# Solve the inequality : $||x|-1|<|1-x|$

Solve the following inequality: $$||x|-1|<|1-x|$$

My Attempt:

I tried expanding this inequality by considering $$8$$ cases, but I am having trouble finding the range of the each of the solutions I got in the $$8$$ cases.

I am sorry I couldn't explain what I did properly. This is my third question on Math.SE and am still learning about how to properly frame questions.

• Note: according to the reverse triangle inequality, $||x|-1|\mathbf\le|1-x|$ for all $x$ – J. W. Tanner Apr 28 at 20:02

Well, if $$x\geqslant 0$$, we have $$|x|=x$$, so the inequality is $$|x-1|<|1-x|$$, but as $$|t|=|-t|$$ always, there can be no solution here.

Remaining case is $$x< 0$$, where the inequality is $$|1+x|<|1-x|$$, which says the distance of $$x$$ from $$-1$$ is less than the distance of $$x$$ from $$1$$, which is true for all negatives, so all such $$x$$ is in the solution set.

If $$x>0$$ there is no solution since 0 is not strictly less than 0. Suppose $$x<0$$. You have to solve then: $$|-x-1|=|x+1|<1-x$$ since $$1-x>0$$. Squaring both side yields: $$(x+1)^2<(1-x)^2.$$ I let you finish it...

The best I think to solve such problems is to draw the graphs. I am attaching an image below stating the steps to draw the LHS. Leaving out drawing the RHS as an exercise to you and figuring out the answer.

Hope this helps.

Cheers!

• This is a good strategy to figure out the answer but does not provide rigorous mathematical proof – J. W. Tanner Apr 28 at 20:04
• @J.W.Tanner Why? It's an application of translations and reflections, is geometry inferior to algebra? – Eureka May 1 at 9:45

It can be interesting to notice that , in case x>0 or equal to 0, the two numbers inside the absolute value function are opposites.

We know that, as a general rule: it is impossible for the absolute value of a number to be less than the absolute value of its opposite ( additive inverse). If A and B are opposites, then , their absolute value is necessarily equal. ( In other words, two opposite numbers are necessarily at the same distance from 0).

We also know as a general rule that : a-b and b-a are opposites. For example : 9-3 and 3-9. So (x-1) and (1-x) are opposites, in other words (1-x) = - (x-1)

Suppose x>0 or x=0. In that case |x| = x , and therefore

| |x|-1 | < | 1- x |

means

|x-1| < |1-x|

which implies

|x-1| < |- (x-1) |

The case " x>0 or x = 0 " leads to an impossibility for all values of x. The reason is that it is impossible the absolute value of A and of -A not to be equal.

So only the case in which x<0 has to be treated ( with its subordinate cases).

In this case the best strategy is the graphic one(knowing that absolute value is a reflection of the negative $$y$$ semiplane on the positive one). Here there is the right side in blue and the leftside in red:

We can easily notice that the equality is true only for $$x \geq 0$$

:)