Prove that for every real number $x$, if $|x-3|>3$ then $x^2>6x$.

Question

Not a duplicate of

Prove that for every real number $x$, if $|x − 3| > 3$ then $x^2 > 6x$.

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 $|x-3|>3$ then $x^2>6x$.

Here is my proof:

Let $x$ be an arbitrary element of $\Bbb R$. Suppose $|x-3|>3$. Now we consider two different cases.

Case $1.$ Suppose $x-3\geq0$. Ergo $x-3>3$ and so $x>6$. Multiplying both sides of $x>6$ by $x$ we obtain $x^2>6x$.

Case $2.$ Suppose $x-3<0$. Ergo $3-x>3$ and so $x<0$. Since $x<0$, then clearly $x<6$. Multiplying both sides of $x<6$ by $x$ we obtain $x^2>6x$.

Since the above cases are exhaustive, $x^2>6x$. Therefore if $|x-3|>3$ then $x^2>6x$. Since $x$ is arbitrary, $\forall x(|x-3|>3\rightarrow x^2>6)$. $Q.E.D.$

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 $x^2 > 6x$ then $|x − 3| > 3$ $\Rightarrow$ $x^2 > 6x$.

But the cases must be exhaustive and from both of them we should reach the conclusion that $x^2>6x$. 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.

Answer 1

\begin{align} |x-3|>3 & \Rightarrow (x-3)^2>9 \\ & \Rightarrow x^2-6x+9>9 \\ & \Rightarrow x^2-6x>0 \\ & \Rightarrow x^2>6x \\ \end{align}

Answer 2

$$f(x)=x-3, x>3; -(x-3), x \le 3$$ So $$|x-3|>3 \implies x>6 ~or~ x<0.$$ $$x^2 >6x \implies x(x-6) >0 \implies x<0 ~or~ x>6.$$ So both the equations are the same.

Also note that $$|x-3|>3 \implies x^2-6x+9>9 \implies x^2>6x,$$ which is more transparent.

Answer 3

For the second case we can say that $x$ is negative, $x^2$ is positive and $6x$ is negative so $x^2> 6x$