Proof using triangle inequality

Question

Fix a real number x and a positive number $\epsilon$. If $\lvert x-1\rvert \le \epsilon$, show that $\lvert 2-x\rvert \ge 1 - \epsilon$

Pf: Fix $x \in R$ and let $\epsilon>0$,

We know $-\epsilon\le x-1 \le \epsilon$ or $-\epsilon\le 1-x \le \epsilon$

$\lvert 2 -x \rvert \ge \lvert 2 \rvert - \lvert x \rvert = 2 - x \ge 1+(1-x) \ge 1-\epsilon $

Thus proving that $\lvert 2-x \rvert \ge 1-\epsilon $

Answer 1

$|x-1| \leq \epsilon$

Definition of inequality and absolute value.

$-\epsilon \leq x-1 \leq \epsilon$

Multiple by $-1$$.

$-\epsilon \leq 1-x \leq \epsilon$

Add $1$.

$1 -\epsilon \leq 2-x \leq 1+ \epsilon$

Focus on the left inequality.

$1 -\epsilon \leq 2-x $

We know that

$1 -\epsilon \leq 2-x \leq |2-x|$

Drop the middle.

$1 -\epsilon \leq |2-x|$