Follow the steps outlined here to prove the triangle inequality: $|a+b| \leq|a|+|b|$ for any real numbers $a$ and $b$ :
(a) Argue that for any real number $x,|x|=\sqrt{x^{2}}$.
(b) Show that $(a+b)^{2} \leq(|a|+|b|)^{2}$. (Hint: Start on the left-hand side, multiply out the expression, and use the fact that $a \leq|a|$ and $b \leq|b| .)$
(c) Take the square root of both sides of the inequality from part (b) (this is valid since both sides are positive), and use part (a) to show that $|a+b| \leq|a|+|b|$.