Example 1 If the inequality about $x$
$$
a x^{2}-|x+1|+2 a<0
$$

has a solution set of $\varnothing$, find the range of real number $a$.
Explanation: First, convert the condition to "for any $x \in \mathbf{R}$,

there is $a x^{2}-|x+1|+2 a \geqslant 0$, then separate the parameter $a$, transforming it into a function problem.
By the problem, for any $x \in \mathbf{R}$, we have
$$
a x^{2}-|x+1|+2 a \geqslant 0 \text {, }
$$

which means $a \geqslant f(x)=\frac{|x+1|}{x^{2}+2}$.
Thus, $a \geqslant f(x)_{\max }$.
Let $t=x+1$, then
$$
f(x)=g(t)=\frac{|t|}{t^{2}-2 t+3} \text {. }
$$
(1) When $t=0$, $g(0)=0$;
(2) When $t>0$, $g(t)_{\max }=g(\sqrt{3})=\frac{\sqrt{3}+1}{4}$;
(3) When $t<0$, $g(t)_{\max }=g(-\sqrt{3})=\frac{\sqrt{3}-1}{4}$.

Therefore, $f(x)_{\max }=g(t)_{\max }=\frac{\sqrt{3}+1}{4}$.
Hence, $a \in\left[\frac{\sqrt{3}+1}{4},+\infty\right)$.
Note: To find the range of the real number $a$, it is only necessary to find the maximum value of the function $f(x)$.