Solving $2|x+1|>|x+4|$ I'm trying to solve the following equations and inequalities for $x\in\mathbb R$:
$$2|x+1|>|x+4|$$
I know I'm supposed to consider the intervals $(-\infty,-4), [-4,-1]$ and $(-1,\infty)$ but don't know how each affects the equation and how to conclude a result.
I think considering interval $(-\infty,-4)$: $2|1+x|>|4+x|$ $\Rightarrow x>2$ but if I'm right how will it work for the other intervals?
 A: What is the sign of $x+1$ and $x+4$ in each interval?
Then use $|y|=y$ if $y\geq 0$, and $|y|=-y$ if $y\leq 0$.
For instance, if $x\in [-4,-1]$, you have $x+4\geq 0$ so $|x+4|=x+4$ and $x+1\leq 0$ so $|x+1|=-x-1$. In this interval:
$$
2|x+1|>|x+4|\quad\Leftrightarrow\quad 2(-x-1)>x+4\quad\Leftrightarrow\quad 3x<-6\quad\Leftrightarrow\quad x<-2.
$$
So this yields $[-4,-2)$.
Repeat the procedure on the other intervals.
Alternative: draw simultaneously the graphs of $2|x+1|$ and $|x+4|$. Then find where the former is above the latter. It is very easy. You just have to find the two intersections.
See here.
A: The considered intervals are right, use the definition of the absolute value 
$$|x|=\left\{ \begin{array}{rl}
x & x\geq 0 \\
-x & x < 0 \\ 
\end{array}\right.
$$ 
In the Intervalls you can always think which one is the case and so eliminate the absolut value.
To check your results here is a plot, the red one is $2\cdot |x+1|$

For $x \in (-4,-1)$ we have $x+4 >0$ and $x+1<0$
So the inequality is 
\begin{align*}
2 |x+1| & > |x+4| \\
-2\cdot  (x+1) &> x+4 \\
\iff -2-2x &> x+4 \\
\iff   -6 &> 3 x \\
\iff -2 &> x
\end{align*}
A: $$\begin{align}
&4(x+1)^2>(x+4)^2\\
&4x^2+8x+4>x^2+8x+16\\
&3x^2-12>0\\
&x>2\ \text{ or }\ x<-2
\end{align}$$
