<s>[INST] 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 $x>0$.
 [/INST] In fact， we may prove the more stonger one

Let $x_1,x_2,\cdots,x_n$ be positive numbers. Then
  $$\frac{1}{n}(x_1+x_2+\cdots+x_n)\geq \sqrt[n]{x_1x_2\cdots x_n}.\tag1$$

Put $x_1x_2\cdots x_n=1$ into $(1)$. You will get yours.
Inductive Proof
Obviously, $(1)$ holds for $n=1$, which is trivial.
Notice that, from $(\sqrt{x_1}-\sqrt{x_2})^2 \geq 0$, we may have $\dfrac{1}{2}(x_1+x_2) \geq \sqrt{x_1x_2}$, which shows $(1)$ holds for $n=2$.
Assume that $(1)$ holds for $n=k$, namely, $$\frac{1}{k}(x_1+x_2+\cdots+x_k)\geq \sqrt[k]{x_1x_2\cdots x_k}.$$ When $n=k+1$, denote $\dfrac{1}{k+1}(x_1+x_2+\cdots+x_k+x_{k+1})=x.$ Then,
\begin{align*}
x&=\frac{1}{2k}[(x_1+x_2+\cdots+x_k)+x_{k+1}+(k-1)x]\\&=\frac{1}{2}\left[\frac{1}{k}(x_1+x_2+\cdots+x_k)+\frac{1}{k}(x_{k+1}+(k-1)x)\right]\\
&\geq \frac{1}{2}\left(\sqrt[k]{x_1x_2\cdots x_k}+\sqrt[k]{x_{k+1}x^{k-1}}\right)\\
&\geq \sqrt[2k]{x_1x_2\cdots x_k \cdots x_{k+1}x^{k-1}},
\end{align*}
namely, $$x^{2k} \geq x_1x_2\cdots x_kx_{k+1}x^{k-1},$$ i.e. $$x^{k+1} \geq x_1x_2\cdots x_kx_{k+1}.$$Thus, $$\frac{1}{k+1}(x_1+x_2+\cdots+x_k+x_{k+1})\geq \sqrt[k+1]{x_1x_2\cdots x_kx+{k+1}},$$which shows that $(1)$ holds for $n=k+1.$
By induction, $(1)$ holds for all $n=1,2,\cdots$. 
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