Example 7 Given non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}(n \geqslant 3)$ satisfy the inequality: $x_{1}+x_{2}+\cdots+$ $x_{n} \leqslant \frac{1}{2}$, find the minimum value of $\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots\left(1-x_{n}\right)$.
When $x_{1}, x_{2}, \cdots, x_{n-2}, x_{n-1}+x_{n}$ are all constants, since
$$\left(1-x_{n-1}\right)\left(1-x_{n}\right)=1-\left(x_{n-1}+x_{n}\right)+x_{n-1} x_{n}$$

it is evident that the larger $\left|x_{n-1}-x_{n}\right|$, the smaller the value of the above expression. Therefore, for $n \geqslant 3$, let
$$x_{i}^{\prime}=x_{i}, i=1,2, \cdots, n-2, x_{n-1}^{\prime}=x_{n-1}+x_{n}, x_{n}^{\prime}=0$$

Then $x_{n-1}^{\prime}+x_{n}^{\prime}=x_{n-1}+x_{n}, x_{n-1}^{\prime} x_{n}^{\prime}=0 \leqslant x_{n-1} x_{n}$. Therefore, we have
$$\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots\left(1-x_{n}\right) \geqslant\left(1-x_{1}^{\prime}\right)\left(1-x_{2}^{\prime}\right) \cdots\left(1-x_{n-1}^{\prime}\right)$$

where
$$x_{1}^{\prime}+x_{2}^{\prime}+\cdots+x_{n-1}^{\prime}=x_{1}+x_{2}+\cdots+x_{n} \leqslant \frac{1}{2}$$

Performing the smoothing transformation similar to (1) $n-2$ times, we get
$$\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots\left(1-x_{n}\right) \geqslant 1-\left(x_{1}+x_{2}+\cdots+x_{n}\right) \geqslant \frac{1}{2}$$

Equality is achieved when $x_{1}=\frac{1}{2}, x_{2}=x_{3}=\cdots=x_{n}=0$.