6. Let $a$ and $b$ be real numbers such that the equation $x^{4} + a x^{3} + b x^{2} + a x + 1 = 0$ has at least one real root. For all such pairs of real numbers $(a, b)$, find the minimum value of $a$. (15th IMO Problem)
6. First consider the equation
$$x+\frac{1}{x}=y$$

where $y$ is a real number. This equation can be written as a quadratic equation in $x$:
$$x^{2}-y \bar{x}+1=0$$

It has real roots if and only if its discriminant is greater than or equal to 0, i.e.,
$$y^{2}-4 \geqslant 0,|y| \geqslant 2$$

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 $y$ gives
$$y=\frac{-a \pm \sqrt{a^{2}-4(b-2)}}{2}$$

Since the original equation must have at least one real root, from equation (1) we know that at least one root of equation (3) must have an absolute value greater than or equal to 2. Therefore,
$$\begin{array}{l}
|a|+\sqrt{a^{2}-4(b-2)} \geqslant 4 \Rightarrow \sqrt{a^{2}-4(b-2)} \geqslant 4-|a| \Rightarrow \\
8|a| \geqslant 8+4 b \Rightarrow 4 a^{2} \geqslant b^{2}+4 b+4 \Rightarrow 4\left(a^{2}+b^{2}\right) \geqslant \\
5 b^{2}+4 b+4 \Rightarrow a^{2}+b^{2} \geqslant \frac{5}{4}\left(b+\frac{2}{5}\right)^{2}+\frac{5}{4}
\end{array}$$

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