Three. (20 points) Let real numbers $a, b$ satisfy $a^{2}+b^{2} \leqslant \frac{4}{5}$. Determine whether the equation $x^{4}+a x^{3}+b x^{2}+a x+1=0$ has real solutions for $x$.


Three, if the equation has a real root $x_{0}\left(x_{0} \neq 0\right)$, substituting it into the equation, we get
$$
\left(x_{0}+\frac{1}{x_{0}}\right) a+b+\left(x_{0}^{2}+\frac{1}{x_{0}^{2}}\right)=0 .
$$

Considering $a, b$ as variables, the problem transforms into:
whether the line $\left(x_{0}+\frac{1}{x_{0}}\right) a+b+\left(x_{0}^{2}+\frac{1}{x_{0}^{2}}\right)=0$ intersects the disk $a^{2}+b^{2} \leqslant \frac{4}{5}$.

Using the point-to-line distance formula, the distance from the origin $(0,0)$ to the line is
$$
\begin{aligned}
d & =\frac{x_{0}^{2}+\frac{1}{x_{0}^{2}}}{\sqrt{\left(x_{0}+\frac{1}{x_{0}}\right)^{2}+1}} \\
& =\sqrt{x_{0}^{2}+\frac{1}{x_{0}^{2}}+3}-\frac{3}{\sqrt{x_{0}^{2}+\frac{1}{x_{0}^{2}}+3}} .
\end{aligned}
$$

Since $x_{0}^{2}+\frac{1}{x_{0}^{2}} \geqslant 2, f(t)=\sqrt{t+3}-\frac{3}{\sqrt{t+3}}$ is an increasing function on $[2,+\infty)$, thus,
$$
d \geqslant \sqrt{5}-\frac{3}{\sqrt{5}}=\frac{2}{\sqrt{5}} .
$$

Therefore, the equation has real roots if and only if $d=\frac{2}{\sqrt{5}}$, at which point
$$
x_{0}^{2}+\frac{1}{x_{0}^{2}}=2, x_{0}= \pm 1 \text {. }
$$

Substituting into the equation, we get
$$
\left\{\begin{array} { l } 
{ 2 a + b + 2 = 0 , } \\
{ a ^ { 2 } + b ^ { 2 } = \frac { 4 } { 5 } ; }
\end{array} \left\{\begin{array}{l}
b-2 a+2=0, \\
a^{2}+b^{2}=\frac{4}{5} .
\end{array}\right.\right.
$$

Solving, we get $a=-\frac{4}{5}, b=-\frac{2}{5}$ or $a=\frac{4}{5}, b=-\frac{2}{5}$.
The real roots of the original equation are $1, -1$.
Additionally, the equation has no other real solutions.