<s>[INST] Solution for this inequality $a^2 - 4b + 8 > 0$ I dont know where I messed up, this is what I have done:
This is the isolated x of a function 
$$x = \frac{1}{2}\biggl(\pm\sqrt{a^2 - 4 b + 8} - a - 4\biggr)$$
The root above must be positive and greater than $0$ so a function could have inflection points.
So: $a^2 - 4 b + 8 > 0$ 
$b$ isolated:
$b<\frac{1}{4}\ (a^2 + 8)$
And $a$ isolated:
$a> \sqrt{4b-8}$
$b$ must be $\geq 2$ so that the root above could be real
With this in mind
if $a=1$ 
then $b<\frac{9}{4}\ $
And here is where I dont know what's wrong because It works even if I use $b$ values greater than $\frac{9}{4}\ $
 [/INST] To guarantee that exist
$$\sqrt{a^2 - 4 b + 8} $$
you have to set
$$a^2 - 4 b + 8 \geq 0\implies a^2\geq4b-8$$
and consider 2 cases


*

*$$4b-8<0 \implies a^2\geq0>4b-8$$

*$$4b-8\geq0 \implies -\sqrt{4b-8}\leq a^2\leq \sqrt{4b-8}$$


To solve
$$x = \frac{1}{2}\biggl(\pm\sqrt{a^2 - 4 b + 8} - a - 4\biggr)>0 \iff\pm\sqrt{a^2 - 4 b + 8} - a - 4>0$$
consider two cases
CASE 1
$$\sqrt{a^2 - 4 b + 8} - a - 4>0 \implies\sqrt{a^2 - 4 b + 8}>a + 4$$
The system to solve is
$$\begin{cases}\sqrt{a^2 - 4 b + 8}>a + 4\\ a^2 - 4 b + 8 \geq 0\end{cases}$$
If:
$a+4<0$ you only have to check that $a^2 - 4 b + 8 \geq 0$
$a+4\geq0$ you can square and the system becomes
$$\begin{cases}a^2 - 4 b + 8>a^2+8a + 16\\ a^2 - 4 b + 8 \geq 0\end{cases}\implies \begin{cases}2a+b+2<0\\ a^2 - 4 b + 8 \geq 0\end{cases}$$
CASE 2
$$-\sqrt{a^2 - 4 b + 8} - a - 4>0 \implies a+4<-\sqrt{a^2 - 4 b + 8}$$
The system to solve is
$$\begin{cases}a+4<-\sqrt{a^2 - 4 b + 8}\\ a^2 - 4 b + 8 \geq 0\end{cases}$$
If:
$a+4\geq0$ no solutions
$a+4<0$ you can square reversing the sign and the system becomes
$$\begin{cases}a^2+8a + 16>a^2 - 4 b + 8\\ a^2 - 4 b + 8 \geq 0\end{cases}\implies \begin{cases}2a+b+2>0\\ a^2 - 4 b + 8 \geq 0\end{cases}$$
</s>