|x| ≤ 1 ⇒ |x^2 − x − 2| ≤ 3|x + 1|
im guessing you factor it and cancel out the x+1 then im left with this
|x-2| ≤ 3
What do i do from there?
$\displaystyle |x - 2| = |x + (-2)|$.
Now
$\displaystyle |x + (-2)| \leq |x| + |(-2)|$ by the triangle inequality.
$\displaystyle |x + (-2)| \leq |x| + 2$.
And since $\displaystyle |x| \leq 1$
that means $\displaystyle |x| + 2 \leq 3$.
Therefore $\displaystyle |x - 2| \leq 3$
and thus
$\displaystyle |x| \leq 1 \implies |x^2 - x - 2| \leq 3|x + 1|$.