$\text{Min} \left ( \text{Max} \right )$

Ideally, all three products will equal $\sqrt[3]{9!}\approx 71.32$ . However, since they are all integers, at least one product will be $\ge 72$ .

This also means that the lower bound for the minimum of all maximums is $72$ .

But achieving $72$ is possible: we take the groups $(1,8,9),(3,4,6),(2,5,7)$ . (This gives a product of $(70,72,72)$ .)

Since $72$ is necessary and achievable, then it must be the answer.

My approach is same. To get the minimum of the maximum of the three products $P_{1}$ , $P_{2}$ and $P_{3}$ , our task is to even out the three products. That is to make $P_{1} \approx P_{2} \approx P_{3}$ . My first approach was to match the smallest number $1$ with the largest number $9$ , second smallest $2$ with the second largest $8$ and $3$ with $7$ , and then choose the remaining number for each group to get the minimum maximum. I got the following:

$3 \times 4 \times 7 = 84$ $2 \times 5 \times 8 = 80$ $1 \times 6 \times 9 = 54$

The answer was 84, which is wrong, and I have two more tries.

During a walk in the park, I realized that $P_{1} \approx P_{2} \approx P_{3} \approx \sqrt [ 3 ]{ 9! } \approx 71.33$ . Now still using the previous idea of matching the smallest $1$ with the largest $9$ , the remaining number should be $8$ to make the product $1 \times 8 \times 9 = 72 \approx 71.33$ . Move on the next largest number $7$ , the only possible combination to have the product $< 72$ is $2 \times 5 \times 7 = 70$ . The remaining three numbers give $3 \times 4 \times 6 = 72$ . So the answer is $\boxed{72}$ .

brute force using python:

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l=[]
for p in permute([1,2,3,4,5,6,7,8,9]):
    a,b,c,d,e,f,g,h,k=p
    l.append(max(a*b*c, d*e*f, g*h*k))

print min(l)

with permute is the function I wrote in my note

I actually don't understand how to solve these problems... most of the cases my approaches resemble a brute force approach. I started with three triples (1,9,5), (3,4,7), (2,6,8) , the maximum being 96. I "inspected" that interchanging 5 and 8 will result in a maximum of 84. After that, I interchanged 5 and 7, obtained the final result. It is more of a "local search" than logic.