Proof

From the definition of absolute value for real numbers it follows $x \le |x|$ and $y\le|y|$ for all $x,y\in\mathbb R$.
Together with the rules of calculation with inequalities (in particular the rule 4), it follows that, if we add both inequalities, we get $x+y\le |x|+|y|.$
Using the same argument, we get from the sum of the inequations $-x \le |x|$ and $-y\le|y|$ that $-(x+y)=-x-y\le |x|+|y|.$
Using the definition of absolute value for real numbers once again, we get $|x+y|\le |x|+|y|$ for all real numbers $x,y\in\mathbb R.$

Since the absolute value of complex numbers defines a distance of complex numbers, it fulfills the properties of a metric, in particular the triangle inequality.
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Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983