4. $\frac{7}{8}$.

Let's assume $A B=2, A C=3, B C=x(1< x < 5)$. Then, $\frac{B E}{C F}=\sqrt{\frac{2 x^{2}-1}{2 x^{2}+14}}=\sqrt{1-\frac{15}{2 x^{2}+14}}$.
Since $1<x<5$, we have $t \geqslant \frac{7}{8}$.
4. In $\triangle A B C$, $3 A B=2 A C, E 、 F$ are the midpoints of $A C 、 A B$ respectively. If $B E<t C F$ always holds, then the minimum value of $t$ is $\qquad$ .