How to get from $\vert\vert3x-1\vert-5\vert$ to $\le$ than this $\vert3x-6\vert$

Question

My textbook's workings show $\vert\vert3x-1\vert-5\vert\le\vert3x-6\vert$ .

I was wondering how they got from the $\vert\vert3x-1\vert-5\vert$ to the $\vert3x-6\vert$.

For context it was a question asking to prove the limit of $\vert1-3x\vert$ as x approaches 2 is equal to 5, using $\delta$ and $\epsilon$.

Answer 1

For $x\geq\frac{1}{3}$ it's obvious.

Let $x\leq\frac{1}{3}$.

Hence, we need to prove that $$|-3x+1-5|\leq6-3x$$ or $$-6+3x\leq 3x+4\leq6-3x,$$ which is just $x\leq\frac{1}{3}$.

Done!

Answer 2

This is $||a|-5|\le|a-5|$ for suitable $a$. If $a\ge0$ then this says nothing. The hard case is $a<0$. In that case write $a=-b$ with $b>0$. Then we want to prove $|b-5|\le |-b-5|$. That is $|b-5|\le|b+5|=b+5$. But $|b-5|=|b+(-5)|\le|b|+|-5|=b+5$ (triangle inequality).