Proof: (related to "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 $n\ge 1.$

| | | | created: 2020-06-25 09:51:29 | modified: 2020-06-25 09:51:55 | by: | references: [8311]