Absolute Value Proof By Cases

Question

I'm currently working through D. Velleman's How to Prove it. I have a question regarding an absolute value proof by cases (#10; section 3.5).

The question asked is to prove that: $$ \forall x\in\mathbb{R} \left(\lvert x-3 \rvert \gt 3 \rightarrow x^2 \gt 6x \right) $$

The book recommends a proof by cases, using the following as cases:

1. if $ x-3 \geq 0 \rightarrow \lvert x-3 \rvert = x-3 $
2. if $ x-3 \lt 0 \rightarrow \lvert x-3 \rvert = 3 - x$

My attempt:

let $x \in \Bbb{R}$ be arbitrary real number
For case 1 where $\lvert x-3 \rvert = x-3 $ (because $x-3 \geq 0)$

$x-3 \gt 3$
$x \cdot \left(x-3 \gt 3\right)$
$x^2 - 3x \gt 3x$
$x^2 > 6x$

For case 2, $ \left(x-3\right) < 0 $ so $ \lvert x-3 \rvert = 3 - x$

$3-x \gt 3$

but then I'm not quite sure. I tried multiplying both sides by $3-x$, giving

$ \left( 3-x \right) \cdot \left( 3-x \right) \gt 3 \cdot \left( 3-x \right) $
$9 -6x + x^2 \gt 9 - 3x$
$-6x + x^2 \gt -3x$
$x^2 \gt 3x$

Which looks similar to $x^2 \gt 6x$ but simply isn't.

If anyone could advise regarding where to proceed, I'd be very grateful.

Thanks!

Answer 1

HINT:

$$|x-3|>3\to x-3>3\to x>6$$ OR $$|x-3|>3\to-(x-3)>3\to x<0$$

You just have to show that $x^2>6x$ holds for $x>6$ and $x<0$. If you get stuck with this, I'm happy to give a further hint, but give it a go first.

Answer 2

If $3-x <3$ then $x <0$ so $3 <6$ and $3x>6x $.

So $x^2>3x>6x$