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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.

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

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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...

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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.

enter image description here

Hope this helps.

Cheers!

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

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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: enter image description here

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

:)

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