10. (20 points) Given the function
$$
f(x)=x^{4}+a x^{3}+b x^{2}+a x+1(a, b \in \mathbf{R})
$$

has at least one root. Find the minimum value of $a^{2}-b$.
10. Let the zero of $f(x)$ be $m$.

Then $m^{4}+a m^{3}+b m^{2}+a m+1=0$.
Dividing both sides by $m^{2}$, we get
$$
\begin{aligned}
b & =-\left(m^{2}+\frac{1}{m^{2}}\right)-a\left(m+\frac{1}{m}\right) \\
& =2-\left(m+\frac{1}{m}\right)^{2}-a\left(m+\frac{1}{m}\right) \\
\Rightarrow & a^{2}-b=a^{2}+a\left(m+\frac{1}{m}\right)+\left(m+\frac{1}{m}\right)^{2}-2 \\
& =\left(a+\frac{m+\frac{1}{m}}{2}\right)^{2}+3\left(\frac{m+\frac{1}{m}}{2}\right)^{2}-2 \\
& \geqslant 3\left(\frac{m+\frac{1}{m}}{2}\right)^{2}-2 \geqslant 1 .
\end{aligned}
$$

When $m=1, a=-1, b=0$, the equality holds. Thus, the minimum value of $a^{2}-b$ is 1.