Let $O_1$, $O_2$, $R_1$, $R_2$ be the centres and radii of $\omega_1$ and $\omega_2$ respectively. Let $M$ and $N$ be the midpoints of $AC$ and $AB$ respectively. To calculate $R_1$, see that $\angle NBO_1 = 90^{\circ} - \angle B$. Notice that triangle $NBO_1$ has a right angle at $N$, so: \[\sin (\angle B) = \cos (90^{\circ} - \angle B)=\frac{c}{2R_1} \implies R_1=\frac{c}{2 \sin (\angle B)}\]Similarly, $R_2=\frac{b}{2 \sin (\angle C)}$ The product of the areas is then: \[\pi^2R_1^2R_2^2=\pi^2 \left( \frac {b \cdot c}{2\sin {\angle B}\cdot 2\sin {\angle C}}\right)^2 = \pi^2 \left( r \cdot r\right)^2 = \pi^2 r^4\]by sine rule, which is constant as it only depends on the fixed radius $r$. The sum of the areas is: \[\pi R_1^2 + \pi R_2^2 = \pi (R_1^2 + R_2^2) = \pi \left( \frac{c^2}{4\sin ^2(\angle B)} + \frac{b^2}{4\sin ^2(\angle C)}\right) \ge 2\pi \left( \frac {b \cdot c}{2\sin {\angle B}\cdot 2\sin {\angle C}}\right) = 2\pi r^2\]by AM-GM and sine rule. Equality holds when $R_1=R_2$, or equivalently $b\sin (\angle B)=c\sin (\angle C)$. Again through sine rule, this is equivalent to $\frac{b^2}{2r}=\frac{c^2}{2r}$, which yields $b=c$. Thus, equality holds when $ABC$ is an isosceles triangle with apex $A$.