Example 5 Let the function $f(x)$ be defined for all $x>0$, and satisfy the following conditions:
(1) For $x>0$, $f(x) f\left[f(x)+\frac{1}{x}\right]=1$; (2) $f(x)$ is increasing on $(0,+\infty)$. Find $f(1)$.
Solution: Let $f(1)=a$. When $x=1$, from (1) we have $a f(a+1)=1$.
Let $x=a+1$ and substitute into (1) to get $f(a+1) f\left[f(a+1)+\frac{1}{a+1}\right]=1$.
Thus, we derive $\frac{1}{a} f\left(\frac{1}{a}+\frac{1}{a+1}\right)=a=f(1)$.
Since $f(x)$ is increasing on $(0,+\infty)$, we have $\frac{1}{a}+\frac{1}{a+1}=1$, leading to $a=\frac{1 \pm \sqrt{5}}{2}$.
If $a=\frac{1+\sqrt{5}}{2}$, then $1<a=f(1)<f(1+a)=\frac{1}{a}<1$, which is a contradiction. Therefore, $f(1)=\frac{1-\sqrt{5}}{2}$.