5. The inequality $x^{2}+25+\left|x^{3}-5 x^{2}\right| \geqslant a x$ holds for all $x$ in the interval $[1,12]$. Then the maximum value of the real number $a$ is $\qquad$.
5. 10 Explanation: For all $x \in [1,12]$, we have $a \leqslant x + \frac{25}{x} + x|x-5|$, thus $a_{\max} = \left(x + \frac{25}{x} + x|x-5|\right)_{\text{min}}$. $\because x + \frac{25}{x} \geqslant 10$, the equality holds when $x=5; x|x-5| \geqslant 0$, the equality holds when $x=5, \therefore x + \frac{25}{x} + x|x-5| \geqslant 10$, the equality holds when $x=5$, hence $a_{\max} = \left(x + \frac{25}{x} + x|x-5|\right)_{\min} = 10$.