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 of $a^{2}+b^{2}$.
［Solution］First, consider the equation
$$x+\frac{1}{x}=y$$

Here $y$ is a real number. This equation can be written as a quadratic equation in $x$: $x^{2}-y x+1=0$. It has real roots if and only if 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
$$y^{2}+a y+(b-2)=0$$

Solving for $y$, we get
$$y=\frac{-a \pm \sqrt{a^{2}-4(b-2)}}{2}$$

Since the original equation has at least one real root, by (1), at least one root in (2) must have an absolute value $\geqslant 2$, so
$$|a|+\sqrt{a^{2}-4(b-2)} \geqslant 4 \text {, }$$

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 $\frac{4}{5}$.