1 Let $x_{i} \geqslant 0(1 \leqslant i \leqslant n), \sum_{i=1}^{n} x_{i} \leqslant \frac{1}{2}, n \geqslant 2$. Find the minimum value of $F=\left(1-x_{1}\right)(1-$ $\left.x_{2}\right) \cdots\left(1-x_{n}\right)$.
1. When $n=2$, $x_{1}+x_{2} \leqslant \frac{1}{2},\left(1-x_{1}\right)\left(1-x_{2}\right)=1+x_{1} x_{2}-\left(x_{1}+x_{2}\right) \geqslant 1-x_{1}-x_{2} \geqslant \frac{1}{2}$. The equality holds when $x_{1}+x_{2}=\frac{1}{2}, x_{1} x_{2}=0$, i.e., $x_{1}=\frac{1}{2}, x_{2}=0$. When $n=3$, $x_{1}+x_{2}+x_{3} \leqslant \frac{1}{2}$, using the above transformation, we have $\left(1-x_{1}\right)\left(1-x_{2}\right)(1-\left.x_{3}\right) \geqslant\left(1-x_{1}\right)\left[1-\left(x_{2}+x_{3}\right)\right] \geqslant 1-x_{1}-\left(x_{2}+x_{3}\right) \geqslant \frac{1}{2}$. Note that $0 \leqslant x_{i} \leqslant 1$. The equality holds when $x_{1}+x_{2}+x_{3}=\frac{1}{2}$, and $x_{2} x_{3}=x_{1}\left(x_{2}+x_{3}\right)=0$, i.e., $x_{1}=\frac{1}{2}, x_{2}=x_{3}=0$. From the above, we can conjecture that the extremal point in the general case is: $\left(\frac{1}{2}, 0,0, \cdots, 0\right)$. Using the smoothing transformation. First, prove the lemma: If $0 \leqslant x, y \leqslant 1$, then $(1-x)(1-y) \geqslant 1-x-y$. This inequality is directly proven by expanding the left side. This smoothing tool is equivalent to $(x, y) \rightarrow (x+y, 0)$. Suppose $n \geqslant 2$, for the variable set $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$, without loss of generality, assume $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}$, then $F=\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots\left(1-x_{n}\right) \geqslant\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots\left(1-x_{n-2}\right)(1-\left.x_{n-1}-x_{n}\right) \geqslant\left(1-x_{1}\right)\left(1-x_{2}\right) \cdots\left(1-x_{n-3}\right)\left(1-x_{n-2}-x_{n-1}-x_{n}\right) \geqslant \cdots \geqslant 1-x_{1}-x_{2}-\cdots-x_{n} \geqslant \frac{1}{2}$. The equality holds when the variable set becomes $\left(\frac{1}{2}, 0,0, \cdots, 0\right)$, at this point, $F$ reaches its minimum value $\frac{1}{2}$.