Let $G$ be the centroid of $\triangle{ABC}$. Points $M$ and $N$ are on side $AB$ and $AC$, respectively such that $\overline{AM} = m\cdot\overline{AB}$ and $\overline{AN} = n\cdot\overline{AC}$ where $m$ and $n$ are two positive real numbers. Find the minimal value of $mn$.
Extend $AG$ and let it intersect $BC$ at point $D$. Because $G$ is the centroid, $D$ must be the middle point of $BC$. Furthermore, we have $$\overrightarrow{AG} = \frac{2}{3}\overrightarrow{AD} = \frac{2}{3}\times\frac{1}{2}(\overrightarrow{AB}+\overrightarrow{AC}) = \frac{1}{3}(\frac{1}{m}\cdot\overrightarrow{AM} +\frac{1}{n}\overrightarrow{AN}) = \frac{1}{3m}\cdot\overrightarrow{AM} + \frac{1}{3n}\cdot\overrightarrow{AN}$$
Because $M$, $G$, and $N$ are collinear, we find $\frac{1}{3m}+\frac{1}{3n}=1$, or $\frac{1}{m}+\frac{1}{n}=3$, where $m$ and $n$ are both positive. It then follows that $$\sqrt{\frac{1}{m}\cdot\frac{1}{n}}\le \frac{\frac{1}{m}+\frac{1}{n}}{2}=\frac{3}{2}$$ Hence, $mn\ge \boxed{\frac{4}{9}}$.