Real Analysis introductory absolute value proof

Question

let $x \in R$

Prove that $\vert x\vert \leq 2$ implies $\vert x^2 -4 \vert \leq 4 \vert x-2 \vert$

Here is my work:

$\vert x-2 \vert \vert x + 2 \vert \leq 4 \vert x-2 \vert$

By the first multiplicative property this implies $\vert x + 2 \vert \leq 4$

$\vert x \vert + \vert 2 \vert \leq 4$

$\vert x \vert + 2 \leq 4$

$\vert x \vert \leq 2$

Which is our hypothesis therefor $\vert x \vert \leq 2$ implies $\vert x^2-4 \vert \leq 4 \vert x-2 \vert$

I am going over the book myself and want to know if my reasoning was correct

Answer 1

It seems that you are doing things in the opposite order as is required.

You should start with $|x|\le 2$. As you wrote, this means $|x|+2\leq 4$, or $|x|+|2|\leq 4$. Then use the triangle inequality, which implies that $|x+2|\leq |x|+|2|\leq 4$. Finally, multiply both sides by $|x-2|$.