A19 (15-3, Sweden) Let $a, b$ be real numbers, and $x^{4}+a x^{3}+b x^{2}+a x+1=0$ has at least one real root. Find the minimum value of $a^{2}+b^{2}$.


It is known that $x=0$ is not a root of the original equation. The original equation can thus be transformed into
$$
\left(x+\frac{1}{x}\right)^{2}+a\left(x+\frac{1}{x}\right)+(b-2)=0 .
$$

Let $y=x+\frac{1}{x}$, then it is easy to prove that $|y| \geqslant 2$.
Solving the equation $y^{2}+a y+(b-2)=0$ yields
$$
y=\frac{-a \pm \sqrt{a^{2}-4(b-2)}}{2}
$$

At least one of the roots has an absolute value $\geqslant 2$, which is equivalent to
$$
|a|+\sqrt{a^{2}-4(b-2)} \geqslant 4 .
$$

That is,
$$
\sqrt{a^{2}-4(b-2)} \geqslant 4-|a| .
$$
(i) When $|a| \geqslant 4$, $a^{2}+b^{2} \geqslant 16$.
(ii) When $|a|<4$, squaring both sides and simplifying, we get
$$
2|a| \geqslant 2+b .
$$
If $b \leqslant -2$, then $a^{2}+b^{2} \geqslant 4$. Otherwise, if $b \geqslant -2$, squaring both sides again, we get
$$
4 a^{2} \geqslant 4+4 b+b^{2} \text {, }
$$

or
$$
\begin{aligned}
4\left(a^{2}+b^{2}\right) & \geqslant 5 b^{2}+4 b+4 \\
& =5\left(b+\frac{2}{5}\right)^{2}+\frac{16}{5} \\
& \geqslant \frac{16}{5} .
\end{aligned}
$$

Thus, when $b=-\frac{2}{5}$ and $a= \pm \frac{4}{5}$, 
$a^{2}+b^{2}$ reaches its minimum value $\frac{4}{5}$. Upon verification, the original equation has real roots $x=\mp 1$.