 

$$klm=\frac{abc\Delta}{s^2}=\frac{4R\Delta^2}{s^2}=4Rr^2$$