7. In $\triangle A B C$, $E, F$ are the midpoints of $A C, A B$ respectively, $A B=\frac{2}{3} A C$. If $\frac{B E}{C F}<t$ always holds, then the minimum value of $t$ is $\qquad$ .
Given $A B=2, A C=3, B C=x(1<x<5)$. By the median formula, we have $2\left(B E^{2}+C E^{2}\right)=A B^{2}+B C^{2} \Rightarrow B E^{2}=\frac{4+x^{2}}{2}-\frac{9}{4}=\frac{2 x^{2}-1}{4}$, and similarly, $C F^{2}=\frac{9+x^{2}}{2}-1=\frac{x^{2}+7}{2}$. Therefore, $\frac{B E^{2}}{C F^{2}}=\frac{2 x^{2}-1}{2 x^{2}+14}=1-\frac{15}{2 x^{2}+14}$. Since $2 x^{2}+14 \in(16,64)$, it follows that $\frac{B E^{2}}{C F^{2}} \in\left(\frac{1}{16}, \frac{49}{64}\right)<t^{2}$ always holds $\Rightarrow t \geqslant \frac{7}{8}$,
so the minimum value of $t$ is $\frac{7}{8}$.