Example 8 Given non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy the inequality $x_{1}+x_{2}+\cdots+x_{n} \leqslant \frac{1}{2}$, find the minimum value of $f\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\prod_{i=1}^{n}\left(1-x_{i}\right)$.
When $x_{1}, x_{2}, \cdots, x_{n-2}, x_{n-1}+x_{n}$ are all constants, since
$$
\left(1-x_{n-1}\right)\left(1-x_{n}\right)=1-\left(x_{n-1}+x_{n}\right)+x_{n-1} x_{n} \text {, }
$$

it is evident that the larger $|x_{n-1}-x_{n}|$ is, the smaller the value of the above expression. For this reason,
let $x^{\prime}{ }_{i}=x_{i}(i=1,2, \cdots, n-2), x^{\prime}{ }_{n-1}=x_{n-1}+x_{n}, x^{\prime}{ }_{n}=0$, then $x_{n-1}^{\prime}+x^{\prime}{ }_{n}=x_{n-1}+x_{n}, x_{n-1}^{\prime} \cdot x^{\prime}{ }_{n}=01-\left(x_{1}+x_{2}+\cdots+x_{n}\right) \geqslant \frac{1}{2}$, where the equality holds when $x_{1}=\frac{1}{2}$, $x_{2}=x_{3}=\cdots=x_{n}=0$.
Therefore, the minimum value sought is $\frac{1}{2}$.