# Prove that for all $x\in\mathbb{R}\enspace \lvert x\rvert + \lvert x-6 \rvert\geq 6$

I am beginning proofs in analysis. I am reading Kane's book, but I am not sure if this proof counts as a proof in analysis. I have tried proving by contradiction, but I failed. I also tried using the triangle inequality ($$\lvert x - 6\rvert\leq\lvert x\rvert + \vert\text{-}6\rvert$$), but it just does not follow from it. The farthest I got was showing that $$\lvert 2x -6\rvert\leq \rvert x\lvert + \rvert x - 6\lvert$$ by letting $$y = x-6$$ and using the triangle inequality. If anybody has any idea how to proceed, I would greatly appreciate some guidance. Thanks.

• Use triangular inequality: $a,b\in\mathbb{R}\;\implies\;|a+b|\le |a|+|b|$, and notice that $6=x+(-x+6)$ Oct 9 '18 at 22:43
• @ÁngelMarioGallegos Thanks for your comment. It was what I needed to figure it out. Oct 9 '18 at 22:52

Use the triangle inequality $$|a|+|b|\geq |a+b|$$ and the fact that $$|-a|=|a|$$.