# Thread: Proof concerning absolute values and Triangle Inequality

1. ## Proof concerning absolute values and Triangle Inequality

Hey guys ...I haven't been online for six months. I need your help with the following question.

Prove that $\mid \mid a \mid - \mid b \mid \mid \leq \mid a-b \mid$

Thank You

2. \begin{aligned}
\left| a-b \right|^{2}&=(a-b)^{2} \\
& =a^{2}-2ab+b^{2} \\
&\ge \left| a \right|^{2}-2\left| a \right|\left| b \right|+\left| b \right|^{2} \\
& =\Big[ \left| a \right|-\left| b \right| \Big]^{2} \\
\end{aligned}

3. Originally Posted by hercules
Hey guys ...I haven't been online for six months. I need your help with the following question.

Prove that $\mid \mid a \mid - \mid b \mid \mid \leq \mid a-b \mid$

Thank You
I like Kriz's method! here's an alternate method:

$|a| = |(a - b) + b| \le |a - b| + |b|$ (The $\triangle$-inequality)

Thus, we have $|a - b| \ge |a| - |b|$

similarly, we can start with $|b|$ and deduce that $|a - b| \ge |b| - |a| \implies -|a - b| \le |a| - |b|$

putting them together we have $-|a - b| \le |a| - |b| \le |a - b| \implies ||a| - |b|| \le |a - b|$ by the definition of absolute values

4. Originally Posted by Jhevon

(The $\triangle$-inequality)
This can be proved in the same fashion as I did above.

5. Originally Posted by Krizalid
This can be proved in the same fashion as I did above.

yes, i know. that's the kind of proof i have for it

i just used this method because he mentioned triangle inequality in the original post

6. Thank You both replies were really helpful.