# The interval determined by absolute value inequality $|1-2x| < |1+x|$

I'm working on a series convergence problem and am stuck on this part:

The series converges when $|1-2x| < |1+x|$.

How can I proceed from here to pick the values of $x$ that satisfy this inequality? I tried using the triangle inequalities, but it wasn't helpful and gave silly upper bounds.

• Draw the functions $x \mapsto |1+x|$ and $x \mapsto |1-2x|$. Commented Dec 30, 2014 at 4:13
• i see that you are back now.
– abel
Commented Dec 30, 2014 at 4:15
• yes, @abel :) trying to move a little faster, though, since the exams are coming up soon.. Commented Dec 30, 2014 at 4:19

Note that $$|1-2x|\lt |1+x| \quad\text{if and only if}\quad (1-2x)^2\lt (1+x)^2.$$ If we expand, the inequality on the right turns out to be equivalent to $$3x^2-6x\lt 0.$$ This inequality holds precisely if $x$ is (strictly) between the two roots of $3x^2-6x=0$. So the inequality holds whenever $0\lt x\lt 2$.

• Agree with @edwardjiang - very cool solution :) Commented Dec 30, 2014 at 10:58
• Thanks so much, @andrenicolas - have a great night. Commented Dec 30, 2014 at 10:59
• You are welcome. Commented Dec 30, 2014 at 15:27

Suppose, $x\leq 0$. Then $1-2x>0$, $|1-2x|=1-2x$.
$1-2x < |1+x|$ means that either $$1+x>1-2x$$ or $$1+x<2x-1.$$ And it could be simplified to $$3x>0$$ or $$x>2,$$ which is not true for $x\leq 0$.

Suppose $x>0$. Then $1+x>0$, $|1+x|=1+x$.
$|1-2x| < 1+x$ means that $$-1-x<1-2x<1+x,$$ which is equal to $$-2<-x<2x.$$ $-x<2x$ is obviously true for all $x>0$, but $-2<-x$ means $x<2$.

So, $0<x<2$.

Edit: Corrected a mistake (turned a $4$ part solution into a $2$ part one). I apologize for the error.

The standard way would be to just consider all cases. That is, $$1-2x<1+x$$ $$2x-1<1+x$$ This will give a $2$-part solution to the inequality.

$0<x<2$

• I was actually working on the 4-part solution and thought it would be good to do it the "long way" but found some contradictions. I see your edited answer, @EdwardJiang -- how come we don't have to consider the other value, -1-x? Since, by definition, |1+x| = (1+x) when this number is positive, and equal to -(1+x), when the number (1+x) is negative. Is there a canceling effect since 1-2x is also in absolute value? Commented Dec 30, 2014 at 4:39
• It's because having $1-2x<1+x$ is the same as having $2x-1>-x-1$, which is obviously in contradiction with $2x-1<-x-1$. Commented Dec 30, 2014 at 4:49
• Ok got it - thanks @edwardjiang :) Commented Dec 30, 2014 at 11:15

I find drawing the graphs to be the quickest.

If I have had a few glasses of wine, or the computation has some life threatening possibilities, I do the following:

If you look at the quantities in the absolute values it will be clear that the two points of 'interest' are $-1, {1 \over 2}$.

If $x \in(-\infty,-1]$, we have $1-2x < -1-x$, or $2 < x$, so no solutions here.

If $x \in (-1,{1 \over 2}]$, we have $1-2x < 1+x$, or $0 < x$, so the solutions here are $(0, {1 \over 2}]$.

Otherwise, we have $2x-1 < 1+x$, or $x<2$, so the solutions here are $({1 \over 2}, 2)$.

Combining gives $(0,2)$.

• Ok, thanks so much @copper.hat :) Commented Dec 30, 2014 at 11:27