Prove that for every real number $x$, if $|x-3|>3$ then $x^2>6x$. Not a duplicate of
Prove that for every real number $x$, if $|x − 3| &gt; 3$ then $x^2 &gt; 6x$.
This is exercise $3.5.10$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:
Prove that for every real number $x$, if $|x-3|&gt;3$ then $x^2&gt;6x$.
Here is my proof:
Let $x$ be an arbitrary element of $\Bbb R$. Suppose $|x-3|&gt;3$. Now we consider two different cases.
Case $1.$ Suppose $x-3\geq0$. Ergo $x-3&gt;3$ and so $x&gt;6$. Multiplying both sides of $x&gt;6$ by $x$ we obtain $x^2&gt;6x$.
Case $2.$ Suppose $x-3&lt;0$. Ergo $3-x&gt;3$ and so $x&lt;0$. Since $x&lt;0$, then clearly $x&lt;6$. Multiplying both sides of $x&lt;6$ by $x$ we obtain $x^2&gt;6x$.
Since the above cases are exhaustive, $x^2&gt;6x$. Therefore if $|x-3|&gt;3$ then $x^2&gt;6x$. Since $x$ is arbitrary, $\forall x(|x-3|&gt;3\rightarrow x^2&gt;6)$. $Q.E.D.$
Is my proof valid$?$
One other question: In the above linked-post, there is a clear mistake in the proof:

Since by one of the cases we have $x^2 &gt; 6x$ then $|x − 3| &gt; 3$
$\Rightarrow$ $x^2 &gt; 6x$.

But the cases must be exhaustive and from both of them we should reach the conclusion that $x^2&gt;6x$. Is my reasoning correct$?$ If yes, then why no one mentioned it in the comments or the answers to the above post$?$
Thanks for your attention.
\begin{align}
|x-3|&gt;3 &amp; \Rightarrow (x-3)^2&gt;9 \\
 &amp; \Rightarrow x^2-6x+9&gt;9 \\ 
 &amp; \Rightarrow x^2-6x&gt;0 \\
 &amp; \Rightarrow x^2&gt;6x \\ 
\end{align}
For the second case we can say that $x$ is negative,
$x^2$ is positive and $6x$ is negative so $x^2&gt; 6x$