Hey guys ...I haven't been online for six months. I need your help with the following question.
Prove that $\displaystyle \mid \mid a \mid - \mid b \mid \mid \leq \mid a-b \mid$
Thank You
$\displaystyle \begin{aligned}
\left| a-b \right|^{2}&=(a-b)^{2} \\
& =a^{2}-2ab+b^{2} \\
&\ge \left| a \right|^{2}-2\left| a \right|\left| b \right|+\left| b \right|^{2} \\
& =\Big[ \left| a \right|-\left| b \right| \Big]^{2} \\
\therefore\quad \left| a-b \right|&\ge \Big| \left| a \right|-\left| b \right| \Big|.\quad\blacksquare
\end{aligned}$
I like Kriz's method! here's an alternate method:
$\displaystyle |a| = |(a - b) + b| \le |a - b| + |b|$ (The $\displaystyle \triangle$-inequality)
Thus, we have $\displaystyle |a - b| \ge |a| - |b|$
similarly, we can start with $\displaystyle |b|$ and deduce that $\displaystyle |a - b| \ge |b| - |a| \implies -|a - b| \le |a| - |b|$
putting them together we have $\displaystyle -|a - b| \le |a| - |b| \le |a - b| \implies ||a| - |b|| \le |a - b|$ by the definition of absolute values