9. If the solution set of the inequality with respect to $x$
$$
a x^{2}-|x+1|+2 a<0
$$

is an empty set, then the range of values for $a$ is $\qquad$.
9. $\left[\frac{\sqrt{3}+1}{4},+\infty\right)$.

From the problem, for all real numbers $x$, we have
$$
a x^{2}-|x+1|+2 a \geqslant 0 \text {, }
$$

which means $a \geqslant f(x)=\frac{|x+1|}{x^{2}+2}$.
Therefore, $a \geqslant \max f(x)$.
Let $t=x+1$, then
$$
f(x)=\frac{|t|}{(t-1)^{2}+2}=\frac{|t|}{t^{2}-2 t+3} \text {. }
$$
(1) When $x \geqslant-1$, i.e., $t \geqslant 0$, we have
$$
f(x)=\frac{t}{t^{2}-2 t+3} \leqslant \frac{t}{2 \sqrt{3} t-2 t}=\frac{\sqrt{3}+1}{4} \text {. }
$$

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
$$
f(x)=\frac{-t}{t^{2}-2 t+3} \leqslant \frac{-t}{-2 \sqrt{3} t-2 t}=\frac{\sqrt{3}-1}{4} \text {. }
$$

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