Using standard notation of $ \triangle A'B'C'$: $ a' = B'C', b' = C'A', c' = A'B',$ $ [A'B'C']$ is triangle area, $ p'$ semiperimeter, $ r' = \frac {_r}{^2}$ inradius. $ A'I \ge 2r' \Longleftrightarrow \frac {b'c' (p' - a')}{p'} \ge 4r'^2 \Longleftrightarrow \frac {2[A'B'C'] r'}{\sin A' \tan \frac {A'}{2}} \ge 4r' [A'B'C] \Longleftrightarrow$ ${ \sin A' \tan \frac {_{A'}}{^2}} \le \frac {_1}{^2} \Longleftrightarrow \sin^2 \frac {_{A'}}{^2} \le \frac {_1}{^4} \Longleftrightarrow \sin \frac {_{A'}}{^2} \le \frac {_1}{^2} \Longleftrightarrow A' \le 60^\circ$ All $ A', B', C'$ are outside the incircle $ (I)$ of $ \triangle ABC$ or some coincide with the incircle tangency points. $ A'I, B'I, C'I \ge r = 2r' \Longrightarrow$ $ A' = B' = C' = 60^\circ \Longrightarrow$ $ A'I, B'I, C'I = 2r' = r \Longrightarrow$ $ \triangle A'B'C'$ is equilateral and coincides with the intouch triangle of $ \triangle ABC$ $ \Longrightarrow$ $ \triangle ABC$ is equilateral.

IMHO, it's easier to use $\frac{r'}{\overline{A'I}}=\sin\left(\frac{\alpha'}{2}\right)$.^^ Moreover, $\sin\left(\frac{\alpha'}{2}\right)\leqslant \frac{1}{2}$ doesn't necessarily yield $\alpha'\leqslant 60^{\circ}$, but the case $\alpha' \geqslant 150^{\circ}$ is easy to handle.

Suppose $\angle {C'A'B'}=A,\angle{A'B'C'}=B,\angle{A'C'B'}=C,A'B'=c,B'C'=a,C'A'=b$ and similarly define every thing adding $'$ to all of $a,b,c,S$ for $ABC$. Now $\frac {A'I}{c}=\frac {Sin(\frac {B}{2})}{Cos(\frac {C}{2})}$ so that implies basically $AI'=\sqrt{\frac {(S-a)bc}{S}}$. Now $AI'^2\geq 4r^2\implies bc\geq (S-b)(S-c)$ and now so multiplying those we get $abc\geq (S-a)(S-b)(S-c)\geq abc$ , and so we must have $a=b=c$. So we get $A'B'C'$ is equilateral.Now for instance, suppose $\angle{BA'C'}=x$ such that $x\leq 60^0$.Now easy application of sine rule gives $Sin(x+30^0)=2Sin(30^0)=1\implies x=60^0$. So we get in circle of $ABC$ meets $BC,CA,AB$ at $A',B',C'$.Now basically that implies $(S'-b')Sin(\frac {B'}{2})$ is fixed for all $a',b',c'$ and now suppose, $f(x)=x(S'-x)$. Clearly now we get $f(a')=f(b')=f(c')$ but $f(x)$ can't have at most two distinct roots, so wlog suppose $a'=b'$ now we have $a'(S'-a')=c(S'-c')\implies a'=c'$ , and so ultimately we get $a'=b'=c'$ and we are done.

Here's a synthetic solution: Let $2R$ be the inradius of $ABC$ and $R$ be the inradius of $A'B'C'$ (and $\omega$, $\omega_1$ be the incircles). Then, since $B, D, A', C$ are collinear, we take its Pole in $\omega_1$, note that $ID = 2R$ so, $ID' = \frac{R}{2}$ where $D'$ is inverse of $D$ in $\omega$. But, the Polar of $A'$ in $\omega_1$ is the intouch chord, $E'F'$, so we have that $D'$ lies on $E'F'$. So distance of $I$ from $E'F'$ is atmost $R$ (since a point on $E'F'$ is at distance $R$ from $I$). So, we reflect $I$ over $E'F'$ (which is inside $\omega_1$, so $\angle E'IF' + \angle E'D'F' \geq 180^{\circ}$), and conclude that $\angle E'D'F' \geq 60^{\circ}$, and similarly for $D'F'E'$ and $F'E'D'$. So, due to angle sum property, each of them is $60^\circ$ and therefore, $A'B'C'$, $B'C'A'$ and $C'A'B'$ are all $60^{\circ}$. Now, since each angle is $60^\circ$, the distance from $I$ to the sides of $D'E'F'$ is $\frac{R}{2}$, so the inverse of $D$ in $\omega_1$, actually coincides with the foot from $I$ onto $E'F'$, which is $IA' \cap E'F'$, and $ID \cap E'F'$ is the inverse of $D$, so $D$ and $A'$ conicide and similarly, $E$ and $B'$, $F$ and $C'$ coincide thus giving us that $DEF$ is equilateral implying $ABC$ is equilateral.