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I added my solution, but I'm not sure I've got it right. I'd like to know what you think.

The question:

Solve the equation: $$\lfloor |x+1|-|x| \rfloor \ge x^2.$$

enter image description here

the left and right symbols aren't square brackets, they are 'floors' more info: https://en.wikipedia.org/wiki/Floor_and_ceiling_functions

My Solution:

enter image description here

Thanks.

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    $\begingroup$ Hint: break $|x+1|-|x|$ where $x<-1$, $-1\leq x<0$ and $x\geq 0$ your answer is for $x\geq 0$ $\endgroup$
    – R.N
    Oct 24 '15 at 9:38
  • $\begingroup$ Computing for a small number of specific $x$ will show that the answer in the picture is not right. It may also tell you enough about what is going on to lead to a complete analysis. $\endgroup$ Oct 24 '15 at 9:49
  • $\begingroup$ @RaziehNoori The question is not with square brackets, it is with a floor function. en.wikipedia.org/wiki/Floor_and_ceiling_functions $\endgroup$
    – Yakirbu
    Oct 24 '15 at 10:05
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At first notice that for $x=-1$, we have $\lfloor |-1+1|-|-1| \rfloor = -1$ however, $x^2=1$, so the inequality does not hold.

Thus a better approach is to consider different cases to treat the absolute value.

  1. If $x< - 1 \Rightarrow x+1 <0 \text{ and } x<0 \Rightarrow \lfloor |x+1|-|x| \rfloor =\lfloor -x-1+x \rfloor = -1 $ and as $x ^2 \geq 0$, so in this case no $x$ satisfies the inequality.
  2. If $-1 \leq x <0 \Rightarrow x+1 \geq 0 \text{ and } x<0 \Rightarrow \lfloor |x+1|-|x| \rfloor =\lfloor x+1 + x \rfloor = \lfloor 2x+1 \rfloor $ . Let $ n =\lfloor 2x+1 \rfloor$ but $-1 \leq x <0 \Rightarrow -1 \leq 2x +1<1 $ so $n=-1 \text{ or } 0$. If $n=-1<0$ the inequqlity clearly does not hold, and if $n=0$ then $0 \geq x^2 \Rightarrow x=0 $ , but $x <0$.So in this case also we have no solution.
  3. If $x \geq 0 \Rightarrow x+1 \geq 0 \text{ and } x<0 $, then $\lfloor |x+1|-|x| \rfloor= \lfloor x+1-x \rfloor = \lfloor 1 \rfloor =1$, so $1 \geq x^2 \Rightarrow -1 \leq x \leq 1 $ but in this case $x \geq 0 $ so the set of solution is $ x \in [ 0,1]$.

Thus the set of solution is $ x \in [ 0,1]$.

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  • $\begingroup$ Thank you! Extremely detailed. $\endgroup$
    – Yakirbu
    Oct 24 '15 at 12:49

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