Let $f(x)$ have two (not necessarily distinct) real roots, $x = 1 - a, -1 + b$ ($a, b \ge 0$) such that: $-1 + b \le 1 - a$, or $0 \le a + b \le 2$. Setting $f(x) = [x - (1 -a)][x - (-1 + b)] = x^2 - (1 - a - 1 + b)x + (1 - a)(-1 + b)$ (i), the minimum value occurs at: $f'(x) = 2x - (b - a) = 0 \Rightarrow x = (b - a)/2$ (and $f''(x) = 2 > 0$, a global minimum) or $f((b-a)/2) = [(b - a)/2]^2 - (b - a)*(b - a)/2 + (a + b - ab - 1) = -(1/4)(b - a)^2 + a + b - ab - 1$; or $(-1/4)(a + b)^2 + (a + b) - 1$; or $-[(a + b)/2 -1]^2$; and $[(a + b)/2 - 1]^2$ is the maximum value of $|f(x)|$ over $[-1 + b, 1 - a]$. The maximum values of $|f(x)|$ over $[-1, -1 + b]$ and $[1 - a, 1]$ respectively occur at $x = \pm 1$ since $|f(x)|$ is respectively decreasing (increasing) on these intervals. This yields the values $|f(-1)| = 2a - ab$ and $|f(1)| = 2b - ab$. Now, if the maximum value of $|f(x)|$ over $[-1, 1]$ equals 1, then we require non-negative $(a, b)$ that satisfies the feasible region: $0 \le a, b$ $0 \le a + b \le 2$; $0 \le [(a + b)/2 - 1]^2 \le 1$; $0 \le 2a - ab \le 1$; $0 \le 2b - ab \le 1$ which solves as $R_{1} \cup R_{2}$, where: $R_{1} = \{(a, b) : 0 \le a \le 1/2, 0 \le b \le 1/(2 - a)\}; R_{2} = \{(a, b) : 1/2 < a \le 1, 2 - 1/a \le b \le 1/(2 - a)\}$ that has extreme points $(a, b) = (0, 0); (0, 1/2); (1/2, 0); (1, 1)$. Finally, we check each extreme point against $|f'(x)| = |2x - (b - a)| = 2|x - (b - a)/2|$ (ii) over $-1 \le x \le 1$. Notice that the vertex of (ii) lies entirely within $[-1, 1]$, which implies that the maximum of (ii) occurs at $x = \pm 1$: $(a, b) = (0, 0): 2|\pm 1 - (0 - 0)/2| = 2|\pm1| =$ $\fbox{2}$; $(a, b) = (0, 1/2): 2|\pm 1 - (1/2 - 0)/2| = 2|-1/4 \pm 1| =$ $\fbox{3/2, 5/2}$; $(a, b) = (1/2, 0): 2|\pm 1 - (0 -1/2)/2| = 2|1/4 \pm 1| =$ $\fbox{5/2, 3/2}$; $(a, b) = (1, 1): 2|\pm 1 - (1 - 1)/2| = 2|\pm 1| =$ $\fbox{2}$ which shows that $|f'(x)| \ge 1$ for all $x \in [-1, 1]$. QED

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