I am not sure on where to start to prove $∀x∈R |x−a|≤1 ⇒ |x+a|≤1 + 2|a|$

Question

I don't know if the epsilon-delta definition of limit will help here. $\forall_{x \in \mathbb{R}}$ we have $|x-a| \leq 1 \implies |x+a| \leq 1 + 2|a|$

Answer 1

Hint:

By the triangle inequality,

$|x+a|\le|x|+|a|,$

and $|x|\le|x-a|+|a|.$

Answer 2

For all $x\in R$

$$|x-a| \leq 1 \implies |x+a|=|x-a+2a|\leq |x-a|+|2a|\leq 1 + 2|a|$$

Answer 3

You have to use the following triangle inequality: \begin{eqnarray} |x|-|y| \le |x-y| \tag{1}\label{1} \end{eqnarray} Suppose $|x-a| \le 1$. Writing: $$|x-a| = |x-2a+a|$$ call $x+a = z$. Then, by (\ref{1}): $$|z| -2|a| \le |z-2a| = |x-a| \le 1 \Rightarrow |x+a| \le 1+2|a|$$