(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 $$\begin{align}\left|\sum_{k=1}^{n+1} a_k\right|&\le \left|\sum_{k=1}^{n} a_k\right|+|a_{n+1}|\nonumber\\ &\le\sum_{k=1}^{n} |a_k|+|a_{n+1}|\quad\text{(by assumption)}\nonumber\\ &= \sum_{k=1}^{n+1} |a_k|.\nonumber\end{align}$$