Let $\overrightarrow{A B}=\vec{b}, \overrightarrow{A C}=\vec{c}$
$$
\begin{array}{l}
\overrightarrow{A M}=\lambda \vec{b} \\
\overrightarrow{A N}=m \vec{c}
\end{array}
$$
Let $G$ divides $\mathrm{MN}$ in the ratio $\mathrm{K}: 1$
So $\frac{k \mu \vec{c}+\lambda \vec{b}}{k+1}=\frac{\vec{b}+\vec{c}}{3}$
$$
\begin{array}{l}
\Rightarrow \frac{\mathrm{k} \mu}{\mathrm{k}+1}=\frac{1}{3} \quad \frac{\lambda}{\mathrm{k}+1}=\frac{1}{3} \\
\Rightarrow \mathrm{k}=\frac{\lambda}{\mu} \\
\Rightarrow \frac{1}{\lambda}+\frac{1}{\mu}=3 \\
\mathrm{AM} \geq \mathrm{GM} \\
\frac{1}{\lambda}+\frac{1}{\mu}=\frac{1}{2} \geq \frac{2}{\sqrt{\lambda \mu}} \Rightarrow\left(\frac{2}{3}\right)^{2} \leq \lambda \mu...(1)
\end{array}
$$
Now, $\frac{\text { area of } \triangle \mathrm{AMN}}{\text { area of } \mathrm{ABC}}=\frac{\frac{1}{2} \lambda \mu|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}|}{\frac{1}{2}|\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}|}$
$=\lambda \mu$
using $\frac{1}{\lambda}+\frac{1}{\mu}=3 \Rightarrow$ Ratio $=\frac{\lambda}{3 \lambda-1} \lambda \in[0,1]$ maximum value of ratio $=\frac{\lambda^{2}}{3 \lambda-1}$ attain when $\lambda=1$ using derivative but $\lambda$ is not 1 becuase $M$ is an interior point.
so $\frac{4}{9} \leq$ ratio $ < \frac{1}{2}$