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.
 A: By AM-GM inequality, 
$$(1+x_i) \geq 2 \sqrt x_i$$  
Taking product from $i=1$ to $n$, we get the desired result.
A: $$LHS \geqslant 2\sqrt{x_1}\cdot 2\sqrt{x_2}\cdots 2\sqrt{x_n}=2^n.$$
A: Not substantially different form other answers, but perhaps an interesting manipulation: multiply the LHS by $$1=1^{-1/2}=\prod_{k=1}^{n}x_k^{-1/2}$$ obtaining $$\prod_{j=1}^k(1+x_j)=\prod_{j=1}\left(\frac1{x_j^{1/2}}+x_j^{1/2}\right)$$
And then use the fact that (for the reason you prefer) $$\min\left\{ \frac1x+x\,:\,x>0\right\}=2$$
A: 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:
$$\prod_{S\subseteq \{1,\dots,n\}}\prod_{i\in S} x_i=x_1^{2^{n-1}}\cdots x_n^{2^{n-1}}$$
A: By Holder
$$(1+x_1)(1+x_2)...(1+x_n)\geq\left(1+\sqrt[n]{x_1x_2...x_n}\right)^n=2^n$$
