I know how to solve if I have something like this: $$|2x + 3| – |1 – x| \geq 3$$
But I don't know how to solve something like this: $$|x + |3x + 9|| \geq 3$$
Anyone can show me how to start solving this type of absolute value in inequalities?
[Math] How to solve inequalities which have absolute value inside absolute value
inequality
Related Solutions
We need to solve the following system $$\frac{x^2-5x+4}{x^2-4}\leq1$$ and
$$\frac{x^2-5x+4}{x^2-4}\geq-1$$ For the first we need to solve $$\frac{x^2-5x+4}{x^2-4}-1\leq0$$ or $$\frac{8-5x}{(x-2)(x+2)}\leq0,$$ which by the intervals method gives $x>2$ or $-2<x\leq1.6$.
The second inequality gives $$\frac{x^2-5x+4}{x^2-4}+1\geq0$$ or $$\frac{2x^2-5x}{x^2-4}\geq0$$ or $$\frac{x(2x-5)}{(x-2)(x+2)}\geq0,$$ which by the intervals method again gives $x\geq2.5$ or $0\leq x<2$ or $x<-2$.
Thus, after solving of this system we'll get the answer: $$[0,1.6]\cup[2.5,+\infty)$$
Note that the absolute value of a number equals its distance from $0$ which is always a non-negative number.
So the statement that $\vert x\vert<a$ tells us that $a$ is a positive number and that $x$ is closer to $0$ than $a$ is. From this we conclude that $x$ lies somewhere between $-a$ and $a$. In terms of inequalities
$$ -a<x\text{ and }x<a$$
A pair of inequalities connected by the word "and" [but not the word "or"] can be united in a combined inequality provided both inequalities are in the same direction. So the pair of inequalities above can be written in combined form as
$$-a<x<a$$
which is read as "negative $a$ is less than $x$ and $x$ is less than $a$" or more simply $x$ is between negative $a$ and positive $a$.
Now for the case where $|f(x)|<|g(x)|$.
Note that for non-negative numbers, the function $y=x^2$ is increasing. So it follows that $0\le a<b$ if and only if $a^2<b^2$. Therefore
$$ |f(x)|<|g(x)|\text{ iff }f^2(x)<g^2(x) $$
which is equivalent to
$$ g^2(x)-f^2(x)>0 $$
which is equivalent to
$$ [g(x)-f(x)]\cdot[g(x)+f(x)]>0 $$
which is true if and only if $g(x)-f(x)$ and $g(x)+f(x)$ are both the same sign--both positive or both negative.
Ultimately, it means that $f(x)$ must always lie between $g(x)$ and $-g(x)$. That is
Either
$-g(x)<f(x)<g(x)$ or
$g(x)<f(x)<-g(x)$.
So to find the solution set of $|f(x)|<|g(x)|$ you would find the solution of both 1 and 2 and take their union since the two inequalities are connected by the word "or."
Best Answer
One simple way: split it into cases.
$$|x+|3x+9||\geqslant 3$$
Case 1
If $x+|3x+9| \geqslant 0$ then $x+|3x+9|\geqslant 3$.
$$|3x+9| \geqslant -x \implies x\in(-\infty;-\tfrac92\rangle \cup \langle -\tfrac94;+\infty)=A$$ $$|3x+9|\geqslant 3 - x \implies x\in(-\infty;-6\rangle \cup \langle -\tfrac32;+\infty)=B$$ $$\Downarrow$$ $$x\in A\cap B=(-\infty;-6\rangle \cup \langle -\tfrac32;+\infty) = S_1$$ The set above contains the solutions from this case.
Case 2
If $x+|3x+9| < 0$ then $-x-|3x+9|\geqslant 3$.
$$|3x+9| < -x \implies x\in(-\tfrac92;-\tfrac94)=A$$ $$-x-|3x+9|\geqslant 3 \implies x\in\{-3\}=B$$ $$\Downarrow$$ $$x\in A\cap B=\{-3\}=S_2$$
Summary
$$x\in S_1\cup S_2=(-\infty;-6\rangle \cup \{-3\}\cup \langle -\tfrac32;+\infty)$$