4- 47 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. Try to find the minimum value that $a^{2}+b^{2}$ can take.
［Solution］First, consider the equation
$$
x+\frac{1}{x}=y,
$$

where $y$ is a real number. This equation can be rewritten as a quadratic equation in $x$: $x^{2}-y x+1=0$. The necessary and sufficient condition for it to have real roots is that its discriminant is not less than 0, i.e.,
$$
y^{2}-4 \geqslant 0 \text { or }|y| \geqslant 2 \text {. }
$$

The original equation can be written as
$$
\left(x+\frac{1}{x}\right)^{2}+a\left(x+\frac{1}{x}\right)+(b-2)=0 .
$$

Let $y=x+\frac{1}{x}$, then we get
$$
y^{2}+a y+(b-2)=0 \text {, }
$$

Solving for $y$ gives $y=\frac{-a \pm \sqrt{a^{2}-4(b-2)}}{2}$. Since the original equation has at least one real root, by (1), at least one of the roots in (2) must have an absolute value $\geqslant 2$, so
$$
|a|+\sqrt{a^{2}-4(b-2)} \geqslant 4,
$$

which implies
$$
\sqrt{a^{2}-4(b-2)} \geqslant 4-|a|,
$$

Removing the square root, we get $\quad 8|a| \geqslant 8+4 b$,
Squaring both sides, we get $4 a^{2} \geqslant b^{2}+4 b+4$,
Thus,
$$
4\left(a^{2}+b^{2}\right) \geqslant 5 b^{2}+4 b+4,
$$

Therefore,
$$
a^{2}+b^{2} \geqslant \frac{5}{4}\left(b+\frac{2}{5}\right)^{2}+\frac{4}{5} .
$$

From this, we can see that when $b=-\frac{2}{5}$, $a^{2}+b^{2}$ achieves its minimum value $\frac{4}{5}$.