Example 3 Find the smallest positive number $c$, such that for all infinite positive sequences $\left\{x_{n}\right\}$ satisfying $x_{1}+x_{2}+\cdots+x_{n} \leqslant x_{n+1}$, the inequality
$$
\begin{array}{l}
\sqrt{x_{1}}+\sqrt{x_{2}}+\cdots+\sqrt{x_{n}} \\
\leqslant c \sqrt{x_{1}+x_{2}+\cdots+x_{n}}(n=1,2, \cdots) .
\end{array}
$$
holds.
Analysis: To find the smallest positive number $c$, it is necessary to find a constant $A$ such that $c \geqslant A$. This can be achieved by selecting an appropriate infinite sequence of positive numbers $\left\{x_{n}\right\}$ that satisfies $x_{1}+x_{2}+\cdots+x_{n} \leqslant x_{n+1}$, and substituting it into inequality (1) to obtain the result.

What kind of positive sequence $\left\{x_{n}\right\}$ makes $\sqrt{x_{1}}+\sqrt{x_{2}}+\cdots+\sqrt{x_{n}}$ and $x_{1}+x_{2}+\cdots+x_{n}$ easy to compute?

Notice that when $\left\{x_{n}\right\}$ is a geometric sequence, $\left\{\sqrt{x_{n}}\right\}$ is also a geometric sequence. Therefore, we can take the geometric sequence $x_{n}=q^{n}$ for testing. Substituting it into the original inequality, we get
$$
\begin{array}{l}
c \geqslant \frac{\sqrt{x_{1}}+\sqrt{x_{2}}+\cdots+\sqrt{x_{n}}}{\sqrt{x_{1}+x_{2}+\cdots+x_{n}}} \\
=\frac{\sqrt{q}\left[(\sqrt{q})^{n}-1\right]}{\sqrt{q}-1} \cdot \frac{1}{\sqrt{\frac{q(q)}{q-1}}} \\
=\frac{\sqrt{q-1}}{\sqrt{q-1}} \cdot \frac{(\sqrt{q})^{n}-1}{\sqrt{q^{n}-1}} .
\end{array}
$$

Moreover, the chosen $q$ should satisfy $x_{1}+x_{2}+\cdots+x_{n} \leqslant x_{n+1}$, then $\frac{x_{1}\left(q^{n}-1\right)}{q-1} \leqslant x_{1} q^{n}$. Solving this, we get $q \geqslant 2-\frac{1}{q^{n}}$. Letting $n \rightarrow \infty$, we get $q \geqslant 2$. Therefore, we can take $x_{n}=2^{n}$ to obtain the corresponding inequality.
Solution: Let $x_{n}=2^{n}$. Substituting it into the original inequality, we get
$$
\begin{aligned}
& c \geqslant \frac{\sqrt{x_{1}}+\sqrt{x_{2}}+\cdots+\sqrt{x_{n}}}{\sqrt{x_{1}+x_{2}+\cdots+x_{n}}} \\
= & \frac{\sqrt{2}\left[(\sqrt{2})^{n}-1\right]}{\sqrt{2}-1} \cdot \frac{1}{\sqrt{\frac{2\left(2^{n}-1\right)}{2-1}}} \\
= & \frac{1}{\sqrt{2}-1} \cdot \frac{(\sqrt{2})^{n}-1}{\sqrt{2^{n}-1}} \\
= & (\sqrt{2}+1) \frac{1-\frac{1}{(\sqrt{2})^{n}}}{\sqrt{1-\frac{1}{2^{n}}}} .
\end{aligned}
$$

Letting $n \rightarrow \infty$, we get $c \geqslant \sqrt{2}+1$.
Next, we prove that when $c=\sqrt{2}+1$, the inequality holds, i.e., for all infinite sequences of positive numbers $\left\{x_{n}\right\}$ that satisfy $x_{1}+x_{2}+\cdots+x_{n} \leqslant x_{n+1}$, we have
$$
\begin{array}{l}
\sqrt{x_{1}}+\sqrt{x_{2}}+\cdots+\sqrt{x_{n}} \\
\leqslant(\sqrt{2}+1) \sqrt{x_{1}+x_{2}+\cdots+x_{n}}(n=1,2, \cdots) .
\end{array}
$$

We prove this by mathematical induction.
When $n=1$, the conclusion is obviously true.
Assume the conclusion holds for $n=k$.
When $n=k+1$, by the induction hypothesis, we have
$$
\begin{array}{l}
\sqrt{x_{1}}+\sqrt{x_{2}}+\cdots+\sqrt{x_{k}}+\sqrt{x_{k+1}} \\
\leqslant(\sqrt{2}+1) \sqrt{x_{1}+x_{2}+\cdots+x_{k}}+\sqrt{x_{k+1}} .
\end{array}
$$

Thus, we only need to prove
$$
\begin{array}{l}
(\sqrt{2}+1) \sqrt{x_{1}+x_{2}+\cdots+x_{k}}+\sqrt{x_{k+1}} \\
\leqslant(\sqrt{2}+1) \sqrt{x_{1}+x_{2}+\cdots+x_{k+1}} \cdots
\end{array}
$$

Notice that
$$
\begin{array}{l}
(\sqrt{2}+1) \sqrt{x_{1}+x_{2}+\cdots+x_{k}}+\sqrt{x_{k+1}}- \\
(\sqrt{2}+1) \sqrt{x_{1}+x_{2}+\cdots+x_{k+1}} \\
=(\sqrt{2}+1)\left(\sqrt{x_{1}+x_{2}+\cdots+x_{k}}-\right. \\
\left.\sqrt{x_{1}+x_{2}+\cdots+x_{k+1}}\right)+\sqrt{x_{k+1}} \\
=(\sqrt{2}+1)
\end{array}
$$
$$
\begin{aligned}
& \frac{-x_{k+1}}{\sqrt{x_{1}+x_{2}+\cdots+x_{k}}+\sqrt{x_{1}+x_{2}+\cdots+x_{k+1}}}+\sqrt{x_{k+1}} \\
\leqslant & (\sqrt{2}+1) \frac{-x_{k+1}}{\sqrt{x_{k+1}}+\sqrt{x_{k+1}+x_{k+1}}}+\sqrt{x_{k+1}} \\
= & (\sqrt{2}+1) \frac{-\sqrt{x_{k+1}}}{1+\sqrt{2}}+\sqrt{x_{k+1}}=0 .
\end{aligned}
$$

Therefore, when $n=k+1$, the conclusion holds.
In summary, the minimum value of $c$ is $\sqrt{2}+1$.