How to deal with this inequation with 3 absolute values?

Question

Solve:$$|x-2|+|3x+2|-x-2|x+4|\le -x-4$$

I am having trouble even starting with this inequation. Do I find out first the zeroes of each absolute value and then I would get intervals for each absolute value?

Answer 1

One way to go about this is to explicitly plot the graph of $$f(x) = |x-2|+|3x+2|-x-2|x+4| + x + 4$$ With some skill$\color{red}{^1}$, you should be able to see that in fact $f(x)\ge 0$ for all $x\in \Bbb R$. Thus, to solve $f(x) \le 0$ would be the same as finding the zeros of $f$, i.e. solving $f(x) = 0$. See the graph of $f$ below.

The set of values of $x$ which satisfy $f(x) = 0$ constitute the closed interval $$\left[-\frac{2}{3}, 2\right]$$ on the real line.

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$\color{red}{1.}$ How would you plot the graph? Just consider what $f(x)$ looks like when $x \ge 2$, $-\frac23 \le x\le 2$ and $-4\le x\le -\frac23$. You should be able to obtain piecewise definitions of $f(x)$, and plot the graph. Even without plotting the graph it should be clear that $f(x)\ge 0$ for all $x\in \Bbb R$. Needless to mention, you can solve $f(x) = 0$ from the piecewise definitions too - quite easily.

If you need more details, or have any questions, please let me know.

Answer 2

Rewrite your equation as $$|x-2| + |3x+2| + 4 = 2|x+4| \iff |-x+2| + |3x+2| + 4 = |2x+8|$$ Recall in the triangle inequality among $n$ real numbers $u_1,u_2,\ldots, u_n$,

$$|u_1| + \cdots + |u_n| \ge |u_1 + \cdots + u_n|,$$

the equality is achieved when and only when all non-zero $u_k$ have the same sign. Since $$(-x + 2) + (3x +2) + 4 = 2x+8\quad\verb/and/\quad 4 > 0$$ the equation you have is equivalent to $$-x + 2 \ge 0\;\;\verb/and/\;\; 3x + 2 \ge 0 \quad\iff\quad x \in \left[ -\frac23, 2\right]$$