12. If the inequality $x^{3}-a x+1 \geqslant 0$ holds for all $x \in[-1,1]$, then the range of the real number $a$ is $\qquad$
When $x=0$, $a \in \mathbf{R} ; 0<x \leqslant 1$  when, $a \leqslant x^{2}+\frac{1}{x} ; -1 \leqslant x<0$  when, $a \geqslant x^{2}+\frac{1}{x}$. Since $x^{2}+\frac{1}{x}=x^{2}+\frac{1}{2 x}+\frac{1}{2 x} \geqslant 3 \sqrt[3]{\frac{1}{4}}=\frac{3 \sqrt[3]{2}}{2}$, then $a \leqslant \frac{3 \sqrt[3]{2}}{2}$; also, $(-1)^{2}+\frac{1}{-1}=0 \Rightarrow a \geqslant 0$. Therefore, $a \in\left[0, \frac{3 \sqrt[3]{2}}{2}\right]$.