Prove $|a−b| \ge |a|−|b|$

Question

I can't go anywhere with this. Thanks. Prove that $|a - b| \ge |a| - |b|$. Hint: apply the triangle inequality to $|a| = |(a - b) + b|$. Using the triangle inequality: $$ |a + b| \le |a| + |b| \Rightarrow $$ $$ |a + b| \le |(a - b) + b| + |b| $$

Read the hint carefully. Perhaps it would help to rewrite the triangle inequality with different letters. — Ted Shifrin, Mar 01 '23 at 04:57
Does this answer your question? Triangle inequality for subtraction? - found using an Approach0 search. Note this answer there shows how to use your provided hint. There's also other "reverse triangle" questions, e.g., Similar to triangle (and reverse triangle) inequality, proof that $||x|-|y|| ≤ |x+y|$. — John Omielan, Mar 01 '23 at 06:16

score 0 · Answer 1 · answered Mar 02 '23 at 05:22

Prove that $|a - b| \ge |a| - |b|$ $$|a| = |(a - b) + b|$$ Apply the triangle inequality $|a + b| \le |a| + |b|$ by substituting $(a - b)$ for $|a|$: $$|a| = |(a - b) + b| \le |a - b| + |b| \Rightarrow$$ $$|a| - |b| \le |a - b| \Rightarrow$$ $$|a - b| \ge |a| - |b|$$

Prove $|a−b| \ge |a|−|b|$

1 Answers1

Linked