<s>[INST] How to solve double absolute value inequality? This question comes from Spivak Calculus Chapter 1.
How can we algebraically solve $|x − 1|+|x − 2| > 1?$
I know that if we 2 absolute values and no constants, we can square both sides, but I'm pretty sure this is not the case here. My attempt was to split this into different sections:
$|x − 1|+|x − 2| > 1 \rightarrow |x − 1| > 1 - |x − 2|$.
So we would have:
$x − 1 > 1 - |x − 2|$
$x − 1< -1 +| x − 2|$
Then we can split this into 4 equations more equations based on the absolute value on $(x-2)$.
However, after doing this, I obtained conflicting solutions and unsolvable expressions (i.e $2<-2$).
That being said, how would I go about algebraically solving this inequality?
Thanks!
 [/INST] "However, after doing this, I obtained conflicting solutions and unsolvable expressions"
Those are cases with no solutions.  Nothing wrong with that.
Do cases be keep track of you initial assumptions.
Case 1:  $x-1 \ge 0; x-2 \ge 0$.  Thus $x\ge 1$ and $x \ge 2$.  This is the case that $x \ge 2$.
Okay $|x-1| + |x-2|> 1$ so
$(x-1) + (x-2) > 1$ so
$2x - 3 > 1$ so $2x > 4$ and $x >2$.  And we restrict this to $x \ge 2$ to get
$x > 2$ AND $x \ge 2$ so
Conclusion $x > 2$.
Case 2:  $(x-1) \ge 0$ and $(x-2) < 0$.  That is $x \ge 1$ and $x < 2$ so this is the case that $1 \le x < 2$.
We get $(x-1) -(x-2) > 1$ so
$1 > 1$.  This is never the case so there are no solutions where $1 \le x < 2$.
If we want to be thurough we would say.
We must restrict to where $1 > 1$ AND $1\le x < 2$.  There are no cases where both are true.
Case 3:  $(x-1) < 0$ and $x -2 \ge 0$. This means $x < 1$ and $x \ge 2$.  This is impossible.  There are no such $x$ and so no such $x$ can be a solution (as there are no such $x$!).
If we want to be thorough (which we don't but let's pretend we do) we would solve
$-(x-1) + (x-2) > 1$ so $-1 > 1$ and or solution occurs when $-1 > 1$ and $x< 1$ and $x \ge 2$.  As those three conditions are never concurrently true we have no solution in this interval which doesn't exist in the first place.
Case 4: $(x-1) < 0$ and $(x-2) < 0$.  This means $x < 1$ and $x < 2$ so is the case when $x < 1$.
So $-(x-1) -(x-2) > 1$ so $-2x + 3> 1$ so $-2x > -2$ so $x < 2$.
So these solutions occur when $x < 2$ AND $x < 1$
Conclusion: so these solutions occur whenever $x < 1$
Combining Case 1, and Case 4 (and 2 and 3 although those had no result) we have final solution
$|x-1| + |x-2| >1 $ if
$x >2$ OR $x < 1$ or $x \in (-\infty, 1)\cup (2, \infty)$.
If we want to be thorough (which be now you should know we don't)
We could so we have solutions when:
$x > 2$ OR $1 < 1$ OR ($x < 1$ AND $x\ge 2$) OR $x < 1$ or
$x \in (2, \infty) \cup \emptyset \cup \emptyset \cup (-\infty, 1)=$
$(-\infty, 1)\cup (2, \infty)$.
=====
Familiarity and common sense and we can allow ourselve to consider then intervals $(-\infty, 1], [1,2],$ and $[2,\infty)$.
If $x \in (-\infty 1]$ then $(x-1)\le 0; x-2 < 0$ so $|x-1|+|x-2|=-(x-1)-(x-2)=-2x+3 > 1$ so $x < 1$.
If $x \in [1,2]$ then $x-1 \ge 0$ and $x-2\le 0$ so $|x-1|+|x-2| = (x-1)-(x-2) = 1 > 1$ which is impossible.
If $x \in [2,\infty)$ then $x-1>0$ and $x -2\ge 0$ so $|x-1| + |x-2| = x-1 + x-2=2x -3 >1$ so $x > 2$.
So $x< 1$ or $x > 2$ and $x \in (-\infty,1)\cup (2, \infty)$.
....
this way we know $x-1 <0$ while $x-2 \ge 0$ was absurd from the start and never needed to be considered in the first place.
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