Inequality in Algebra: $1 \leq x_1 x_2 \cdots x_n$ implies that $2^{n} \leq (1 + x_1)(1+x_2) \cdots (1 + x_n).$ How do I prove that if $x_1, \ldots, x_n$ are positive real numbers, then
$$1 \leq x_1 x_2 \cdots x_n \text{ implies that } 2^{n} \leq (1 + x_1)(1+x_2) \cdots (1 + x_n).$$
I attempted a proof by induction but am not able to nail the inductive step. Any help would be appreciated!
 A: Hint 
$$x_{i}+1\ge 2\sqrt{x_{i}}$$
A: $$
(1+x_1)(1+x_2)\dotsb(1+x_n)=(1+x_1+x_2+\dotsb+x_{n-1}+x_n)+(x_1x_2+x_1x_3+\dotsb+x_{n-1}x_n)+\dotsb+x_1x_2\dotsb x_n
$$
Now use AM-GM inequality for the right side of the above equation with $2^n$ terms.
A: By Holder 
$$(1+x_1)(1+x_2)...(1+x_n)\geq\left(1+\sqrt[n]{x_1x_2...x_n}\right)^n\geq(1+1)^n=2^n.$$
Done!
A: Hint:
Use the following form of induction:


*

*If $P(n)$, then $P(2n)$.

*If $P(n)$ then $P(n-1)$.

A: HINT: Expand the RHS and collect the terms with equal degree. Then use AM-GM on these collections separately. For example, the first such collection would be $$\frac{\sum_{i=1}^nx_i}{\binom n1}\ge(x_1x_2x_3...x_n)^{\frac 1n}\ge 1$$ Hence $$\sum_{i=1}^nx_i\ge \binom n1$$ Add all these results and use the fact that $$\sum_{i=0}^n\binom ni=2^n$$That leads you to the answer.
A: We are given that $x_1, x_2, \cdots, x_n$ are positive real numbers and $x_1x_2\cdots x_n\geq 1\forall n\in \mathbb N$. 
We have to show $(1+x_1)(1+x_2)\cdots (1+x_n)\geq 2^n$.
For $n=1$ the result is trivial as $x_1\geq 1\Rightarrow (1+x_1)\geq 1+1=2^1.$
Let the result be true for $n=m$ viz $x_1x_2\cdots x_m\geq 1$ implies 
$$(1+x_1)(1+x_2)\cdots (1+x_m)\geq 2^m$$ 
Multiplying both sides by the inequality $1+x_{m+1}\geq 2$ we see that 
$$(1+x_1)\cdots(1+x_m)(1+x_{m+1})\geq 2^{m+1}$$ which shows that the inequality is true for $n=m+1$ whenever it is true for $n=m$.
Induction now completes. 
