【答案】 $\frac{4}{9}$

【解析】 解：如图,

\begin{aligned} & \because \overrightarrow{B E}=\lambda \overrightarrow{B C}, \overrightarrow{D F}=\mu \overrightarrow{D C}, \text { 且 } \lambda+\mu=\frac{2}{3}, \\ & \therefore \overrightarrow{A E} \cdot \overrightarrow{A F}=(\overrightarrow{A B}+\overrightarrow{B E}) \cdot(\overrightarrow{A D}+\overrightarrow{D F}), \\ & =(\overrightarrow{A B}+\lambda B C) \cdot(\overrightarrow{A D}+\mu \overline{D C})=(\overrightarrow{A B}+\lambda \overrightarrow{A D}) \cdot(\overrightarrow{A D}+\mu \overrightarrow{A B}) \\ & =(1+\lambda \mu) \overrightarrow{A B} \cdot \overrightarrow{A D}+\lambda|\overrightarrow{A D}|^2+\mu|\overrightarrow{A B}|^2 \\ & =(1+\lambda \mu) \times 2 \times 2 \times\left(-\frac{1}{2}\right)+4(\lambda+\mu)=-2(1+\lambda \mu)+\frac{8}{3} . \end{aligned}

\begin{aligned} & \because \lambda+\mu=\frac{2}{3}, \\ & \therefore \lambda \mu \leqslant\left(\frac{\lambda+\mu}{2}\right)^2=\frac{1}{9}, \text { 则 }-2(1+\lambda \mu) \geqslant-\frac{20}{9}, \\ & \therefore-2(1+\lambda \mu)+\frac{8}{3} \geqslant \frac{4}{9} \text { (当且仅当 } \lambda=\mu=\frac{1}{3} \text { 时等号成立), } \\ & \therefore \overrightarrow{A E} \cdot \overrightarrow{A F} \text { 的最小值为 } \frac{4}{9} . \end{aligned}