How to solve double absolute value inequality?

Question

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!

Answer 1

"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.

Answer 2

Hint:

As $|y|=|-y|,$

$$|x-1|+|x-2|=|x-1|+|-(x-2)|\ge|x-1-(x-2)|$$

The equality occurs if $1-x=x-2$

Answer 3

Think geometrically, x verifies the inequality if and only if it lies outside the closed interval $[1,2]$

Answer 4

For $x>2$ or for $x<1$ it's obviously true.

But for $1\leq x\leq 2$ we need $1<|x-1|+|x-2|=x-1+2-x=1,$ which is wrong, which gives the answer: $$(-\infty,1)\cup(2,+\infty).$$

Answer 5

The best way to "try to avoid" errors is to consider the following intervals

• $x<1\implies |x − 1|+|x − 2| > 1 \iff 1-x+2-x>1 \iff 2x<2 \iff x<1$

• $1\le x<2\implies x − 1+2-x > 1 \iff 1>1 $

• $x\ge2\implies |x − 1|+|x − 2| > 1 \iff x-1+x-2>1 \iff x>2 $

Answer 6

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 $1$.)

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This technique works for every sum of absolute values of linear binomials.

Answer 7

A different method:

Note that: $|x-1|+|x-2|=||x-1|+|x-2||$

then we have:

$$\left(|x-1|+|x-2|\right)^2>1$$

$$2x^2-6x+4+2|(x-1)(x-2)|>0$$

$$(x-1)(x-2)+|(x-1)(x-2)|>0$$

Case $-1$ $$\begin{cases} (x-1)(x-2)≥0 \\ 2(x-1)(x-2)>0\end{cases} \Longrightarrow (x-1)(x-2)>0 \Longrightarrow x\in (-\infty, 1)∪(2,+\infty)$$

Case $-2$ $$\begin{cases} (x-1)(x-2)≤0 \\ (x-1)(x-2)-(x-1)(x-2)>0 \end{cases} \Longrightarrow x\in {\emptyset}$$

So, we get $$x\in (-\infty, 1)∪(2,+\infty).$$