Taiwan IMO 1J, 2020

Denote the original inequality by (*). Without loss of generality, we assume $u_i \le u_{i+1}$, $i = 1, ..., n-1$. Let
$$P = \{i : u_i > 0\}, \quad Q = \{i : u_i \le 0\}.$$
By the given conditions, it is clear that $u_n \in P$, $u_1 \in N$, $u_1 < 0$, and
$$\max_{i \in P} u_i = u_n, \quad \max_{i \in N} |u_i| = |u_1|. \quad (1)$$

First, from $\sum_{i=1}^{n} u_{i}^{2019} = 0$ we obtain
$$\sum_{i \in P} u_{i}^{2019} = \sum_{i \in N} |u_i|^{2019}. \quad (2)$$
(Obtaining this result independently earns one point. The cumulative total so far is two points.)
Hence, we obtain
$$\begin{aligned} 0 < \sum_{i \in P} u_i^{2020} &\le u_n \sum_{i \in P} u_i^{2019} = u_n \sum_{i \in N} |u_i|^{2019} \implies u_n \ge \frac{\sum_{i \in P} u_i^{2020}}{\sum_{i \in N} |u_i|^{2019}}, \\ 0 < \sum_{i \in N} u_i^{2020} &\le |u_1| \sum_{i \in N} |u_i|^{2019} = |u_1| \sum_{i \in P} u_i^{2019} \implies |u_1| \ge \frac{\sum_{i \in N} u_i^{2020}}{\sum_{i \in P} u_i^{2019}}. \end{aligned}$$

Furthermore, using (1) we can obtain
$$\begin{align} u_n|u_1| &\ge \frac{\sum_{i \in P} u_i^{2020}}{\sum_{i \in N} |u_i|^{2019} \sum_{i \in P} u_i^{2019}} \sum_{i \in N} u_i^{2020} \nonumber \\ &= \frac{\sum_{i \in P} u_i^{2020}}{\left(\sum_{i \in P} u_i^{2019}\right)^2} \sum_{i \in N} u_i^{2020} \ge \frac{\sum_{i \in N} u_i^{2020}}{\sum_{i \in P} u_i^{2018}}. \tag{3} \end{align}$$
Here, we have used the Cauchy inequality to obtain $(\sum_{i \in P} u_i^{2019})^2 \le \sum_{i \in P} u_i^{2018} \sum_{i \in P} u_i^{2020}$ to confirm that the last estimate holds. From (3), we deduce that
$$u_n|u_1| \sum_{i \in P} u_i^{2018} \geq \sum_{i \in N} u_i^{2020}. \quad (4)$$
Similarly, we can also obtain
$$u_n|u_1| \sum_{i \in N} u_i^{2018} \geq \sum_{i \in P} u_i^{2020}. \quad (5)$$
From (4)–(5) together with the equality $\sum_{i=1}^{n} u_i^{2018} = 1$, we obtain inequality (*), where $(u_{k_1}, u_{k_2}) = (u_1, u_n)$.