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I am having trouble finding what the answer to this inequalitie is: $|x+4|x + |x+2| > -2$

first i did this:

$|x+4|=\left\{ \begin{align} x+4 & \text{ , if }x\geq -4 \\ -(x+4) & \text{ , if }x <-4 \end{align} \right\}$

$|x+2|=\left\{ \begin{align} x+2 & \text{ , if }x\geq -2 \\ -(x+2) & \text{ , if }x <-2 \end{align} \right\}$

Then i got this:

$1. x<-4: $

$ -(x+4)x + (-(x+2)) > -2$

$2. -4 $\le$x<-2:$

$ (x+4)x + (-(x+2)) > -2$

$3. x$\geq$-2: $

$ (x+4)x+x+2 > -2$

After i calculated top 3 inequalities i got this answers:

$1. x$^2$ + 5x < 0$

$2. x$^2$ + 3x > 0$

$3. x$^2$+5x+4 > 0$

I think i calculated everything correctly, but now i do not know how to find the answer to this absolute value inequalitie.

Thank you for your help.

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Guide:

For the first case, we need $x<-4$ and $x^2+5x<0$

that is $x<-4$ and $x(x+5) <0$ which is equivalent to

$x<-4$ and $-5<x<0$. which can be summarized as $-5<x<-4$.

Do the same thing for the other cases as well.

Remark: $x(x+5)<0$ is equivalent to $-5<x<0$ can be seen from its graph:

enter image description here

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