9・36 Find the smallest real number $c$ such that for any positive sequence $\left\{x_{n}\right\}$, if $x_{1}+x_{2}+\cdots+x_{n} \leqslant x_{n+1}, n=1,2,3, \cdots$, then
$$\sqrt{x_{1}}+\sqrt{x_{2}}+\cdots+\sqrt{x_{n}} \leqslant c \sqrt{x_{1}+x_{2}+\cdots+x_{n}}, n=1,2,3, \cdots$$
［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 $n$,
$$\frac{\sqrt{x_{1}}+\sqrt{x_{2}}+\cdots+\sqrt{x_{n}}}{\sqrt{x_{1}+x_{2}+\cdots+x_{n}}}=\frac{\sqrt{2^{n+1}}-\sqrt{2}}{(\sqrt{2}-1) \sqrt{2^{n+1}-2}}$$

Since
$$\lim _{n \rightarrow \infty} \frac{\sqrt{2^{n+1}}-\sqrt{2}}{(\sqrt{2}-1) \sqrt{2^{n+1}-2}}=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$$

Therefore, $c \leqslant \sqrt{2}+1$.
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 $\square$
$$\sqrt{x_{1}}+\sqrt{x_{2}}+\cdots+\sqrt{x_{n}} \leqslant(\sqrt{2}+1) \sqrt{x_{1}+x_{2}+\cdots+x_{n}}$$

For any natural number $n$. In fact, when $n=1$, (1) is obviously true. Assume that when $n=k$,
(1) holds. 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,
$$\begin{aligned}
\sqrt{x_{k+1}} \leqslant & (\sqrt{2}+1)\left(\sqrt{x_{1}+\cdots+x_{k}+x_{k+1}}-\right. \\
& \left.\sqrt{x_{1}+\cdots+x_{k}}\right)
\end{aligned}$$

By the induction hypothesis, we know
$$\sqrt{x_{1}}+\sqrt{x_{2}}+\cdots+\sqrt{x_{k}} \leqslant(\sqrt{2}+1) \sqrt{x_{1}+\cdots+x_{k}} .$$

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

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