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Proof concerning absolute values and Triangle Inequality 
![\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}](http://www.mathhelpforum.com/math-help/latex2/img/e66fd27c27bf32bdf09b17d86002c854-1.gif "\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}")
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