Trick with inequalities

Question

I have that $\forall n >n_0$ I have $|a_n-a|<1$. I want to get to the result that says $\forall n>n_0$ we have $|a_n|<|a|+1$.

I am very doubtful about my working out as I don't know if it works, especilly the highlighted part:

Suppose $a_n-a \geq 0$, then $|a_n-a|=a_n-a<1$ hence $a_n<a+1$.

Suppose $a_n-a <0$ then $|a_n-a|=-a_n+a <1$ hence $-a_n <-a +1$.

Hence since $a+1$ or $-a +1$, which is (is it though?) $|a|+1$, this works for $a_n$ and $-a_n$, we have $|a_n|<|a|+1$

Is this even correct? Or how should I prove that?

Answer 1

Just use the triangle inequality: $\;\lvert a_n \rvert=\lvert (a_n-a) +a\rvert\le \lvert a_n -a\rvert+\lvert a \rvert <1+\lvert a \rvert $.

Answer 2

$$|a\pm b|\ge |a|-|b|\implies |a_n|-|a|\le|a_n-a|<1\implies |a_n|<|a|+1$$