Prove for any real number, $|x-6|+x \gt 3$

Question

Prove for any real number, $|x-6|+x \gt 3$

I think I'm doing the proof wrong but this is what I have on paper:

Case 1: let $x-6=0$, then $x=6$

That implies the following: $|6-6|+6 \gt 3$, which is true.

Case 2: let $x-6 \gt 0$, then $x \gt 6$

That implies $|x-6|+x \gt x\gt 3$ which is true

Case 3: let $x-6 \lt 0$

yeah so this is where I trip up

Answer 1

If $x-6<0$ then $|x-6| = -(x-6) = -x+6$

So, $|x-6|+x = 6>3$

Answer 2

If $x-6<0$, then the absolute value of the expression would be $6-x$. The inequality then becomes $$6-x+x > 3 \implies 6 > 3$$ Which is true.

Here's what this inequality looks like graphically:

The purple line is $y=|x-6|+x$, whereas the grey line is $y=3$.

Answer 3

Because $$3-x<6-x\leq|6-x|=|x-6|.$$

Answer 4

Split it in two cases: $$x-6+x>3, ~x-6\geq0\implies x>9/2,~ x\geq 6~~~~(x\in[6,+\infty))$$ $$-(x-6)+x>3,~x-6<0\implies-x+6+x=6>3,~x<6 ~~~~(x\in\left(-\infty,6)\right)$$ Now, $(-\infty,6)\cup[6,+\infty)=\mathbb{R}$