II. (25 points) Let $a, b$ be real numbers that make the equation $x^{4}+a x^{3}+b x^{2}+a x+1=0$ have at least one real root. For all such pairs $(a, b)$, find the minimum value of $a^{2}+b^{2}$.

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Obviously, $x \neq 0$, dividing both sides of the original equation by $x^{2}$, we get
$$
\left(x+\frac{1}{x}\right)^{2}+a\left(x+\frac{1}{x}\right)+(b-2)=0 \text {. }
$$

By the inequality $(a+b)^{2} \geqslant 4 a b$, we know
$$
\left|x+\frac{1}{x}\right| \geqslant 2 \text {. }
$$

Let $y=x+\frac{1}{x}$, then equation (1) becomes
$$
y^{2}+a y+(b-2)=0 .(|y| \geqslant 2) \text {. }
$$

From $|y| \geqslant 2$, we can get
$$
\begin{array}{l}
\left|\frac{a}{2}\right|+\left|\frac{\sqrt{a^{2}-4(b-2)}}{2}\right| \\
\geqslant\left|\frac{-a \pm \sqrt{a^{2}-4(b-2)}}{2}\right|
\end{array}
$$
$\geqslant 2$.
Thus, $\sqrt{a^{2}-4(b-2)} \geqslant 4-|a|$.
To find the minimum value of $a^{2}+b^{2}$, without loss of generality, assume $0 \leqslant a \leqslant 4$, then
$$
\sqrt{a^{2}-4}-(b-2) \geqslant \geqslant 4--a>0 \text {. }
$$

When $b+2 \geqslant 0$, we have
$$
4 a^{2} \geqslant b^{2}+4 b+4 \text {. }
$$

Completing the square, we get
$$
a^{2}+b^{2} \geqslant \frac{5}{4}\left(b+\frac{2}{5}\right)^{2}+\frac{4}{5} \text {. }
$$
$\therefore$ When $b=-\frac{2}{5}$, $a^{2}+b^{2}$ takes the minimum value $\frac{4}{5}$.
When $b+2\frac{4}{5}$, the above conclusion still holds.