Example 7 (2006 Zhejiang Province High School Mathematics Summer Camp Problem) If the solution set of the inequality about $x$
$$
a x^{2}-|x+1|+2 a<0
$$

is an empty set, then the range of values for $a$ is $\qquad$ .
From the problem, we know that for all real numbers $x$, we have $a x^{2}-|x+1|+2 a \geqslant 0$,
which means $a \geqslant f(x)=\frac{|x+1|}{x^{2}+2}$.
Therefore, $a \geqslant \max f(x)$.
(1) When $x \geqslant-1$, i.e., $t \geqslant 0$, we have $f(x)=\frac{t}{t^{2}-2 t+3} \leqslant \frac{t}{2 \sqrt{3} t-2 t}=\frac{\sqrt{3}+1}{4}$.

The equality holds if and only if $t=\sqrt{3}$, i.e., $x=\sqrt{3}-1$.
(2) When $x<-1$, i.e., $t<0$, we have $f(x)=\frac{-t}{t^{2}-2 t+3} \leqslant \frac{-t}{-2 \sqrt{3} t-2 t}=\frac{\sqrt{3}-1}{4}$.

The equality holds if and only if $t=-\sqrt{3}$, i.e., $x=-\sqrt{3}-1$.
In summary, $\max _{\lrcorner \in \mathbb{K}} f(x)=\frac{\sqrt{3}+1}{4}$.
Therefore, the range of $a$ is $\left[\frac{\sqrt{3}+1}{4},+\infty\right)$.