Example 3 (2006 Shanghai Volume for Science) Three students proposed their own ideas for the problem "The inequality $x^{2}+25+\left|x^{3}-5 x^{2}\right| \geqslant a x$ holds for all $x$ in $[1,12]$, find the range of the real number $a$."
Student A says: "We only need the minimum value of the left side of the inequality to be no less than the maximum value of the right side."
Student B says: "Transform the inequality into a function of variable $x$ on the left and a constant on the right, then find the maximum and minimum values of the function."
Student C says: "Treat both sides of the inequality as functions of $x$, and plot the function graphs."
Referring to the above problem-solving ideas, what do you think is the correct conclusion they discussed, i.e., the range of $a$ is $\qquad$.
Analyzing, this problem is about finding the range of a parameter for which an inequality always holds. The problem setter simplifies the difficulty, setting up a scenario where three students provide their approaches to solving the problem, testing the ability to gather, analyze, and process information in practical operations.

Solution: Transform the original inequality into:
\[
\left|x^{3}-5 x^{2}\right| \geqslant -x^{2} + a x - 25
\]
which must hold for all \( x \) in the interval \([1, 12]\). We need to find the range of the real number \( a \).

Using the approach provided by Student A, the function \(\left|x^{3}-5 x^{2}\right|\) is non-negative on \([1, 12]\). Therefore, it is sufficient that the function \(-x^{2} + a x - 25\) is non-positive on \([1, 12]\). That is, \(-x^{2} + a x - 25 \leqslant 0\) must hold for all \( x \) in \([1, 12]\).

Using the approach provided by Student B, we can transform the above statement into: \(x + \frac{25}{x} \geqslant a\) for all \( x \) in \([1, 12]\).

Then, using the approach provided by Student C. By plotting the graph of the function \(f(x) = x + \frac{25}{x}\) on \([1, 12]\) (details omitted), it is easy to see that the function \(f(x) = x + \frac{25}{x}\) is monotonically decreasing on \([1, 5]\) and monotonically increasing on \([5, 12]\). Therefore, \(f(x)_{\min} = f(5) = 10\), so \(a \leqslant 10\), i.e., \(a \in (-\infty, 10]\).

Explanation: The parameter separation method can also be used to solve the problem directly.
Transform the original inequality into:
\[
x + \frac{25}{x} + \left|x^{2} - 5 x\right| \geqslant a.
\]

Let \(f(x) = x + \frac{25}{x} + \left|x^{2} - 5 x\right|\). Since \(x \in [1, 12]\), we have \(x + \frac{25}{x} \geqslant 2 \sqrt{x \cdot \frac{25}{x}} = 10\), with equality when \(x = 5\).
\(\left|x^{2} - 5 x\right| \geqslant 0\), with equality when \(x = 5\) or \(0\).
Therefore, \(f(x) \geqslant 10 + 0 = 10\), with equality when \(x = 5\). Hence, \(f(x)_{\text{min}} = 10\). Combining this with (1), we get \(a \leqslant 10\), i.e., \(a \in (-\infty, 10]\).