Proof by Induction: If $x_1x_2\dots x_n=1$ then $x_1 + x_2 + \dots + x_n\ge n$ 
If $x_1,x_2,\dots,x_n$ are positive real numbers and  if $x_1x_2\dots x_n=1$ then $x_1 + x_2 + \dots + x_n\ge n$

There is a step in which I am confuse. My proof is as follows (it must be proven using induction).
By induction, for $n=1$ then $x_1=1$ and certainly $x_1\ge1$.
Suppose $x_1 + x_2+\dots+x_n\ge n$ (does this mean that $x_1x_2\dots x_n = 1$ also hold?) and $x_1x_2\dots x_n x_{n+1} = 1$ hold. Then
\begin{align}
x_1 + x_2 + \dots+x_n+x_{n+1} &\ge n + x_{n+1} \\
                              &=n+2-1/x_{n+1} \\
                              &=n+2-x_1x_2\dots x_n.
\end{align}
My problem is in the last step. As I wrote before, I don't think $x_1x_2\dots x_n = 1$  should hold, because if this is the case then $x_{n+1}=1$.
EDIT:
In the itermediate step I used that $x + 1/x \ge 2$, where $x>0$.

Here is a way to prove AM-GM using backwards induction and then using this to prove your proposition:
We will start with $n=2$: $$(x_1+x_2)^2/4-x_1x_2=\frac14(x_1^2+2x_1x_2+x_2^2-4x_1x_2)=\frac14(x_1-x_2)^2\ge0\\\implies\frac{x_1+x_2}{2}\ge\sqrt{x_1x_2}$$
Assume it is true for $2^k$ for some $k$ natural number we will show it is true for $2^{k+1}$;
$$\begin{align}&\frac{x_1+x_2+\ldots+x_{2^k}+x_{2^k+1}+\ldots+x_{2^{k+1}}}{2^{k+1}}\\=&\frac{\frac{x_1+x_2+\ldots+x_{2^k}}{2^{k}}+\frac{x_{2^k+1}+\ldots+x_{2^{k+1}}}{2^{k}}}{2}\\\overset{\mbox{assumption}}{\ge}&\frac{({x_1+x_2+\ldots+x_{2^k}})^{2^{-k}}+({x_{2^k+1}+\ldots+x_{2^{k+1}}})^{2^{-k}}}{2}\\\overset{\mbox{base case(n=2)}}{\ge}&\sqrt{({x_1x_2\ldots x_{2^k}})^{2^{-k}}({x_{2^k+1}\ldots x_{2^{k+1}}})^{2^{-k}}}\\=&({x_1x_2\ldots x_{2^k}}{x_{2^k+1}\ldots x_{2^{k+1}}})^{2^{-(k+1)}}\end{align}$$
Now let's prove that if it is true for $n$ it is true for all natural numbers less than $n$, take $n<2^k$ for some $k$:$$\mbox{denote }a=\frac{x_1+x_2+\ldots+x_n}{n}$$Hence$$a=\frac{x_1+x_2+\ldots+x_n+(2^k-n)a}{2^k}=\frac{x_1+x_2+\ldots+x_n+\overbrace{a+\ldots+a}^{2^k-n~times}}{2^k}$$Because we know that the inequality holds for $2^k$ we have$$a\ge(x_1x_2\ldots x_n\overbrace{a\ldots a}^{2^k-n~times})^{2^{-k}}=(x_1x_2\ldots x_n{a}^{2^k-n})^{2^{-k}}\\\implies a^{2^k}\ge x_1x_2\ldots x_n{a}^{2^k-n}\implies a^{n}\ge x_1x_2\ldots x_n\\\implies a=\frac{x_1+x_2+\ldots+x_n}{n}\ge (x_1x_2\ldots x_n)^{\frac1n}$$

Therefore we showed that the AM-GM inequality holds if $n=2^k$ and also holds if $n<2^k$ for some $k$, therefore it is true for all $n$.
Now we can put $x_1x_2\dots x_n=1$ and by the inequality we have $$\frac{x_1+x_2+...+x_n}{n}\geq \sqrt[n]{x_1 x_2\ldots x_n}=1\\\implies x_1+x_2+...+x_n\ge n$$

Maybe it is a little over the top to do it like this but I like this way.
