The sixth question: Given real numbers $x_{1}, x_{2}, \ldots, x_{2021}$ satisfy: $\sum_{i=1}^{2021} x_{i}^{2}=1$, try to find the maximum value of $\sum_{i=1}^{2020} x_{i}^{3} x_{i+1}^{3}$.
The sixth question,
Solution: First, when $x_{1}=x_{2}=\frac{\sqrt{2}}{2}, x_{3}=x_{4}=\ldots=x_{2021}=0$, $\sum_{i=1}^{2020} x_{i}^{3} x_{i+1}^{3}=\frac{1}{8}$.
Next, we prove: $\sum_{i=1}^{2020} x_{i}^{3} x_{i+1}^{3} \leq \frac{1}{8}$.
We prove a more general case: For any integer $n \geq 2, x_{i} \in R(1 \leq i \leq n)$, we have $\sum_{i=1}^{n-1} x_{i}^{3} x_{i+1}^{3} \leq \frac{\left(\sum_{i=1}^{n} x_{i}^{2}\right)^{3}}{8}$.
We first prove a lemma: For real numbers $a_{i} \geq 0(1 \leq i \leq n)$, then $\left(\sum_{i=1}^{n} a_{i}^{3}\right)^{2} \leq \left(\sum_{i=1}^{n} a_{i}^{2}\right)^{3}$.
Proof of the lemma: When $n=1$, the conclusion is obviously true. Below, assume $n \geq 2$, then
\[
\begin{array}{l}
\left(\sum_{i=1}^{n} a_{i}^{2}\right)^{3} \geq \sum_{i=1}^{n} a_{i}^{6} + 3\left(\sum_{1 \leq i < j \leq n} a_{i}^{2} a_{j}^{2}\right) \sum_{i=1}^{n} a_{i}^{2} \geq \left(\sum_{i=1}^{n} a_{i}^{3}\right)^{2}.
\end{array}
\]
According to homogeneity, without loss of generality, assume $\sum_{i=1}^{n} a_{i}^{r}=1$, then we only need to prove: $\sum_{i=1}^{n} a_{i}^{s} \leq 1$.
In fact, for any $1 \leq i \leq n$, we have $a_{i}^{r} \leq \sum_{i=1}^{n} a_{i}^{r}=1$, hence $a_{i} \leq 1$, and thus $a_{i}^{s} \leq a_{i}^{r}$, equality holds if and only if $a_{i} \in \{0,1\}$. Therefore, $\sum_{i=1}^{n} a_{i}^{s} \leq \sum_{i=1}^{n} a_{i}^{r}=1$, equality holds if and only if each $a_{i} \in \{0,1\}$. That is, among $a_{1}, a_{2}, \ldots, a_{n}$, there are exactly $n-1$ zeros and one 1.
In summary, the proposition is proved!