As shown in the figure, the number filled in " $O$ " is equal to the product of the numbers in the three connected " $\triangle$ ", and the number filled in $\square$ is equal to the sum of the numbers in the three connected " $\bigcirc$ ". Now, fill the numbers $5, 6, 7, 8, 9$ into the five " $\triangle$ ". Then, the largest number in $\square$ is $\qquad$ .
【Analysis and Solution】
As shown in the figure, fill in the letters $A \sim I$ respectively;
Since the number filled in $\square$ is equal to the sum of the three "middle numbers" connected to it;

So $I=F+G+H$; Because the number filled in “ $\bigcirc$ ” is equal to the product of the three " $\triangle$ " numbers connected to it; so $F=A \times B \times C ， G=B \times C \times D, H$ $=C \times D \times E$;

Here, $C$ appears 3 times, $B$ and $D$ each appear 2 times, and $A$ and $E$ each appear 1 time;
We want $I$ to be as large as possible, so $C=9 ，\{B, D\}=\{7,8\} ，\{A, E\}=\{5,6\}$;
Let's assume $B=7 ， D=8 ；$

Then $G=7 \times 9 \times 8=504, F=A \times 7 \times 9, H=9 \times 8 \times E$; Therefore,
take $A=5, E=6$ ；

Then $F=5 \times 7 \times 9=315 ， G=7 \times 9 \times 8=504 ， H=9 \times 8 \times 6=432 ， I=315+504+432=1251$ ；
The number in $\square$ is maximized to 1251 .