Example 3 Let $x_{i} \geqslant 0(1 \leqslant i \leqslant n), \sum_{i=1}^{n} x_{i}=1, n \geqslant 2$. Find the maximum value of $F=\sum_{1 \leq i<j \leqslant n} x_{i} x_{j}\left(x_{i}+\right.$ $x_{j}$ ). (32nd IMO Shortlist Problem)
Analysis and Solution: Since $F$ is continuous on a closed region, the maximum value of $F$ must exist, which suggests using local adjustments. Without loss of generality, let's try it. If we fix $x_{2}, x_{3}, \cdots, x_{n}$, then $x_{1}$ is also fixed, so we can only fix $n-2$ numbers. Let's fix $x_{3}, x_{4}, \cdots, x_{n}$, then
$$\begin{aligned}
F= & x_{1} x_{2}\left(x_{1}+x_{2}\right)+\cdots+x_{1} x_{n}\left(x_{1}+x_{n}\right)+x_{2} x_{3}\left(x_{2}+x_{3}\right)+\cdots \\
& +x_{2} x_{n}\left(x_{2}+x_{n}\right)+\sum_{3 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right)
\end{aligned}$$

where $x_{1}+x_{2}=1-\left(x_{2}+x_{3}+\cdots+x_{n}\right)=p$ (a constant).
Noting that $x_{2}=p-x_{1}$, $F$ is a quadratic function of $x_{1}$: $f\left(x_{1}\right)=A x_{1}^{2}+$ $B x_{1}+C\left(0 \leqslant x_{1} \leqslant p\right)$. However, the coefficients of this function and the range of the variable are very complex, making the form of its extremum also very complex, so we need to find another way to find the extremum.

Let's first guess the extremum point. For this, we use the technique of specialization.
When $n=2$, $F=x_{1} x_{2}\left(x_{1}+x_{2}\right)=x_{1} x_{2} \leqslant \frac{\left(x_{1}+x_{2}\right)^{2}}{4}=\frac{1}{4}$.
But $n=2$ is too special and lacks generality. Let's try again.
When $n=3$, $F=x_{1} x_{2}\left(x_{1}+x_{2}\right)+x_{1} x_{3}\left(x_{1}+x_{3}\right)+x_{2} x_{3}\left(x_{2}+x_{3}\right)$.
Let's fix $x_{3}$, then
$$\begin{aligned}
F & =x_{1} x_{2}\left(x_{1}+x_{2}\right)+\left(x_{1}^{2}+x_{2}^{2}\right) x_{3}+\left(x_{1}+x_{2}\right) x_{3}^{2} \\
& =x_{1} x_{2}\left(1-x_{3}\right)+\left[\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}\right] x_{3}+\left(1-x_{3}\right) x_{3}^{2} \\
& =x_{1} x_{2}\left(1-3 x_{3}\right)+\left(1-x_{3}\right)^{2} x_{3}+\left(1-x_{3}\right) x_{3}^{2},
\end{aligned}$$

To use the inequality $x_{1} x_{2} \leqslant \frac{\left(x_{1}+x_{2}\right)^{2}}{4}$ to amplify $x_{1} x_{2}\left(1-3 x_{3}\right)$, we need to ensure $1-3 x_{3} \geqslant 0$. For this, we can optimize the assumption: $x_{1} \geqslant x_{2} \geqslant x_{3}$. Then
$$\begin{aligned}
F & =x_{1} x_{2}\left(1-3 x_{3}\right)+\left(1-x_{3}\right)^{2} x_{3}+\left(1-x_{3}\right) x_{3}^{2} \\
& \leqslant\left(1-3 x_{3}\right) \frac{\left(x_{1}+x_{2}\right)^{2}}{4}+\left(1-x_{3}\right)^{2} x_{3}+\left(1-x_{3}\right) x_{3}^{2} \\
& =\left(1-3 x_{3}\right) \frac{\left(1-x_{3}\right)^{2}}{4}+\left(1-x_{3}\right)^{2} x_{3}+\left(1-x_{3}\right) x_{3}^{2} \\
& \leqslant \frac{1}{4}, \text { (using derivatives we can see) }
\end{aligned}$$

where the equality holds when $x_{1}=x_{2}=\frac{1}{2}, x_{3}=0$.
From this, we can conjecture that the extremum point in the general case is $\left(\frac{1}{2}, \frac{1}{2}, 0,0, \cdots, 0\right)$.
Using the smoothing transformation: When $n \geqslant 3$, for the variable group: $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$, assume without loss of generality that $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}$, let $x_{1}^{\prime}=x_{1}, x_{2}^{\prime}=x_{2}, \cdots, x_{n-2}^{\prime}=x_{n-2}, x_{n-1}^{\prime}=x_{n-1}+x_{n}$, $x_{n}^{\prime}=0$, to get a new variable group $\left(x_{1}, x_{2}, \cdots, x_{n-2}, x_{n-1}+x_{n}, 0\right)$, the corresponding value of $F$ is
$$\begin{aligned}
F^{\prime}= & \sum_{1 \leq i<i \leq n-2} x_{i} x_{j}\left(x_{i}+x_{j}\right)(\text { this expression is denoted as } A) \\
& +\sum_{i=1}^{n-2} x_{n-1}^{\prime} x_{i}\left(x_{n-1}^{\prime}+x_{i}\right)+\sum_{i=1}^{n-1} x_{n}^{\prime} x_{i}\left(x_{n}^{\prime}+x_{i}\right) \\
= & A+\sum_{i=1}^{n-2}\left(x_{n-1}+x_{n}\right) x_{i}\left(x_{n-1}+x_{n}+x_{i}\right) \\
= & A+\sum_{i=1}^{n-2}\left(x_{n-1}^{2}+2 x_{n-1} x_{n}+x_{n-1} x_{i}+x_{n}^{2}+x_{n} x_{i}\right) x_{i} \\
= & A+\sum_{i=1}^{n-2} x_{n-1} x_{i}\left(x_{n-1}+x_{i}\right)+\sum_{i=1}^{n-2} x_{n} x_{i}\left(x_{n}+x_{i}\right)+2 x_{n-1} x_{n} \sum_{i=1}^{n-2} x_{i} \\
= & A+\sum_{i=1}^{n-2} x_{n-1} x_{i}\left(x_{n-1}+x_{i}\right)+\sum_{i=1}^{n-1} x_{n} x_{i}\left(x_{n}+x_{i}\right) \\
& -x_{n-1} x_{n}\left(x_{n-1}+x_{n}\right)+2 x_{n-1} x_{n} \sum_{i=1}^{n-2} x_{i} \\
= & F+x_{n-1} x_{n}\left(2 \sum_{i=1}^{n-2} x_{i}-x_{n-1}-x_{n}\right)=F+\left[2-3\left(x_{n-1}+x_{n}\right)\right] x_{n-1} x_{n}
\end{aligned}$$

Since $\sum_{i=1}^{n} x_{i}=1$, and $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}$, we have
$$\frac{x_{n-1}+x_{n}}{2} \leqslant \frac{x_{1}+x_{2}+\cdots+x_{n}}{n}=\frac{1}{n}$$

So $x_{n-1}+x_{n} \leqslant \frac{2}{n}$. And $n \geqslant 3$, so $x_{n-1}+x_{n} \leqslant \frac{2}{n} \leqslant \frac{2}{3}$, so $2-3\left(x_{n-1}+x_{n}\right) \geqslant$ 0, so $F^{\prime} \geqslant F$.

As long as the number of non-zero components in the variable group $n \geqslant 3$, the above transformation can continue, and after at most $n-2$ adjustments, $n-2$ components can be set to 0. Using the case when $n=2$, we know that $F$ reaches its maximum value $\frac{1}{4}$ at $\left(\frac{1}{2}, \frac{1}{2}, 0,0, \cdots, 0\right)$.