Proof: By Induction

(related to Proposition: Generalized Bernoulli's Inequality)

We provide a proof by induction for natural numbers $n.$ * Base case: $n=1$ $$\prod_{k=1}^1(1+x_k)=1+x_1\ge 1+\sum_{k=1}^1 x_k=1+x_1.$$ * Induction step $n\to n+1$ * Assume, $$\prod_{k=1}^n(1+x_k)\ge 1+\sum_{k=1}^n x_k$$ holds for some $n\ge 1.$ * Then $$\begin{align}\prod_{k=1}^{n+1}(1+x_k)&=\prod_{k=1}^{n}(1+x_k)\cdot (1+x_{n+1})\nonumber\\ &\ge \left(1+\sum_{k=1}^n x_k\right)\cdot (1+x_{n+1})\nonumber\\ &= 1+x_{n+1} +\sum_{k=1}^n x_k+x_{n+1}\cdot \sum_{k=1}^n x_k\nonumber\\ &= 1+\sum_{k=1}^{n+1} x_k+x_{n+1}\cdot \sum_{k=1}^n x_k\nonumber\\ &\ge 1+\sum_{k=1}^{n+1} x_k\nonumber\end{align}$$ * Thus, the above product holds for all natural numbers $n\ge 1.$

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  1. Modler, F.; Kreh, M.: "Tutorium Analysis 1 und Lineare Algebra 1", Springer Spektrum, 2018, 4th Edition