Proving Inequalities with Absolute values

Question

Say the following expression is to be proven for all real $x$:

$$|x-1| + |x+1| \geq 2$$

Is it sufficient to just look at the cases where $|x-1|$ and $|x+1|$ are minimum? Or is it necessary to show that the expression it true using other algrebraic manipulations?

Answer 1

This is just triangle inequality (remember that $|a-b|=|b-a|$):

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

Answer 2

Hint: Distinguish the following cases: $$x\geq 1$$ so $$x-1+x+1=2x\geq 2$$ $$-1\le x<1$$ so $$-x+1+x+1\geq 2$$ $$x<.1$$ so $$-x+1-x-1=-2x\geq 2$$