OMG, I hope I post it in the correct forum, I don't want to get infracted

Heyah, I have this problem to solve, I tried a bit and came to the point where I don't know how to continue. I think I just have to prove my results.. Dunno. I just want to add,that english is not my mother tongue, and I might name some things incorrectly, so just correct my mistake but don't friggin kill me :P . Here it is:

Define all $\displaystyle x\in \mathbb{R}$ with $\displaystyle x\neq 1,2$ and:

$\displaystyle \frac {(x^{16}+5)max\left \{ x^{2}+x-2,2x+4 \right \}}{(x-2)^{2}(x-1)}\geq \frac {(4x+8)(x^{16}+5)}{(x-2)(x^{2}-4)}$

First I have to simplyfy , so I checked this first :

$\displaystyle (x^{16}+5)\geq 1\geq 0$

so I don't have to change the inequality sign:

$\displaystyle \frac {max...}{(x-2)(x-2)(x-1)}\geq \frac {(4x+8)}{(x-2)(x^{2}-4)}$

$\displaystyle (x-2)< 0 \Rightarrow \frac {max...}{(x-2)(x-1)}\leq \frac {(4x+8)}{(x^{2}-4)}$

Here I've changed the inequality sign and I also noticed that:

$\displaystyle x^{2}-4=(x-2)(x+2)\Rightarrow \frac {max...}{(x-1)}\geq \frac {4(x+2)}{(x+2)}\Leftrightarrow \frac {max...}{(x-1)}\geq 4$

My next step was this :

$\displaystyle max\left \{ .... \right \}=x^{2}+x-2\Leftrightarrow x^{2}+x-2\geq 2x+4\Leftrightarrow x^{2}-x-6=(x+2)(x-3)\geq 0$

So :

$\displaystyle x+2\geq 0 \Rightarrow x\leq -2$

and

$\displaystyle x-3\geq 0 \Rightarrow x\geq 3$

So it follows:

$\displaystyle max\left \{ x^{2}+x-2,2x+4 \right \}=\left\{\begin{matrix} x^{2}+x-2 &for & x\leq-2 & or & x\geq 3 \\ 2x+4&for &-2\leq x\leq 3 \end{matrix}\right.$

and this is the point where I don't know how to continue. I mean, I have an example of it in my script but I don't know how to do that. Can someone help?