China Mathematical Competition

$f'(x) = 3a x^2 + 2b x + c$. We have
$$\begin{cases} f'(0) = c, \\ f'\left(\frac{1}{2}\right) = \frac{3}{4}a + b + c, \\ f'(1) = 3a + 2b + c. \end{cases}$$
Then
$$3a = 2f'(0) + 2f'(1) - 4f'\left(\frac{1}{2}\right).$$
We get
$$\begin{aligned} 3 | a | &= \left| 2f'(0) + 2f'(1) - 4f'\left(\frac{1}{2}\right) \right| \\ &\le 2 |f'(0)| + 2 |f'(1)| + 4 \left| f'\left(\frac{1}{2}\right) \right| \le 8. \end{aligned}$$
Therefore, $a \le \frac{8}{3}$. Furthermore, it is easy to find that $f(x) = \frac{8}{3}x^3 - 4x^2 + x + m$ (where $m$ is any constant) satisfies the given condition. Therefore, the maximum value of $a$ is $\frac{8}{3}$.

Let $g(x) = f'(x) + 1$. Then $0 \le g(x) \le 2$ for $0 \le x \le 1$. Let $z = 2x - 1$. Then $x = \frac{z+1}{2}$ and $-1 \le z \le 1$. Let
$$h(z) = g\left(\frac{z+1}{2}\right) = \frac{3a}{4}z^2 + \frac{3a+2b}{2}z + \frac{3a}{4} + b + c + 1.$$
It is easy to check that $0 \le h(z) \le 2$ and $0 \le h(-z) \le 2$ for $-1 \le z \le 1$.
Therefore, $0 \le \frac{h(z) + h(-z)}{2} \le 2$ for $-1 \le z \le 1$. And that is
$$0 \le \frac{3a}{4}z^2 + \frac{3a}{4} + b + c + 1 \le 2.$$
Then we have $\frac{3a}{4} + b + c + 1 \ge 0$ and $\frac{3a}{4}z^2 \le 2$. From $0 \le z^2 \le 1$ we get $a \le \frac{8}{3}$.
As $f(x) = \frac{8}{3}x^3 - 4x^2 + x + m$ (where $m$ is any constant) satisfies the given condition. We obtain that the maximum value of $a$ is $\frac{8}{3}$. $\square$