Vietnamese Mathematical Olympiad

Using Cauchy-Schwarz inequality, we get
$$x_1 + x_2 + \dots + x_n \le |x_1| + |x_2| + \dots + |x_n| \\ \le \sqrt{n(x_1^2 + x_2^2 + \dots + x_n^2)}.$$

a) Without loss of generality, we can assume $a \le b \le c$ and the inequality turns out to
$$|a - b| + |b - c| + |c - a| = (a - b) + (b - c) + (c - a) \\ = 2(a - c) \le 2\sqrt{2(a^2 + c^2)} \le 2\sqrt{2}.$$
The equality holds when $(a, b, c) = \left(-\frac{\sqrt{2}}{2}, 0, \frac{\sqrt{2}}{2}\right)$.

b) Without loss of generality, suppose that $a_1$ is the smallest number. We investigate the following cases:

* If the sequence $a_1, a_2, \dots, a_{2019}$ is not decreasing then $S$ can be simplified as
$$S = 2|a_1 - a_{2019}| \le 2\sqrt{2(a_1^2 + a_{2019}^2)} \le 2\sqrt{2}.$$
* Otherwise, there must exist $k$ such that $1 < k < 2019$, $a_k \ge a_{k-1}$ and $a_k \le a_{k+1}$. Then $|a_k - a_{k-1}| + |a_{k+1} - a_k| = |a_{k+1} - a_{k-1}|$. Hence, we only need to consider 2018 real numbers and can restate this problem as follows: Given 2018 real numbers $b_1, b_2, \dots, b_{2018}$ that satisfy $b_1^2 + b_2^2 + \dots + b_{2018}^2 \le 1$. Find the maximum value of
$$S = |b_1 - b_2| + |b_2 - b_3| + \dots + |b_{2018} - b_1|.$$
We obtain that
$$S \le (|b_1| + |b_2|) + (|b_2| + |b_3|) + \dots + (|b_{2018}| + |b_1|) \\ = 2 \sum_{1 \le i \le 2018} |b_i| \le 2\sqrt{2018}.$$
The equality holds when all absolute values are equal and the adjacent numbers have different signs. Note that 2018 is even then we choose $b_1 = -b_2 = b_3 = -b_4 = \dots = -b_{2018} = \frac{\sqrt{2018}}{2018}$.

Back to the original problem, we also get $\max S = 2\sqrt{2018}$ and the equality holds in many cases, such as
$$a_1 = -a_2 = a_3 = -a_4 = \dots = -a_{2018} = \frac{\sqrt{2018}}{2018},\ a_{2019} = 0.$$