Sum of three absolute values less than 4 (|x|+|x-1|+|x-2| < 4)

Statement

Solve the inequality $|x|+|x-1|+|x-2|<4$. (Hint: the sum is minimum at $x=1$ and equals $2$.)

Solution

Sign table: the arguments change sign at $x=0$, $x=1$, $x=2$.

• $x<0$: $-x-(x-1)-(x-2)=-3x+3<4 \iff x>-\frac13$, that is $-\frac13<x<0$;
• $0\le x<1$: $x-(x-1)-(x-2)=-x+3<4 \iff x>-1$, always true, the whole of $[0,1)$;
• $1\le x<2$: $x+(x-1)-(x-2)=x+1<4 \iff x<3$, the whole of $[1,2)$;
• $x\ge2$: $x+(x-1)+(x-2)=3x-3<4 \iff x<\frac73$, that is $2\le x<\frac73$.

Union: $\boxed{-\dfrac13<x<\dfrac73}$.

Links

Topics: Parabola
Concepts: Absolute value inequalities · Sign table · Absolute value
Skills: Reasoning by cases · Solving inequalities
Exercise type: Solving an inequality