I was looking into AM-GM inequalities recently and how it could be used to solve and give proof to some problems. I have been faced by this problem that i needed to look up its solution. The solution was very understandable, but a few bits. I would really be grateful if someone can offer me help.
The problem is as follows :-
Let $x_1,x_2,...,x_n > 0$ such that $\frac{1}{1+x_1} + \frac{1}{1+x_2} + ...+\frac{1}{1+x_n} =1$. Prove that $$x_1x_2...x_n \ge (n-1)^n$$
and the solution was :-
Let $y_i = \frac{1}{1+x_1}$, then $x_i = \frac{1}{y_i} - 1 = \frac{1-y_i}{y_i}$. Observe that $y_1+y_2+...+y_n = 1$ implies that $1-y_i = \sum_{j\neq i}y_i$.And by AM-GM $\sum_{j\neq i}y_i \ge (n-1)(\prod_{j\neq i }y_j)^\frac{1}{n-1}$ and $$\prod_i x_i = \prod_i (\frac{1-y_i}{y_i}) = \frac{\prod_i (\sum_{j\neq i} y_j)}{\prod_i y_i} \ge \frac{(n-1)^n\prod_i (\prod_{j\neq i} y_j)^\frac{1}{n-1}}{\prod_i y_i} = (n-1)^n$$
The first two lines are easy to grasp but i have some questions on the rest of the proof :-
1)At the step $\prod_i (\frac{1-y_i}{y_i}) = \frac{\prod_i (\sum_{j\neq i} y_j)}{\prod_i y_i}$ should it not be $\frac{\prod_i (\sum_{j\neq i} y_i)}{\prod_i y_i}$ since $1-y_i = \sum_{j\neq i} y_i$.Is it my bad or just a misprint.
2)I would like an explanation for the numerator change in $\frac{\prod_i (\sum_{j\neq i} y_j)}{\prod_i y_i} \ge \frac{(n-1)^n\prod_i (\prod_{j\neq i} y_j)^\frac{1}{n-1}}{\prod_i y_i} $. I do understand the AM-GM bit however, I don't get where the exponent of $(n-1)$ came from.
3)I also do not understand how $\frac{(n-1)^n\prod_i (\prod_{j\neq i} y_j)^\frac{1}{n-1}}{\prod_i y_i}$ was just simplified down to $(n-1)^n$.
I know that my question may seem very basic to some of you but I have been stuck for more than a day, Thanks.