Two inequalities by induction on the number of variables

1. For $n=1$, $x_1=1$. Assume it holds for $n$ numbers and take $x_1,\dots,x_{n+1}$ with product $1$. Not all are $<1$ and not all $>1$; reorder so that $x_n\ge1\ge x_{n+1}$. The $n$ numbers $x_1,\dots,x_{n-1},x_nx_{n+1}$ have product $1$, so by induction
$$x_1+\dots+x_{n-1}+x_nx_{n+1}\ge n.$$
Therefore
$$\sum_{i=1}^{n+1}x_i\ \ge\ n+x_n+x_{n+1}-x_nx_{n+1}=n+1+(x_n-1)(1-x_{n+1})\ \ge\ n+1.$$

2. For $n=1$ it is an equality. Suppose it holds for $n$, and let $S_n=x_1+\dots+x_n$. Since $1+x_{n+1}>0$,
$$\prod_{i=1}^{n+1}(1+x_i)\ge(1+x_{n+1})(1+S_n)=1+S_{n+1}+x_{n+1}S_n\ge1+S_{n+1},$$
because $x_{n+1}$ and $S_n$ have the same sign, so $x_{n+1}S_n\ge0$.

AM-GM. For positive $a_1,\dots,a_n$ with geometric mean $g$, set $x_i=a_i/g$: the product of the $x_i$ is $1$, so $\sum a_i/g\ge n$, i.e. $\frac{a_1+\dots+a_n}n\ge g$.

Lesson. Induction on the number of variables works when two variables can be merged into one while keeping the constraint: choose the pair on opposite sides of $1$.