9-36 Find the smallest real number $c$ such that for any positive sequence $\left\{x_{n}\right\}$, if
$$
\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{n} \leqslant x_{n+1}, n=1,2,3, \cdots \text { then } \\
\sqrt{x_{1}}+\sqrt{x_{2}}+\cdots+\sqrt{x_{n}} \leqslant c \sqrt{x_{1}+x_{2}+\cdots+x_{n}}, n=1,2,3, \cdots
\end{array}
$$
[Solution] Let $x_{n}=2^{n}$, then
$$
x_{1}+x_{2}+\cdots+x_{n}=2^{n+1}-2 \leqslant 2^{n+1}=x_{n+1}, n=1,2, \cdots
$$

For any natural number $n$ we have
$$
\frac{\sqrt{x_{1}}+\sqrt{x_{2}}+\cdots+\sqrt{x_{n}}}{\sqrt{x_{1}+x_{2}+\cdots+x_{n}}}=\frac{\sqrt{2^{n+1}}-\sqrt{2}}{(\sqrt{2}-1) \sqrt{2^{n+1}-2}} \text {. }
$$

Since $\lim _{n \rightarrow \infty} \frac{\sqrt{2^{n+1}}-\sqrt{2}}{(\sqrt{2}-1) \sqrt{2^{n+1}-2}}=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$,
it follows that $c \leqslant \sqrt{2}+1$.
The following uses induction to prove that if the positive sequence $\left\{x_{n}\right\}$ satisfies
$$
x_{1}+x_{2}+\cdots+x_{n} \leqslant x_{n+1}, n=1,2,3, \cdots
$$

then
$$
\sqrt{x_{1}}+\sqrt{x_{2}}+\cdots+\sqrt{x_{n}} \leqslant(\sqrt{2}+1) \sqrt{x_{1}+x_{2}+\cdots+x_{n}}
$$

for any natural number $n$. In fact, when $n=1$, (1) is obviously true. Assume that (1) holds for $n=k$. By the assumption, we have
$$
\begin{aligned}
& \sqrt{x_{1}+x_{2}+\cdots+x_{k}+x_{k+1}}+\sqrt{x_{1}+\cdots+x_{k}} \\
\leqslant & (\sqrt{2}+1) \sqrt{x_{k+1}},
\end{aligned}
$$

thus
$$
\begin{array}{c}
\sqrt{x_{k+1}} \leqslant(\sqrt{2}+1)\left(\sqrt{x_{1}+\cdots+x_{k}+x_{k+1}}-\right. \\
\left.\sqrt{x_{1}+\cdots+x_{k}}\right) .
\end{array}
$$

By the induction hypothesis, we know

Therefore
$$
\begin{aligned}
& \sqrt{x_{1}}+\sqrt{x_{2}}+\cdots+\sqrt{x_{k}} \leqslant(\sqrt{2}+1) \sqrt{x_{1}+\cdots+x_{k}} \\
& \sqrt{x_{1}}+\sqrt{x_{2}}+\cdots+\sqrt{x_{k}}+\sqrt{x_{k+1}} \\
\leqslant & (\sqrt{2}+1) \sqrt{x_{1}+\cdots+x_{k}+x_{k+1}},
\end{aligned}
$$

which means that (1) also holds for $n=k+1$. From (1), we deduce that $c \geqslant \sqrt{2}+1$.
In summary, the required $c=\sqrt{2}+1$.