1. Given the quadratic function
$$
\begin{aligned}
f(x)= & a(3 a+2 c) x^{2}-2 b(2 a+c) x+ \\
& b^{2}+(c+a)^{2}(a, b, c \in \mathbf{R}).
\end{aligned}
$$

Assume that for any $x \in \mathbf{R}$, we have $f(x) \leqslant 1$. Find the maximum value of $|a b|$. (Yang Xiaoming, Zhao Bin)
1. Completing the square for $f(x)$ with respect to $c$ yields
$$
\begin{array}{l}
f(x)=c^{2}+2\left(a x^{2}-b x+a\right) c+3(a x)^{2}- \\
\quad 4 a b x+b^{2}+a^{2} \\
\leqslant 1,
\end{array}
$$

which implies
$$
\begin{array}{l}
\left(c+a x^{2}-b x+a\right)^{2}+\left(1-x^{2}\right)(a x-b)^{2} \leqslant 1 \\
\Rightarrow\left(1-x^{2}\right)(a x-b)^{2} \leqslant 1 . \\
\text { Let } x= \pm \frac{\sqrt{3}}{3} \text {. Then } \\
\frac{4 \sqrt{3}}{3}|a b| \leqslant \frac{3}{2} \Rightarrow|a b| \leqslant \frac{3 \sqrt{3}}{8},
\end{array}
$$

Let $x= \pm \frac{\sqrt{3}}{3}$. Then

When $a=\frac{3 \sqrt{2}}{4}, b=-\frac{\sqrt{6}}{4}, c=-\frac{5 \sqrt{2}}{4}$, equality holds in equation (1). At this point,
$$
\begin{array}{l}
f(x)=-\frac{3}{8} x^{2}+\frac{\sqrt{3}}{4} x+\frac{7}{8} \\
=-\frac{3}{8}\left(x-\frac{\sqrt{3}}{3}\right)^{2}+1 \leqslant 1 .
\end{array}
$$

Therefore, the maximum value sought is $\frac{3 \sqrt{3}}{8}$.