Solve $f(1)=a+b+c=1$. If $x_{1}=x_{2}=\cdots=x_{n}=1$, then $f\left(x_{1}\right) f\left(x_{2}\right) \cdots f\left(x_{n}\right)=1$.

If $x_{1}, x_{2}, \cdots, x_{n}$ are not all 1, then from $x_{1} x_{2} \cdots x_{n}=1$ we know that there must be one less than 1 and one greater than 1. Without loss of generality, let $x_{1}>1, x_{2}<1$.
\end{array}$$$$

Repeatedly performing the above transformation, we get $f\left(x_{1}\right) f\left(x_{2}\right) \cdots f\left(x_{n}\right) \geqslant f(1) f(1) \cdots f(1)=1$. Therefore, the minimum value of $f\left(x_{1}\right) f\left(x_{2}\right) \cdots f\left(x_{n}\right)$ is 1.