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 $x>0$.

An alternative proof  to AM-GM inequality is Lagrange Multiplier method.
We are minimizing $$x_1+x_2+...+x_n$$ subject to $$x_1x_2x_3...x_n=1$$
That gives us$$ <1,1,1,...,1>=\lambda<x_2x_3...x_n, x_1x_3....x_1x_2...x_{n-1}>$$
$$x_1=x_2=x_3=...=x_n=1$$
$$x_1+x_2+...+x_n=n$$
