If $x_i>0 , i=1,..,n$, and $\prod_i x_i=1$, then prove by induction that the $\sum x_i \ge n$.
My progress: First step: n=1,2. for n=1 we have x1>=1 which is true. For n=2 we have x1x2=1, we have to show that x1+x2>=2; So starting from x1,x2>0 and x1x2=1, we suppose that 0<x1<1; x2>1 than we have 0<(1-x1)(x2-1)=x2-1-x1x2+x1=(x1+x2)-1-1=(x1+x2)-2 which means that x1+x2>=2. So it is true for n=2.
Second step: suppose that it is true for all n: x1,...,xn>0; x1x2..*xn=1 then x1+x2+...+xn>=n.
Third step: Show that if is true for all n+1: If x1,..,xn,xn+1>0; x1*...xnxn+1=1 then x1+...+xn+xn+1>=n+1 (we have to prove this one!)
So I took a "k" : there exist a k from the set {1,2,..,n+1} and a xq<1 then suppose that xq+1>1 (1-xk)(xk+1-1)>0 then I rewrote the sequence as: x1+x2+...+xn+xn+1=x1+x2+...+(xk+xk+1 -1 -xkxk+1)+1+xkxk+1 + (xk+2 + xk-1). This is my progress please if you could give a help, because if I prove this one, it will be easily for me to prove the A.M>G.M>H.M. THANK YOU!
For $n = 2$, we have $x_1 + x_2 = x_1 + \frac{1}{x_1} \ge 2$ (obvious)
Assume true for $n$. Also we can assume all $x_i$ not equal. WLOG, assume $x_n > 1, x_{n+1} < 1$
$x_1x_2\cdots (x_nx_{n+1}) = 1$
So by induction, we get $x_1 + x_2 + ... + x_nx_{n+1} \ge n$ $\implies \sum\limits_1^{n+1} x_i \ge n + x_n + x_{n+1} - x_nx_{n+1} = n+1 + (x_n-1)(1 - x_{n+1})$
The last term is greater than $0$ by assumption.