IMO Team Selection Test

Firstly, we will prove one useful statement:
Lemma. If $x_1, x_2, \dots, x_{n-1} \in \mathbb{R}$ are such that $x_1 + x_2 + \dots + x_{n-1} = 0$, then
$$x_1 + |x_1 - x_2| + |x_2 - x_3| + \dots + |x_{n-2} - x_{n-1}| + x_{n-1} \ge 0$$
(1)
Proof. It is clear that if $x_1 + x_{n-1} \ge 0$ the inequality (1) is true.
If $x_1 + x_{n-1} < 0$, then there exists $i$, $2 \le i \le n-2$ such that $x_i > 0$. From the properties of absolute value we have
$$\begin{aligned} & |x_1 - x_2| + |x_2 - x_3| + \dots + |x_{n-2} - x_{n-1}| \ge |x_1 - x_i| + |x_i - x_{n-1}| \\ &= |x_i - x_1| + |x_i - x_{n-1}| \ge x_i - x_1 + x_i - x_{n-1} = 2x_i - x_1 - x_{n-1} \ge \\ & \ge -x_1 - x_{n-1} \end{aligned}$$
From where (1) follows.
Since there are no other cases, we get that (1) is true, which concludes the proof of the lemma.

If we now use the equality $|a-b| = a+b - 2\min\{a,b\}$, the inequality (1) is equivalent to the inequality
$$\begin{align*} x_1 + x_1 + x_2 - 2\min\{x_1, x_2\} + x_2 + x_3 - 2\min\{x_2, x_3\} + \dots + x_{n-2} + x_{n-1} - 2\min\{x_{n-2}, x_{n-1}\} + x_{n-1} \ge 0 \\ x_1 + x_2 + \dots + x_{n-1} \ge \min\{x_1, x_2\} + \min\{x_2, x_3\} + \dots + \min\{x_{n-2}, x_{n-1}\} \\ 0 \ge \min\{x_1, x_2\} + \min\{x_2, x_3\} + \dots + \min\{x_{n-2}, x_{n-1}\}. \tag{2} \end{align*}$$
Now, let $x_1, x_2, \dots, x_{n-1} \in \mathbb{R}$ be arbitrary real numbers (their sum need not be equal to 0). If we homogenize this sequence,
$$x_1 - \frac{1}{n-1} \sum_{i=1}^{n-1} x_i, x_2 - \frac{1}{n-1} \sum_{i=1}^{n-1} x_i, \dots, x_{n-1} - \frac{1}{n-1} \sum_{i=1}^{n-1} x_i,$$
then we have
$$x_1 - \frac{1}{n-1} \sum_{i=1}^{n-1} x_i + x_2 - \frac{1}{n-1} \sum_{i=1}^{n-1} x_i + \dots + x_{n-1} - \frac{1}{n-1} \sum_{i=1}^{n-1} x_i = \sum_{i=1}^{n-1} x_i - \frac{n-1}{n-1} \sum_{i=1}^{n-1} x_i = 0.$$
Therefore, for this sequence the conditions of the lemma as well as inequality (2) are fulfilled.
If we use the equality
$$\min \left\{ x_j - \frac{1}{n-1} \sum_{i=1}^{n-1} x_i, x_{j+1} - \frac{1}{n-1} \sum_{i=1}^{n-1} x_i \right\} = \min \{x_j, x_{j+1}\} - \frac{1}{n-1} \sum_{i=1}^{n-1} x_i,$$
and we substitute it in (2), we get the inequality
$$\frac{n-2}{n-1} \sum_{i=1}^{n-1} x_i \geq \min\{x_1, x_2\} + \min\{x_2, x_3\} + \dots + \min\{x_{n-2}, x_{n-1}\} \quad (3)$$
Now we return to the proof of the problem statement. Without loss of generality we can assume
that $a_n = \min\{a_1, a_2, \dots, a_{n-1}, a_n\}$. If, otherwise, $a_i = \min\{a_1, a_2, \dots, a_{n-i-1}, a_n\}$ for some $i$, $1 \le i \le n-1$
then we consider the sequence $b_1 = a_{i+1}, \dots, b_{n-i} = a_n, b_{n-i+1} = a_{i-1}, \dots, b_n = a_i$.
Therefore, if we choose $x_i = a_i$, for $i = 1, 2, \dots, n-1$, by substituting in (3), we get
$$\begin{gather*} \frac{n-2}{n-1} \sum_{i=1}^{n-1} a_i \ge \min\{a_1, a_2\} + \min\{a_2, a_3\} + \dots + \min\{a_{n-2}, a_{n-1}\}. \text{ But from the condition } \\ a_1 + a_2 + \dots + a_{n-1} = -a_n, \text{ we get} \end{gather*}$$
$$\frac{n-2}{n-1} a_n \ge \min\{a_1, a_2\} + \min\{a_2, a_3\} + \dots + \min\{a_{n-2}, a_{n-1}\}.$$
From the equalities $a_n = \min\{a_{n-1}, a_n\}$ and $a_n = \min\{a_n, a_1\}$, we get
$$2a_n - \frac{n-2}{n-1} a_n \geq \min\{a_1, a_2\} + \min\{a_2, a_3\} + \dots + \min\{a_{n-2}, a_{n-1}\} + \min\{a_{n-1}, a_n\} + \min\{a_n, a_1\}$$
i.e.
$$\frac{n}{n-1} \min\{a_1, a_2, \dots, a_n\} \geq \min\{a_1, a_2\} + \min\{a_2, a_3\} + \dots + \min\{a_n, a_1\},$$
Q.E.D.