The fifth question: If an infinite sequence of positive real numbers $\left\{x_{n}\right\}$ satisfies: $x_{0}=1, x_{i} \geq x_{i+1}(i \in N)$, then the sequence is called a “good sequence”. Find the smallest constant $c$ such that there exists a “good sequence” $\left\{x_{n}\right\}$ satisfying: $\sum_{i=0}^{n} \frac{x_{i}^{2}}{x_{i+1}} \leq c$ for any positive integer $n$.
Fifth question,
Solution method one: On the one hand, for any "good sequence" $\left\{x_{n}\right\}$, according to the mean inequality, we have:
$$
\begin{array}{l}
=2^{n} \cdot\left(\frac{1}{2^{\sum_{i=0}^{n-1}(n-1-i) 2^{n-1-1}}} \cdot x_{0}^{2^{n}}\right)^{\frac{1}{2^{n}}} \\
=2^{n} \cdot\left(\frac{1}{2^{2^{n}(n-2)+2}}\right)^{\frac{1}{2^{n}}}=2^{2 \frac{1}{2^{n-1}}} \\
\end{array}
$$

Thus, for any positive integer $n$, we have $c \geq 2^{2} \frac{1}{2^{n-1}}$, hence $c \geq 4$.
In summary, the minimum value of $c$ is 4.
Solution method two: On the one hand, according to the Cauchy inequality, for any "good sequence" $\left\{x_{n}\right\}$ and any positive integer $n$, we have
$$
\sum_{i=0}^{n} \frac{x_{i}^{2}}{x_{i+1}} \geq \frac{\left(\sum_{i=0}^{n} x_{i}\right)^{2}}{\sum_{i=1}^{n+1} x_{i}}=\frac{\left(1+\sum_{i=1}^{n} x_{i}\right)^{2}}{\sum_{i=1}^{n} x_{i}+x_{n+1}} \geq \frac{4 \sum_{i=1}^{n} x_{i}}{\sum_{i=1}^{n} x_{i}+x_{n+1}}=\frac{4}{1+\frac{x_{n+1}}{\sum_{i=1}^{n} x_{i}}} \geq \frac{4}{1+\frac{1}{n}}
$$

When $n \rightarrow+\infty$, then $\sum_{i=0}^{n} \frac{x_{i}^{2}}{x_{i+1}} \geq \frac{4}{1+\frac{1}{n}} \rightarrow 4$, hence $c \geq 4$.
On the other hand, if we take $x_{i}=\frac{1}{2^{i}}(i \in \mathbb{N})$, then $\sum_{i=0}^{n} \frac{x_{i}^{2}}{x_{i+1}}=\sum_{i=0}^{n} \frac{\left(\frac{1}{2^{i}}\right)^{2}}{\frac{1}{2^{i+1}}}=\sum_{i=0}^{n} \frac{1}{2^{i-1}}=4-\frac{1}{2^{n-1}}$. In summary, the minimum value of $c$ is 4.