One, (20 points) If the inequality about $x$
$$
a x^{2}-4|x+1|+2 a<0
$$

has no real solutions, find the range of values for $a$.


Given the problem, for all real numbers $x$, we have $a x^{2}-4|x+1|+2 a \geqslant 0$.
Clearly, $a \neq 0$, so $a>0$.
Let $y=a x^{2}-4|x+1|+2 a$. Then
$$
y=\left\{\begin{array}{ll}
a x^{2}-4 x+2 a-4, & x \geqslant-1 ; \\
a x^{2}+4 x+2 a+4, & x<-1.
\end{array}\right.
$$
For the function
$$
y=a x^{2}-4 x+2 a-4 \quad (x \geqslant-1)
$$

the vertex (lowest point) of its graph is not below the x-axis.
Thus, $\frac{4 a(2 a-4)-(-4)^{2}}{4 a} \geqslant 0$.
Therefore, $a^{2}-2 a-2 \geqslant 0$.
Solving this, we get $a \leqslant 1-\sqrt{3}$ or $a \geqslant \sqrt{3}+1$.
Since $a>0$, we have $a \geqslant \sqrt{3}+1$.
(2) For the function $y=a x^{2}+4 x+2 a+4 \quad (x<-1)$, we have
$$
\left\{\begin{array}{l}
a(-1)^{2}+4(-1)+2 a+4 \geqslant 0
\end{array}\right.
$$

or $\left\{\begin{array}{l}
-\frac{4}{2 a} \leqslant-1, \\
\frac{4 a(2 a+4)-4^{2}}{4 a} \geqslant 0 .
\end{array}\right.$
Solving this, we get $a>2$ or $\sqrt{3}-1 \leqslant a \leqslant 2$.
Thus, $a \geqslant \sqrt{3}-1$.
In summary, the range of $a$ is $a \geqslant \sqrt{3}+1$.