Use the Triangle Inequality to prove that $|a − 4b| < x + 2y + |L|$

Question

The question was longer than 150 characters, so I couldn't fit it all in the title. Here's the full question:

Let $A$ and $B$ be nonempty sets of real numbers and let $L ∈ \Bbb R$. Suppose that there exist positive real numbers $x$ and $y$ such that $|a − L| < x$ for all $a ∈ A$ and $|2b − L| < y$ for all $b ∈ B$. Use the Triangle Inequality to prove that: $$|a − 4b| < x + 2y + |L|$$ for all $a ∈ A$ and $b ∈ B$.

(HINT: $|a − 4b| = |a − 2L + 2L − 4b|$.)

I know you can use the triangle inequality theorem to separate the hint out to where

$$|a-2L+2L-4b| ≤ |a-2L| + |2L-4b|$$ but I'm not sure where to go after that.

The triangle inequality theorem states: $$|x+y|≤|x|+|y|,$$ $$|x-y|≥||x|-|y||,$$ $$|x-y|≥|x|-|y|,$$ and $$|x-y|≥|y|-|x|.$$

Answer 1

Here's a basic outline:

We start off with what you've done, $$\begin{align}|a-4b| &= |a-2L + 2L - 4b| \tag{From the hint}\\&\le |a-2L| + |2L - 4b| \tag{Triangle Inequality}\\&= |a-2L| + 2|L-2b| \tag{Factor out $2$}\\&= |a-2L| + 2|2b - L|\tag{Absolute Value removes $-1$}\\ &= |(a -L) - L| + 2|2b-L| \tag{Rewrite left quantity}\\&\le|a-L| + |L| + 2|2b-L| \tag{Triangle Inequality}\\& < x + |L| +2y. \tag{Hypothesis}\end{align}$$