Absolute value inside absolute value inequality (from both sides)

Question

I'm a bit rusty so I'm giving this a go.

Looking at $|2x+|x-3||≥|x-2|$ and I'm so confused. I know I should approach this from the inside out - take a look at $x-3≥0$ and $x-3≤0$ and split into two cases and go from there, but I'm getting confused as to what happens to the other absolute values.

Much appreciated.

Answer 1

in the first case you will have if $$x\geq 3$$ then we get $$|2x+x-3|\geq |x-2|$$ this simplifies to $$3|x-1|\geq |x-2|$$ and since $$x\geq 3$$ we have to solve $$3(x-1)\geq (x-2)$$ in the next case we have for $$2\le x<3$$ the inequality $$x+3\geq x-2$$ in the next case we get for $$-3\le x<2$$ the inequality $$x+3\geq -x+2$$ and for $$x<-3$$ we get $$-x-3\geq -x+2$$ which is impossible

Answer 2

Hints: $x\ge 3:$ the left side equals $|2x +(x-3)|=|3x-3|= 3x-3,$ while the right side equals $x-2.$

$x\le 3:$ the left side equals $|2x +3-x|= |x+3|.$ Note $|x+3| \ge |x-2| \iff (x+3)^2 \ge (x-2)^2.$

Answer 3

$|x-2|≤ |2x+|x-3||$ so $-(2x+|x-3|)≤ x-2≤ 2x+|x-3|$

look at the first part $-(2x+|x-3|)≤ x-2$

$-2x-|x-3|≤x-2$ so $|x-3|≥ 2-3x$

$x-3≥ 2-3x$ and $x-3 ≤ 3x-2$

so $x≥ 5/4$ and $x≥-1/2$

Therefore, $x≥-1/2$ for the first part.

Next part, $x-2≤ 2x+|x-3|$

$-x-2≤|x-3|$ so $|x-3|≥-x-2$

$x-3≥-x-2$ and $x-3≤x+2$

$x≥1/2$ and the other is impossible

Therefore, $x≥1/2$

So combining these answers, your final answer will be $x≥-1/2$