Problem: Find all lists $(x_1, x_2, \dots, x_{2020})$ of non-negative real numbers
such that the following three conditions are all satisfied:
\begin{itemize}
\ii $x_1 \le x_2 \le \dots \le x_{2020}$;
\ii $x_{2020} \le x_1  + 1$;
\ii there is a permutation $(y_1, y_2, \dots, y_{2020})$
of $(x_1, x_2, \dots, x_{2020})$ such that
\[ \sum_{i = 1}^{2020} ((x_i + 1)(y_i + 1))^2 = 8 \sum_{i = 1}^{2020} x_i^3.  \]
\end{itemize}

Solution: The main point of the problem is to prove an inequality.
\begin{claim*}
  If $a$ and $b$ are real numbers with $|a-b| \le 1$,
  then \[ (a+1)^2(b+1)^2 \geq 4(a^3+b^3). \]
  Moreover, equality holds only when $\{a,b\}=\{1,0\}$ or $\{a,b\}=\{2,1\}$.
\end{claim*}
\begin{proof}
  Write
  \begin{align*}
    a^3+b^3 &= (a+b)(a^2-ab+b^2) = (a+b)[(a-b)^2 + ab] \\
    &\le (a+b)(1+ab) \\
    &\le \left( \frac{(a+b)+(1+ab)}{2} \right)^2 = \frac{(a+1)^2(b+1)^2}{4}. \qedhere
  \end{align*}
\end{proof}

From this it follows that $(x_n)_n$
should be either $(1,\dots,1,2,\dots,2)$ or $(0,\dots,0,1,\dots,1)$.

\begin{remark*}
  There is a sense in which the problem \emph{must} be an inequality,
  because given a \emph{fixed} permutation of the indices
  of $y_i$ by $x_i$, one gets an equation which \emph{a priori}
  ought to cut out a codimension $1$ surface in $\RR^{2020}$.
  So the fact that the answer set appears finite
  is an indication that some inequality at play here.
\end{remark*}