Subfield: Algebra


Question: Find all lists $\left(x_{1}, x_{2}, \ldots, x_{2020}\right)$ of non-negative real numbers such that the following three conditions are all satisfied:

(i) $x_{1} \leq x_{2} \leq \ldots \leq x_{2020}$;

(ii) $x_{2020} \leq x_{1}+1$;

(iii) there is a permutation $\left(y_{1}, y_{2}, \ldots, y_{2020}\right)$ of $\left(x_{1}, x_{2}, \ldots, x_{2020}\right)$ such that

$$
\sum_{i=1}^{2020}\left(\left(x_{i}+1\right)\left(y_{i}+1\right)\right)^{2}=8 \sum_{i=1}^{2020} x_{i}^{3}
$$

A permutation of a list is a list of the same length, with the same entries, but the entries are allowed to be in any order. For example, $(2,1,2)$ is a permutation of $(1,2,2)$, and they are both permutations of $(2,2,1)$. Note that any list is a permutation of itself.

Solution:
We first prove the inequality

$$
((x+1)(y+1))^{2} \geq 4\left(x^{3}+y^{3}\right) \tag{1}
$$

for real numbers $x, y \geq 0$ satisfying $|x-y| \leq 1$, with equality if and only if $\{x, y\}=\{0,1\}$ or $\{x, y\}=\{1,2\}$.

Indeed,

$$
\begin{aligned}
4\left(x^{3}+y^{3}\right) & =4(x+y)\left(x^{2}-x y+y^{2}\right) \\
& \leq\left((x+y)+\left(x^{2}-x y+y^{2}\right)\right)^{2} \\
& =\left(x y+x+y+(x-y)^{2}\right)^{2} \\
& \leq(x y+x+y+1)^{2} \\
& =((x+1)(y+1))^{2}
\end{aligned}
$$

where the first inequality follows by applying the AM-GM inequality on $x+y$ and $x^{2}-x y+y^{2}$ (which are clearly nonnegative). Equality holds in the first inequality precisely if $x+y=$ $x^{2}-x y+y^{2}$ and in the second one if and only if $|x-y|=1$. Combining these equalities we have $x+y=(x-y)^{2}+x y=1+x y$ or $(x-1)(y-1)=0$, which yields the solutions $\{x, y\}=\{0,1\}$ or $\{x, y\}=\{1,2\}$. Now, let $\left(x_{1}, x_{2}, \ldots, x_{2020}\right)$ be any sequence satisfying conditions (i) and (ii) and let $\left(y_{1}, y_{2}, \ldots, y_{2020}\right)$ be any permutation of $\left(x_{1}, x_{2}, \ldots, x_{2020}\right)$. As $0 \leq \min \left(x_{i}, y_{i}\right) \leq$ $\max \left(x_{i}, y_{i}\right) \leq \min \left(x_{i}, y_{i}\right)+1$, we can applying inequality (1) to the pair $\left(x_{i}, y_{i}\right)$ and sum over all $1 \leq i \leq 2020$ to conclude that

$$
\sum_{i=1}^{2020}\left(\left(x_{i}+1\right)\left(y_{i}+1\right)\right)^{2} \geq 4 \sum_{i=1}^{2020}\left(x_{i}^{3}+y_{i}^{3}\right)=8 \sum_{i=1}^{2020} x_{i}^{3}
$$

Therefore, in order to satisfy condition (iii), every inequality must be an equality. Hence, for every $1 \leq i \leq 2020$ we must have $\left\{x_{i}, y_{i}\right\}=\{0,1\}$ or $\left\{x_{i}, y_{i}\right\}=\{1,2\}$. By condition (ii), we see that either $\left\{x_{i}, y_{i}\right\}=\{0,1\}$ for all $i$ or $\left\{x_{i}, y_{i}\right\}=\{1,2\}$ for all $i$.

If $\left\{x_{i}, y_{i}\right\}=\{0,1\}$ for every $1 \leq i \leq 2020$, this implies that the sequences $\left(x_{1}, x_{2}, \ldots, x_{2020}\right)$ and $\left(y_{1}, y_{2}, \ldots, y_{2020}\right)$ together have 2020 zeroes and 2020 ones. As $\left(y_{1}, y_{2}, \ldots, y_{2020}\right)$ is a permutation of $\left(x_{1}, x_{2}, \ldots, x_{2020}\right)$ this implies that $\left(x_{1}, x_{2}, \ldots, x_{2020}\right)=(0,0, \ldots, 0,1,1, \ldots, 1)$ with 1010 zeroes and 1010 ones. Conversely, note that this sequence satisfies conditions (i), (ii), and (iii) (in (iii), we take $\left.\left(y_{1}, y_{2}, \ldots, y_{2020}\right)=\left(x_{2020}, x_{2019}, \ldots, x_{1}\right)\right)$, showing that this sequence indeed works. The same reasoning holds for the case that $\left\{x_{i}, y_{i}\right\}=\{1,2\}$ for all $i$.

Final answer: there are two solutions: $(\underbrace{0,0, \ldots, 0}_{1010}, \underbrace{1,1, \ldots, 1}_{1010})$ and $(\underbrace{1,1, \ldots, 1}_{1010}, \underbrace{2,2, \ldots, 2}_{1010})$