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!
Best Answer
The LHS is a piecewise linear function and it suffices to evaluate it at the turning points and evaluate the slopes in between
$$f(1)=1\text{ and }f(2)=1$$ while the slopes are $$-2,0,2.$$
Hence $f(x)>1$ outside $[1,2]$. (There is a flat minimum with value $1$.)
This technique works for every sum of absolute values of linear binomials.