Prove that if $x \leq \vert 2 \vert$ then $ \vert x^2-4 \vert \leq 4 \vert x-2 \vert$

Question

This is a problem on my first homework from Real Analysis

Assume that $\vert x \vert \leq 2$

Proof:

$\vert x \vert + \vert2\vert \leq 2 + \vert 2 \vert \to \vert x+2 \vert \leq 4$, add 2 to each side

$\vert x-2 \vert \vert x+2 \vert \leq 4 \vert x-2 \vert$, multiply each side by the absolute value of $x-2$

Since we multiplied by an absolute value, ie positive, the sign doesn't flip.

Does this suffice to prove the implication is true?

Answer 1

Your solution is correct. It would be clearer if you would write $$\vert x + 2 \vert \leq \vert x \vert + \vert 2 \vert \leq 2 + 2 = 4$$ (by using the triangle inequality). That would improve your proof in my opinion. Good job though :)