(related to Proposition: Generalized Triangle Inequality)
We provide a proof by induction for natural numbers n\ge 1 and absolute value of real numbers (respectively absolute value of complex numbers). * Base case: n=1 \left|\sum_{k=1}^1 a_k\right|=|a_1|\le|a_1|=\sum_{k=1}^1|a_k|. * Induction step n\to n+1 * Assume, \left|\sum_{k=1}^n a_k\right|\le\sum_{k=1}^n |a_k| holds for some n\ge 1. * Then, by the triangle inequality \begin{align}\left|\sum_{k=1}^{n+1} a_k\right|&\le \left|\sum_{k=1}^{n} a_k\right|+|a_{n+1}|\nonumber\\ &\le\sum_{k=1}^{n} |a_k|+|a_{n+1}|\quad\text{(by assumption)}\nonumber\\ &= \sum_{k=1}^{n+1} |a_k|.\nonumber\end{align}
