China Mathematical Competition

For function $f(x)$ and real numbers $x_1, x_2, x_3$ satisfying the conditions, let $f(x_1) = f(x_2) = f(x_3) = c$, and then $x_1, x_2, x_3$ are the three real roots of cubic equation $x^3 + a x^2 + b x - c = 0$. By Vieta's formulas, we know that
$$a = -(x_1 + x_2 + x_3), \quad b = x_1 x_2 + x_2 x_3 + x_3 x_1.$$
Further, by the condition we know that $x_2 - x_1 \ge 1$, $x_3 - x_2 \ge 1$, $x_3 - x_1 \ge 2$, and thus
$$\begin{aligned} a^2 - 3b &= x_1^2 + x_2^2 + x_3^2 - x_1 x_2 - x_2 x_3 - x_3 x_1 \\ &= \frac{1}{2}((x_1 - x_2)^2 + (x_2 - x_3)^2 + (x_3 - x_1)^2) \\ &\ge \frac{1}{2}(1 + 1 + 4) = 3, \end{aligned}$$
i.e., $b \le \frac{a^2}{3} - 1$.
If $|a| \ge \sqrt{3}$, then $|a| + 2|b| \ge |a| + \sqrt{3}$.
If $0 \le |a| < \sqrt{3}$, then $b \le \frac{a^2}{3} - 1 < 0$. Notice that at this point
$$\begin{aligned} |a| - \frac{3}{4} &< \sqrt{3} - \frac{3}{4}, \\ |a| + 2|b| &\ge |a| + \frac{2}{3}(3 - a^2) \\ &\quad - \frac{2}{3}\left(|a| - \frac{3}{4}\right)^2 + \frac{19}{8} \\ &\quad - \frac{2}{3}\left(\sqrt{3} - \frac{3}{4}\right)^2 + \frac{19}{8} = \sqrt{3}. \end{aligned}$$
The above shows that $|a| + 2|b| \ge \sqrt{3}$.
On the other hand, when $a = \sqrt{3}$, $b = 0$, real numbers $x_1 = -1 - \frac{\sqrt{3}}{3}$, $x_2 = -\frac{\sqrt{3}}{3}$, $x_3 = 1 - \frac{\sqrt{3}}{3}$ satisfy the condition because the values of $F_1$, $F_2$, $F_3$ are all $\frac{2\sqrt{3}}{9}$. At this point $|a| + 2|b| = \sqrt{3}$.
In conclusion, the minimum of $|a| + 2|b|$ is $\sqrt{3}$.