AM-GM inequality problem - solution explanation 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.
 A: Let's break this down and fill in the missing steps to see what's really going on here:
1) Yes, this is a misprint, but the mistake is where they have $$1-y_i = \sum_{j\neq i}y_i$$ which should actually be $$1-y_i=\sum_{j\neq i}y_j$$ so the line that $\prod_i\left(\frac{1-y_i}{y_i}\right)=\frac{\prod_i(\sum_{j\neq i}y_j)}{\prod_i y_i}$ is correct
2) By the AM-GM inequality we have
$$ \frac{\sum_{j\neq i}y_j}{n-1}\geq \left(\prod_{j\neq i}y_j\right)^{\frac{1}{n-1}}$$
now taking the product of both sides
$$\prod_i \left(\frac{\sum_{j\neq i}y_j}{n-1}\right)=\frac{\prod_i \sum_{j\neq i}y_j}{(n-1)^n}\geq \prod_i \left(\prod_{j\neq i}y_j\right)^{\frac{1}{n-1}}$$
and dividing both sides by the product of $y_i$ and multiplying by $(n-1)^n$ gives us
$$ \prod_i x_i=\frac{\prod_i \sum_{j\neq i}y_j}{\prod_i y_i}\geq(n-1)^n\frac{\prod_i\left(\prod_{j\neq i}y_j\right)^{\frac{1}{n-1}}}{\prod_iy_i}$$
3) If you examine $\prod_i \left(\prod_{j\neq i}y_j\right)^{1/(n-1)}$ you will find that as the first product runs over each $y_i$, the second product skips that $y_i$ and this happens $n$ times (for each $y_i$) so each $y_i$ appears $(n-1)$ times in the second product. A simple example with $n=3$ would look like this -
$$ \prod_{i}^{3}\left(\prod_{j\neq i}y_j\right)^{\frac{1}{2}} = \sqrt{y_2y_3}\sqrt{y_1y_3}\sqrt{y_1y_2}=y_1y_2y_3$$
[edit] In all the products and sums, it is implicit that the indices range from $1$ to $n$ so $$\prod_{i}(n-1) = \prod_{i=1}^{n}(n-1)$$ if we let $a=n-1$ then I think the easiest way to see why we get $(n-1)^n$ is by writing $$\begin{align}\prod_{i=1}^{n}a &= \prod_{i=1}^{n}(i\cdot 0 + a) \\&= (1\cdot 0+a)(2\cdot 0+a)\cdots ((n-1)\cdot 0 + a)(n\cdot 0+a) \\&= a^n\\ &= (n-1)^n\end{align}$$
here is a link explaining Pi notation in more detail.
The only thing I have to add is that the double product in this case is different only because we have a restriction on the index $j$ $$\prod_{i}\left(\prod_{j \neq i}y_j\right) = \prod_{i=1}^{n} \left(\prod_{j=1,\; j\neq i}^{n}y_j\right)$$ becomes $$ (y_2y_3\cdots y_n)(y_1y_3\cdots y_n)\cdots(y_1y_2\cdots y_{k-1}y_{k+1} \cdots y_n)\cdots(y_1y_2\cdots y_{n-1})$$
You do this in much the same way you would with Sigma notation for sums, but with multiplication. Hopefully that helps.
