Problem: Let $\left\{a_{i}\right\} 、\left\{b_{i}\right\} 、\left\{c_{i}\right\}(i=1,2, \cdots, 10)$ be permutations of $1,2, \cdots, 10$. Find the minimum value of $\sum_{i=1}^{10} a_{i} b_{i} c_{i}$.
The minimum value of this problem is
$$
\begin{array}{l}
1 \times 9 \times 10+9 \times 10 \times 1+10 \times 1 \times 9+2 \times \\
6 \times 7+7 \times 7 \times 2+8 \times 2 \times 6+5 \times 5 \times 4+ \\
6 \times 3 \times 5+4 \times 8 \times 3+3 \times 4 \times 8 \\
=930
\end{array}
$$

If the estimation method from problem 2 is used, since the geometric mean of the ten numbers is $\sqrt[10]{(10!)^{3}} \in(92,93)$, let $d_{i}=$ $a_{i} b_{i} c_{i}$, then there must be some $d_{i} \leqslant 90$, and there must also be some $d_{j} \geqslant$ 96, but no matter which number is used, it can only estimate that
$$
\sum_{i=1}^{10} a_{i} b_{i} c_{i} \geqslant 929 .
$$

How to exclude 929? Considering that if $\sum_{i=1}^{10} a_{i} b_{i} c_{i}=$ 929, then there must be an odd number among $d_{i}$ (let's assume it is $d_{1}$), so $d_{1} \leqslant 81$ or $d_{1} \geqslant 105$.
If $d_{1} \leqslant 81$, then since there must be some $d_{j} \geqslant 93$ (let's assume it is $d_{2}$).
$$
\begin{array}{l}
\text { Hence } d_{1}+d_{2} \geqslant 81+\frac{d_{1} d_{2}}{81} \\
\Rightarrow \sum_{i=1}^{10} d_{i} \geqslant 81+\frac{d_{1} d_{2}}{81}+\sum_{i=3}^{10} d_{i} \\
\geqslant 81+9 \sqrt[9]{\frac{\prod_{i=1}^{10} d_{i}}{81}}=81+9 \sqrt[9]{\frac{(10!)^{3}}{81}} \\
\quad>929 \\
\Rightarrow \sum_{i=1}^{10} a_{i} b_{i} c_{i} \geqslant 930 .
\end{array}
$$

Similarly, if $d_{1} \geqslant 105$, it can also be derived that
$$
\sum_{i=1}^{10} a_{i} b_{i} c_{i} \geqslant 930 \text {. }
$$

In summary, the minimum value of $\sum_{i=1}^{10} a_{i} b_{i} c_{i}$ is 930.