8. If the solution set of the inequality $a x^{2}-|x+1|+2 a<0$ with respect to $x$ is an empty set, then the range of values for $a$ is
8. $\left[\frac{\sqrt{3}+1}{4},+\infty\right)$ Detailed Explanation: According to the problem, for all $x$, we have $a x^{2}-|x+1|+2 a \geqslant 0$.

That is, $a \geqslant \frac{|x+1|}{x^{2}+2}$. Therefore, $a \geqslant \max \left\{\frac{|x+1|}{x^{2}+2}\right\}$. Let $t=x+1$, then $\frac{|x+1|}{x^{2}+2}=\frac{|t|}{t^{2}-2 t+3}$.
(1) When $t \geqslant 0$, $\frac{|t|}{t^{2}-2 t+3}=\frac{1}{t-2+\frac{3}{t}} \leqslant \frac{1}{2 \sqrt{3}-2}=\frac{\sqrt{3}+1}{4}$.
(2) When $t<0$, $\frac{|t|}{t^{2}-2 t+3}=\frac{-t}{t^{2}-2 t+3} \leqslant \frac{1}{-t+2-\frac{3}{t}} \leqslant \frac{\sqrt{3}-1}{4}$.

The equality holds if and only if $t=-\sqrt{3}$, i.e., $x=-\sqrt{3}-1$.
In summary, the maximum value of $\frac{|x+1|}{x^{2}+2}$ is $\frac{\sqrt{3}+1}{4}$. Therefore, the range of $a$ is $\left[\frac{\sqrt{3}+1}{4},+\infty\right)$.