Question 2 Let $a_{1}, a_{2}, \cdots, a_{6} ; b_{1}, b_{2}, \cdots, b_{6} ; c_{1}$, $c_{2}, \cdots, c_{6}$ all be permutations of $1,2, \cdots, 6$. Find the minimum value of $\sum_{i=1}^{6} a_{i} b_{i} c_{i}$. ${ }^{[1]}$
The minimum value of this problem is 162, which is achieved when $a_{1}, a_{2}, \cdots, a_{6}$; $b_{1}, b_{2}, \cdots, b_{6}$; $c_{1}, c_{2}, \cdots, c_{6}$ are $1,2,3,4,5,6$; $5,4,3,6,1,2$; $5,4,3,1,6,2$ respectively.

Compared to Problem 1, if the AM-GM inequality is directly applied in Problem 2, we have
$$
\begin{array}{l}
\sum_{i=1}^{6} a_{i} b_{i} c_{i} \geqslant 6 \sqrt[6]{\prod_{i=1}^{6} a_{i} b_{i} c_{i}}=6 \sqrt[6]{(6!)^{3}} \\
=72 \sqrt{5} \in(160,161),
\end{array}
$$

This only implies that $\sum_{i=1}^{6} a_{i} b_{i} c_{i} \geqslant 161$.
How can we improve this? Reviewing the above approach, we find that if the overall geometric mean is used to estimate the arithmetic mean, there will be an error if the values of $a_{i} b_{i} c_{i}$ are not evenly distributed, and such "uneven distribution" is unavoidable: because the geometric mean of six numbers is
$$
\sqrt[6]{(6!)^{3}}=12 \sqrt{5} \in(26,27),
$$

Let $d_{i}=a_{i} b_{i} c_{i}$, then there must be some $d_{i} \leqslant 26$, but $d_{i} \neq 26$, so $d_{i} \leqslant 25$. Assume this $d_{i}$ is $d_{1}$. Similarly, there must be some $d_{j} \geqslant 27$, assume this $d_{j}$ is $d_{2}$. Therefore, $d_{1}+d_{2} \geqslant 25+\frac{d_{1} d_{2}}{25}$, i.e.,
$$
\begin{array}{l}
\frac{1}{25}\left(d_{1}-25\right)\left(25-d_{2}\right) \geqslant 0 . \\
\text { Hence } \sum_{i=1}^{6} d_{i} \geqslant 25+\frac{d_{1} d_{2}}{25}+d_{3}+d_{4}+d_{5}+d_{6} \\
\geqslant 25+5 \sqrt[5]{\frac{d_{1} d_{2} d_{3} d_{4} d_{5} d_{6}}{25}} \\
=25+5 \sqrt[5]{\frac{(6!)^{3}}{25}}>161 . \\
\text { Thus, } \sum_{i=1}^{6} a_{i} b_{i} c_{i} \geqslant 162 .
\end{array}
$$

Thus, $\sum_{i=1}^{6} a_{i} b_{i} c_{i} \geqslant 162$.
Therefore, the minimum value of $\sum_{i=1}^{6} a_{i} b_{i} c_{i}$ is 162.
In this process, $d_{1}+d_{2} \geqslant 25+\frac{d_{1} d_{2}}{25}$ is a more refined handling technique compared to $d_{1}+d_{2} \geqslant 2 \sqrt{d_{1} d_{2}}$, fully utilizing the discreteness of the product distribution. Readers can try to compare if this step is changed to $d_{1}+d_{2} \geqslant 27+\frac{d_{1} d_{2}}{27}$, then it still only yields $\sum_{i=1}^{6} a_{i} b_{i} c_{i} \geqslant 161$ (because 27 is much closer to $12 \sqrt{5}$ than 25, so the improvement is not significant).