Example 7. Let $a, b$ be real numbers, and the equation $x^{4}+a x^{3}+b x^{2}$ $+a x+1=0$ has at least one real solution. Determine the minimum possible value of $a^{2}+b^{2}$.

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Let $x_{0}$ be a real root of the equation, obviously $x_{0} \neq 0$. Therefore, we have $x_{0}^{2}+a x_{0}+b+\frac{a}{x_{0}}+\frac{1}{x_{0}^{2}}=0$.
The line $l:\left(x_{0}+\frac{1}{x_{0}}\right) a+b+\left(x_{0}^{2}+\frac{1}{x_{0}^{2}}\right)=0$,
The circle $O: a^{2}+b^{2}=R^{2}$.
By Theorem 1 (2), (3), we know
$$
\frac{x_{0}^{2}+\frac{1}{x_{0}^{2}}}{\sqrt{\left(x_{0}+\frac{1}{x_{0}}\right)^{2}+1}} \leqslant R \text {, i.e., } R^{2} \geqslant \frac{\left(x_{0}^{2}+\frac{1}{x_{0}^{2}}\right)^{2}}{\left(x_{0}^{2}+\frac{1}{x_{0}^{2}}+3\right)} \text {. }
$$

Therefore,
$$
\begin{array}{l}
\frac{1}{R^{2}} \leqslant \frac{x_{0}^{2}+\frac{1}{x_{0}^{2}}+3}{\left(x_{0}^{2}+\frac{1}{x_{0}^{2}}\right)^{2}}=\frac{1}{x_{0}^{2}+\frac{1}{x_{0}^{2}}}+\frac{3}{\left(x_{0}^{2}+\frac{1}{x_{0}^{2}}\right)^{2}} . \\
\because x_{0}^{2}+\frac{1}{x_{0}^{2}} \geqslant 2, \therefore \frac{1}{R^{2}} \geqslant \frac{1}{2}+\frac{3}{4}=\frac{5}{4} . \text { i.e., } a^{2}+b^{2}
\end{array}
$$
$\geqslant \frac{4}{5}$. When and only when $x_{0}^{2}=\frac{1}{x_{0}^{2}}, x_{0}= \pm 1$, the equality holds. Substituting $x_{0}= \pm 1$ into the equation, we easily get $a= \pm \frac{4}{5}, b=-\frac{2}{5}$.
$\therefore a= \pm \frac{4}{5}, b=-\frac{2}{5}$, when $a^{2}+b^{2}$ takes the minimum value $\frac{4}{5}$.