Show $(1+x_1)(1+x_2)...(1+x_n)\geq2^n$ given $x_1x_2...x_n=1$ 
Show $(1+x_1)(1+x_2)...(1+x_n)\geq2^n$ given $x_1x_2...x_n=1$ and that all $x_i $ are positive reals.

I think simple AM-GM-HM must work, but I am missing on something trivial.

The brute force approach is:
$$(1+x_1)(1+x_2)\cdots(1+x_n)=\sum_{S\subseteq \{1,\dots,n\}}\prod_{i\in S}x_i$$
But by AM/GM, since there are $2^n$ subsets of $\{1,\dots,n\}$, you get:
$$\frac{1}{2^n}\sum_{S\subseteq \{1,\dots,n\}}\prod_{i\in S}x_i \geq \sqrt[2^n]{x_1^{2^{n-1}}\cdots x_n^{2^{n-1}}}=1$$
This is because:
$$\prod_{S\subseteq \{1,\dots,n\}}\prod_{i\in S} x_i=x_1^{2^{n-1}}\cdots x_n^{2^{n-1}}$$
