Prove that for every real number $x$, if $|x − 3| > 3$ then $x^2 > 6x$. This is Velleman's exercise 3.5.10:
Prove that for every real number $x$, if $|x − 3| > 3$ then $x^2 > 6x$.
And here's my proof of it:
Proof. Suppose $|x − 3| > 3$. We now consider two cases:
Case 1. $x - 3 \ge 0$. Then $|x − 3| = x - 3$, so we have $x - 3 > 3$, and therefore $x > 6$. Since $x > 6$ and positive, then multiplying the inequality by $x$, we get $x^2 > 6x$. 
Case 2. $x - 3 < 0$. Then $|x − 3| = 3 - x$, so we have $3 - x > 3$, and therefore $x < 0$. Subtracting 3 from both sides of the inequality $x < 0$, we get $x - 3 < -3$. Multiplying the inequality by ($x - 3$), we get $x^2 - 6x + 9 > -3x + 9$ and therefore $x^2 > 3x$.
Since by one of the cases we have $x^2 > 6x$ then $|x − 3| > 3$ $\Rightarrow$ $x^2 > 6x$.
Is my proof valid?
Thanks in advance.

Your proof is incomplete: you should show in BOTH cases that $x^2>6x$. So in case 2, when $x<0$, at the end, it suffices to sya that "and therefore $x^2>3x>6x$".
However, there is a much shorter way that I warmly recommend. Since $|x − 3|$ and $3$ are both non-negative, the inequality $|x − 3| > 3$ is equivalent to $|x-3|^2>3^2$. Hence $$0<(x-3)^2-9=(x^2-6x+9)-9=x^2-6x$$
and we are done.
