3. For all real numbers $x$, the inequality $x^{4}+(a-1) x^{2}+1 \geqslant 0$ always holds. Then the range of values for $a$ is ().
(A) $a \geqslant-1$
(B) $a \geqslant 0$
(C) $a \leqslant 3$
(D) $a \leqslant 1$
3. A. When $x=0$, $x^{4}+(a-1) x^{2}+1=1 \geqslant 0$ always holds. When $x \neq 0$, the original inequality is equivalent to
$$
x^{2}+\frac{1}{x^{2}} \geqslant 1-a .
$$

Since the minimum value of $x^{2}+\frac{1}{x^{2}}$ is 2, we get $1-a \leqslant 2$, which means $a \geqslant -1$.