由于\( f(x) \)在\( [0,1] \)连续，从而存在\( x_0 \in [0,1] \)使\(f(x)\)在\(x=x_0\)处取最大值。
(1)如果 \( f(x_0)=0 \)，则\(f(x)=0, x\in[0,1]\)，所以这段曲线的弧长\(s=1 \le 3 \)得证
(2)如果\( f(x_0)>0 \)，有\( f'(x_0) = 0 \), 又因为\( f”(x) \le 0 \)，所以有\( f'(x) \ge 0, x \in [0,c] \), \( f'(x) \le 0, x \in [c,1] \)

从而：\[ \begin{align}
s & = \int_0^1 {\sqrt{1+|f'(x)|^2}dx}\\
~ &= \int_0^c {\sqrt{1+|f'(x)|^2}dx} +\int_c^1 {\sqrt{1+|f'(x)|^2}dx} \\
~ & \le \int_0^c { (1+f'(x) )dx} +\int_c^1 {(1-f'(x))dx} \\
~ & = [x+f(x)]^c_0 +[x-f(x)]^1_c \\
~&= 1+2f(c) \le 3\\
\end{align}\]