# Inequality with two moduli

I have a question similar to this, find all $x \in \mathbb{R}$ satisfying

$\displaystyle 3 < \left| x+1 \right| + \left| x - \frac{1}{2} \right| < 7$

which is rather trivial by distinguishing cases, the question suggests an alternative method which I have been struggling to find, I have tried to apply the triangle inequality in various ways but had no way of ensuring I captured the full range. Can anyone suggest an efficient alternative?

I am not sure if this is really an alternative solution. But choose $u=x+\frac 14$. The function $f(x)=\displaystyle \left| x+1 \right| + \left| x - \frac{1}{2} \right|$ turns into $g(u)=\displaystyle \left| u+\frac 34\right| + \left| u - \frac{3}{4} \right|$. So we are interesting in those $u$ satisfying: $$\displaystyle 3 < \left| u+\frac 34 \right| + \left| u - \frac{3}{4} \right| < 7$$ Note that $g(u)$ is even function ($g(u)=g(-u)$). Therefore it is enough to consider the case where $u>0$ and then the answer for $u<0$ is automatically obtained. Also note that $g(u)$ is equal to $\frac 32$ over $[-\frac 34,\frac 34]$ (here is somehow cheating!). Hence it is enough to focus on the case where inside both absolute value is positive: $$\displaystyle 3 <u+\frac 34+u - \frac{3}{4}< 7\implies \displaystyle 3 <2u< 7.$$ Which means that $u\in[\frac 32,\frac 72]$ or (because $g$ is even) $u\in[-\frac 72,-\frac 32]$. Now roll back the variable change and you are fine.