[Example 3.4.4] Given non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy the inequality
$$
x_{1}+x_{2}+\cdots+x_{n} \leqslant \frac{1}{2} \text {. }
$$

Find the minimum value of $\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots\left(1-x_{n}\right)$.
We seek the extremum point, first conduct some experiments:
Experiment 1: Take $x_{1}=x_{2}=x_{3}=x_{4}=\frac{1}{8}, x_{5}=x_{6}=\cdots=x_{n}=0$, at this point we have
$$
\begin{array}{c}
x_{1}+x_{2}+\cdots+x_{n}=\frac{1}{2}, \\
\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots\left(1-x_{n}\right)=\left(\frac{7}{8}\right)^{4}=\frac{2401}{4096} ;
\end{array}
$$

Experiment 2: Take $x_{1}=x_{2}=\frac{1}{4}, x_{3}=x_{4}=\cdots=x_{n}=0$, at this point we have
$$
\begin{array}{c}
x_{1}+x_{2}+\cdots+x_{n}=\frac{1}{2}, \\
\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots\left(1-x_{n}\right)=\left(\frac{3}{4}\right)^{2}=\frac{9}{16} ;
\end{array}
$$

Experiment 3: Take $x_{1}=\frac{1}{2}, x_{2}=x_{3}=\cdots=x_{n}=0$, at this point we have
$$
\begin{array}{c}
x_{1}+x_{2}+\cdots+x_{n}=\frac{1}{2} \\
\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots\left(1-x_{n}\right)=\frac{1}{2} .
\end{array}
$$

Comparing the above results, we find that the extremum point of this problem is different from the previous examples, and it is not when all variables $x_{1}, x_{2}, \cdots, x_{n}$ are equal that $\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots\left(1-x_{n}\right)$ achieves its minimum value. For example, in Experiment 1, four values are equal (all equal to $\left.\frac{1}{8}\right)$, and in Experiment 2, two values are equal $\left(\right.$ both equal to $\left.\frac{1}{4}\right)$, and their corresponding $\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots(1-$ $\left.x_{n}\right)$ values are respectively
$$
\frac{2401}{4096}>\frac{1}{2}, \frac{9}{16}>\frac{1}{2} .
$$

In Experiment 3, only $x_{1}=\frac{1}{2}$, the rest are 0, at this point $\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots(1-$ $\left.x_{n}\right)=\frac{1}{2}$.
We can conjecture that the extremum point of this problem is
$$
\left(\frac{1}{2}, 0,0, \cdots, 0\right) .
$$

Therefore, our goal is to keep the sum $x_{1}+x_{2}+\cdots+x_{n}$ unchanged, and to eliminate $x_{2}$, $x_{3}, \cdots, x_{n}$.
For this, we set
$$
\begin{array}{c}
x_{i}^{\prime}=x_{i}, i=1,2, \cdots, n-2, \\
x_{n-1}^{\prime}=x_{n-1}+x_{n}, x_{n}^{\prime}=0 .
\end{array}
$$

At this point we have
$$
\begin{aligned}
\left(1-x_{n-1}^{\prime}\right)\left(1-x_{n}^{\prime}\right) & =1-\left(x_{n-1}^{\prime}+x_{n}^{\prime}\right)+x_{n-1}^{\prime} x_{n}^{\prime} \\
& =1-\left(x_{n-1}+x_{n}\right)+0 . \\
\left(1-x_{n-1}\right)\left(1-x_{n}\right) & =1-\left(x_{n-1}+x_{n}\right)+x_{n-1} x_{n} .
\end{aligned}
$$

Thus,
$$
\left(1-x_{n-1}^{\prime}\right)\left(1-x_{n}^{\prime}\right) \leqslant\left(1-x_{n-1}\right)\left(1-x_{n}\right),
$$

i.e., $\left(1-x_{1}^{\prime}\right)\left(1-x_{2}^{\prime}\right) \cdots\left(1-x_{n}^{\prime}\right) \leqslant\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots\left(1-x_{n}\right)$.
Further, we let
$$
\begin{array}{c}
x_{i}^{\prime \prime}=x_{i}, i=1,2, \cdots, n-3 . \\
x_{n-2}^{\prime \prime}=x_{n-2}+x_{n-1}+x_{n}, x_{n-1}^{\prime \prime}=0, x_{n}^{\prime \prime}=0,
\end{array}
$$

From the above, we also have
$$
\left(1-x_{1}^{\prime \prime}\right)\left(1-x_{2}^{\prime \prime}\right) \cdots\left(1-x_{n}^{\prime \prime}\right) \leqslant\left(1-x_{1}^{\prime}\right)\left(1-x_{2}^{\prime}\right) \cdots\left(1-x_{n}^{\prime}\right) .
$$

Thus, after at most $n-1$ transformations, we have
$$
\begin{array}{l}
x_{1}^{0}=x_{1}+x_{2}+\cdots+x_{n} \leqslant \frac{1}{2}, \\
x_{2}^{0}=x_{3}^{0}=\cdots=x_{n}^{0}=0 .
\end{array}
$$

Therefore,
$$
\left(1-x_{1}^{0}\right)\left(1-x_{2}^{0}\right) \cdots\left(1-x_{n}^{0}\right) \geqslant \frac{1}{2} .
$$

That is, the minimum value of $\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots\left(1-x_{n}\right)$ is $\frac{1}{2}$.
As previously noted, this problem is not smoothed when the variables are equal, so the equality of variables is not the essential characteristic of the smoothing transformation.