# Generalized Bernoulli's inequality

I was able to prove Bernoulli's inequality, easily by simple induction.
However, I'm not sure how to prove the generalized inequality (generalized = for each sequence of numbers $i=1,\ldots,n$):

$$\prod\limits_{i = 1}^n {(1 + {x_i})} \ge 1 + \sum\limits_{i = 1}^n {{x_i}},\qquad {x_i} \ge 0$$

How do you prove it? Thanks!

• Simply expand the product? – Gottfried Helms Oct 23 '13 at 7:29
• Or induction again... – Macavity Oct 23 '13 at 7:31

Let $$P(n):(1+x_1)(1+x_2)\cdots(1+x_n)\ge1+\sum_{1\le i\le n}x_i$$
Clearly, $P(n)$ holds for $n=1$
Let $P(n)$ holds for $n=m$
$$\implies (1+x_1)(1+x_2)\cdots(1+x_m)\ge1+\sum_{1\le i\le m}x_i$$
For $n=m+1,$
$$(1+x_1)(1+x_2)\cdots(1+x_m)(1+x_{m+1})\ge(1+\sum_{1\le i\le m}x_i)(1+x_{m+1})$$ $$=1+\sum_{1\le i\le m+1}x_i+x_{m+1}\sum_{1\le i\le m}x_i\ge1+\sum_{1\le i\le m+1}x_i$$ as $x_i>0\forall i\in[1,n]$