It seems that the following statement about absolute value inequalities are very useful in real analysis. I can prove it using case analysis, but would expect a more direct proof, such as via the triangle inequality. Is there a direct proof? Is my proof below the best? Is there a name for this statement?
If $a$ is within $e$ of $b$, then $a > b-e$ and $a < b+e$. More formally, $|a-b|<e \implies b-e < a < b+e$.
Proof: $e$ is clearly greater than $0$. If $a > b$, then $a > b-e$. If $a < b$, then $-(a-b)=b-a < e$, so $a > b-e$. A similar argument follows for the other half of the inequality. QED.