TITLE: $x_1 \cdot \dots \cdot x_n=1 \implies x_1 + \dots + x_n \ge n$
QUESTION [1 upvotes]: Suppose $x_1, \dots, x_n$ are positive real numbers such that $x_1 \cdot \ldots \cdot x_n = 1$. Prove $x_1 + \dots + x_n \ge n$. 

I don't want to use the AM-GM inequality to prove this (from which this statement would follow). 
I suppose induction is the way to go. The base case is true. The case for $n=2$ is true by 
$$x_1 + x+2 - 2 \sqrt{x_1 x_2} = (\sqrt{x_1} - \sqrt{x_n})^2 \ge 0.$$
Now suppose the statement holds for $n$, then
$$x_1 \cdot \ldots \cdot x_n = 1 \implies x_1 + \dots + x_n \ge n.$$
So if $x_1 \cdot \ldots \cdot (x_n x_{n+1}) = 1$, then 
$$
x_1 + \dots + x_n x_{n+1} \ge n.
$$
Then we also have
$$
x_1 + \dots + x_{n-1} + x_n + x_{n+1} 
\ge n - x_{n}x_{n+1} + x_n + x_{n+1}.
$$
Now I'm not sure where else to go from here.

REPLY [0 votes]: The inequality follows from AM-GM, indeed: $\;x_1 + \dots + x_n \ge n \cdot \sqrt[n]{x_1 \cdot \ldots \cdot x_n}\,$. Below is a proof of AM-GM by induction, along a different idea than the ones posted already.
Lemma.   $\;(n-1)t^n+1 \ge n t^{n-1}\;$ for all $\;t \ge 1\,$.
Let $f\,(t)=(n-1)t^n - n t^{n-1} + 1\,$, then $\,f(1) = 0\,$ and $\,f'(t)=n(n-1)t^{n-2}(t-1) \ge 0\,$, so $\,f\,$ is increasing from $\,f(1)=0\,$ for $\,t \ge 1\,$ and therefore $\,f(t) \ge 0\,$ for $\,t \ge 1\,$.
(The stronger statement is true that $\,f(t) \ge 0\,$ for all $\,t \ge 0\,$. That follows from the identity $\,(n-1)t^n - n t^{n-1} + 1 = (t-1)^2 \cdot \sum_{k=1}^{n-1} k \cdot t^{k-1}\,$ which has an elementary no-calculus proof.)
Induction step.   The AM-GM inequality is homogeneous, so it can be assumed WLOG that the smallest $x_j=1\,$. Assume again WLOG that $\,j=n\,$, then the inequality reduces to:
$$x_1+x_2+ \cdots + x_{n-1} + 1 \ge n \cdot \sqrt[n]{x_1 \cdot \ldots \cdot x_{n-1}} \quad\style{font-family:inherit}{\text{with}}\quad x_i \ge 1, \;i=1,2,\ldots,n-1$$
By the induction hypothesis:
$$\,x_1+\cdots+x_{n-1} \ge (n-1) \sqrt[n-1]{x_1\cdot \ldots \cdot x_{n-1}} \tag{1}$$
Also, it follows from the lemma with $\,t = \sqrt[n(n-1)]{x_1\cdots x_{n-1}}\ge 1\,$ that: $$\,(n-1)\cdot\sqrt[n-1]{x_1\cdot \ldots \cdot x_{n-1}} + 1 \ge n \cdot \sqrt[n]{x_1\cdot \ldots \cdot x_{n-1}} \tag{2}$$
Therefore:
$$
x_1+ \ldots +x_{n-1}+1 \;\stackrel{(1)}{\ge}\; (n-1)\sqrt[n-1]{x_1\cdot \ldots \cdot x_{n-1}} + 1 \;\stackrel{(2)}{\ge}\; n \sqrt[n]{x_1 \cdot \ldots \cdot x_{n-1}}
$$
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