16. The quartic equation $x^{4}+a x^{3}+b x^{2}+a x+1=0$ has real roots, find the minimum value of $a^{2}+b^{2}$.
16. The original equation can be transformed into $\left(x+\frac{1}{x}\right)^{2}+a\left(x+\frac{1}{x}\right)+b-2=0$, and the solution of the equation can be obtained by solving two quadratic equations:
$$
\begin{array}{c}
y^{2}+a y+b-2=0 \\
x+\frac{1}{x}=y \quad \text { (2) }
\end{array}
$$
(2) has real roots, $x^{2}-y x+1=0$ has real roots, $y^{2}-4 \geqslant 0 \quad|y| \geqslant 2$, from (1), $y=$ $\frac{-a \pm \sqrt{a^{2}-4(b-2)}}{2},|y| \geqslant 2$ when, $|a|+\sqrt{a^{2}-4(b-2)} \geqslant 4$, $\sqrt{a^{2}-4(b-2)} \geqslant 4-|a|$ removing the square root and simplifying gives: $2|a| \geqslant 2+b, 4 a^{2} \geqslant b^{2}+4 b+4$, $4\left(a^{2}+b^{2}\right) \geqslant 5 b^{2}+4 b+4=5\left(b+\frac{2}{5}\right)^{2}+4-\frac{4}{5}, a^{2}+b^{2} \geqslant \frac{5}{4}\left(b+\frac{2}{5}\right)^{2}+$ $\frac{4}{5} \geqslant \frac{4}{5}$, taking $b=-\frac{2}{5}, a=\frac{4}{5}$, then $a^{2}+b^{2}=\frac{4}{5}$, at this time (1) is $y^{2}+\frac{4}{5} y-\frac{12}{5}=$ 0, with roots $y=-2$, (2) has real roots $x=-1$, so $\left(a^{2}+b^{2}\right)_{\min }=\frac{4}{5}$.