12. If the real number $a$ makes the inequality $|x-2 a|+|2 x-a| \geq a^{2}$ hold for any real number $x$, then the range of the real number $a$ is
Solve $\left[-\frac{3}{2}, \frac{3}{2}\right]$.
Analysis Let $x=k a(k \in \mathbf{R})$, then $|k a-2 a|+|2 k a-a| \geq a^{2}$, which means for $\forall k \in \mathbf{R}$, it always holds that
$$
|k-2|+|2 k-1| \geq|a| \Leftrightarrow|k-2|+|2 k-1|_{\min } \geq|a|
$$

And $|k-2|+|2 k-1|_{\min }=\frac{3}{2}$, so $|a| \leq \frac{3}{2} \Leftrightarrow a \in\left[-\frac{3}{2}, \frac{3}{2}\right]$.