Consider $a$ and $b$ as points on the real number line. Then $|a - b|$ represents the distance between the two points. Similarly, $|b - c|$ represents the distance between the points $b$ and $c$, and so on. Hence, consider a path that starts at $a$, then jumps to $b$, then jumps to $c$, and then back to $a$. Then the total distance covered by the path is |a−b|+|b−c|+|c−a|=20. We claim that $|a - b|$ is at most 10. Consider the portion of the path that goes from $b$ to $c$ to $a$. The sum of these segments must add up to at least the distance from $b$ to $a$. (Think of the maxim, "The shortest distance between two points is a straight line." This is a case of the triangle inequality.) In other words, |b−c|+|c−a|≥|b−a|=|a−b|. Hence, |a−b|+|b−c|+|c−a|≥2|a−b|, so $2|a - b| \le 20$, or $|a - b| \le 10$. To show that the maximum possible value of $|a - b|$ is 10, we must provide a set of values in which $|a - b|$ is equal to 10. If we set $a = 10$ and all 2 other variables equal to 0, then the given equation is satisfied and $|a - b| = 10$. Hence, the maximum possible value of $|a - b|$ is 10.
