Problem 1070

[Proof] Let $f(x)=\frac{1}{1+x}$. Clearly, for $0 \leqslant x<y$ we have
$$0<f(x)-f(y)=\frac{y-x}{(1+x)(1+y)}<y-x .$$

From this, it is easy to prove that for any natural number $n$, when $x \geqslant 0$,
$$g_{n}(x)=x+f(x)+f(f(x))+\cdots+f^{(n)}(x) \text { is strictly increasing, }$$

where $f^{(n)}(x)=f(f(\cdots f(x) \cdots))$ is the $n$-fold composition of the function $f$ with itself.
We prove the desired conclusion by induction. When $n=1$, it is clear that $\quad x_{1}<\frac{F_{1}}{F_{2}}=1$.
Assume that the conclusion holds for $1 \leqslant n \leqslant k-1$. Without loss of generality, assume that $x_{1}, x_{2}, \cdots, x_{k}$ are all non-zero, otherwise by the recursive relation of $\left\{x_{n}\right\}$, we have $x_{k}=0$, and the desired conclusion follows from the induction hypothesis. Thus, $0<x_{i}<1, i=1,2, \cdots, k$, and for $i=2,3, \cdots, k$ we have
$$x_{i-1}=\frac{1}{\left[\frac{1}{x_{i-1}}\right]+x_{i}} \leqslant \frac{1}{1+x_{i}}=f\left(x_{i}\right)$$

Let $y_{1}=x_{k}, y_{2}=x_{k-1}, \cdots, y_{k}=x_{1}$, then from (2) we get
$$0<y_{1}<1, y_{i} \leqslant f\left(y_{i-1}\right), i=2,3, \cdots, k .$$

Since $f(x)$ is decreasing for $x \geqslant 0$, from (3) we have
$$\begin{aligned} x_{1}+x_{2}+\cdots+x_{k} & =y_{1}+y_{2}+\cdots+y_{k} \\ & \leqslant y_{1}+y_{2}+\cdots+y_{k-1}+f\left(y_{k-1}\right) . \end{aligned}$$
, and by (1) and (3) we get
$$x_{1}+x_{2}+\cdots+x_{k} \leqslant y_{1}+y_{2}+\cdots+y_{k-2}+f\left(y_{k-2}\right)+f\left(f\left(y_{k-2}\right)\right) .$$

Repeating the use of (1) and (3) we get
$$x_{1}+x_{2}+\cdots+x_{k} \leqslant y_{1}+f\left(y_{1}\right)+\cdots+f^{(k-1)}\left(y_{1}\right) \text {. }$$

Since $0<y_{1}<1$, by (1) we get
$$x_{1}+x_{2}+\cdots+x_{k}<1+f(1)+\cdots+f^{(k-1)}(1) \text {. }$$

Since $\frac{F_{1}}{F_{2}}=1$, and for $i \geqslant 1$ we have
$$f\left(\frac{F_{i}}{F_{i+1}}\right)=\frac{F_{i+1}}{F_{i}+F_{i+1}}=\frac{F_{i+1}}{F_{i+2}},$$

thus
$$1+f(1)+\cdots+f^{(k-1)}(1)=\frac{F_{1}}{F_{2}}+\frac{F_{2}}{F_{3}}+\cdots+\frac{F_{k}}{F_{k+1}} .$$

Therefore,
$$x_{1}+x_{2}+\cdots+x_{k}<\frac{F_{1}}{F_{2}}+\frac{F_{2}}{F_{3}}+\cdots+\frac{F_{k}}{F_{k+1}},$$

which means that the desired conclusion also holds when $n=k$, completing the induction proof.