Let $A_0$, $C_1$, and $B_2$ be the intouch points of $AEF$ opposite $A$, $E$, and $F$, respectively, and define the intouch points of $BFD$ and $CDE$ similarly. Let $I_A$, $I_B$, and $I_C$ be the incenters of $AEF$, $BFD$, and $CDE$, respectively. Let $r$, $r'$, and $r_0$ denote the inradii of $ABC$, $DEF$, and $AEF$, respectively, and let $p$, $p'$, $p_A$, $p_B$, and $p_C$ denote the perimeters of $ABC$, $DEF$, $AEF$, $BFD$, $CDE$, respectively. By equal tangents, $$p'=DC_0+C_0E+EA_0+A_0F+FB_0+B_0D=DA_2+B_1E+EB_2+C_1F+FC_2+A_1D=A_1A_2+B_1B_2+C_1C_2=I_BI_C+I_CI_A+I_AI_B,$$ so $p'$ is the perimeter of $I_AI_BI_C$. We have $$rp=2[ABC]=2([DEF]+[AEF]+[BFD]+[CDE])=r'p'+r_0(p_A+p_B+p_C)=r'p'+r_0(p+p'),$$ so it suffices to show $rp=(r-r_0)p'+r_0(p+p') \Longleftrightarrow \tfrac{p'}{p}=\tfrac{r-r_0}{r}$. But notice that the homothety at $I$ with scale factor $\tfrac{r-r_0}{r}$ maps $ABC$ to $I_AI_BI_C$, as desired. $\square$

Let $X$, $Y$ and $Z$ denote the incenters of $\triangle AEF$, $\triangle BFD$ and $\triangle CDE$, respectively. Let $r$ denote the common inradius of these three triangles. By chasing tangents, the perimeters of $\triangle DEF$ and $\triangle XYZ$ are equal. Let $p$ be this common perimeter, and let $P$ be the perimeter of $\triangle ABC$. We have $$[ABC] - [XYZ] = \frac{r(p+P)}{2} = [ABC] - [DEF],$$ so the areas of $\triangle XYZ$ and $\triangle DEF$ are equal as well. Thus, their inradii are equal. Since the incenter of $\triangle XYZ$ coincides with the incenter of $\triangle ABC$, it is clear that the inradius of $\triangle ABC$ is the sum of the inradius of $\triangle XYZ$ and $r$, which is in turn equal to sum of the inradius of $\triangle DEF$ and $r$.

Remark