<s>[INST] Given positive numbers $x_1,...,x_n$ such that $x_1\cdots x_n=1$ prove that $\frac{1}{n-1+x_1}+\cdots+\frac{1}{n-1+x_n}\le 1$ 
Given positive numbers $x_1,\dots,x_n$ such that $x_1\cdots x_n=1$ prove that $\frac{1}{n-1+x_1}+\cdots+\frac{1}{n-1+x_n}\le 1$.

I tried solving this question through substitution, e.g. $a_1=\sqrt[n]{x_1},...,a_n=\sqrt[n]{x_n}$, hence $a_1...a_n=1$. Also for $k=1,2,...,n$ we have that $x_k=a_k^n=a_ka_k^{n-1}=\frac{a_k^{n-1}}{a_1...a_n}$.
This is where I got stuck. My intuition tells me that I should be able to finish it off from here using AM-GM, however it isn't working out for me. Could you please explain to me how to finish the question off?
 [/INST] The case $x_1=x_2=...=x_n=1$ is trivial, and for $n >2$, is the only case when equality holds. We prove this below.
Suppose the $x_i$s are not all $1$. Then, there exists $i,j \in \{1,2,..,n\}, i \neq j$, such that $x_i<1<x_j$. Replace the pair $(x_i  \ ,x_j)$ with $(x_i'\ ,x_j')$ , such that:
$$x_i'=1, \ x_j'=x_ix_j.$$
$x_i'$ and $x_j'$ have the same product as $x_i$ and $x_j$. Consider the sum
$$\dfrac{1}{n-1+x_i} + \dfrac{1}{n-1+x_j} = \dfrac{x_i+x_j+2(n-1)}{(n-1)^2 + (n-1)(x_i+x_j) + x_ix_j}.$$
It would be ideal if the above sum was strictly smaller than $\dfrac{1}{n-1+x_i'} + \dfrac{1}{n-1+x_j'} = \dfrac{1}{n} + \dfrac{1}{n-1+x_ix_j}.$ (By our replacement, we increase the number of $x_i$s which are equal to $1$, while strictly increasing the L.H.S. of the given inequality and preserving the given constraint. Eventually, all the $x_i$s will equal $1$, and the inequality becomes an equality.) Attempting to proceed along this line,
\begin{align}
& \dfrac{1}{n-1+x_i} + \dfrac{1}{n-1+x_j} < \dfrac{1}{n} + \dfrac{1}{n-1+x_ix_j} \\
& \iff \dfrac{1}{n-1+x_i} - \dfrac{1}{n} < \dfrac{1}{n-1+x_ix_j} - \dfrac{1}{n-1+x_j} \\
& \iff \dfrac{1-x_i}{n(n-1+x_i)} < \dfrac{x_j-x_ix_j}{(n-1+x_ix_j)(n-1+x_j)} \\
& \iff \dfrac{1}{n(n-1+x_i)} < \dfrac{x_j}{(n-1+x_ix_j)(n-1+x_j)} \\ 
& \iff (n-1+x_ix_j)(n-1+x_j) < nx_j(n-1+x_i) \\
& \iff x_ix_j^2 - x_ix_j + nx_j - x_j + n^2-2n+1 - n^2x_j + nx_j < 0 \\
& \iff n^2(1-x_j) - x_ix_j(1-x_j) - 2n(1-x_j) + (1-x_j) < 0 \\
& \iff n^2-x_ix_j -2n + 1 >0 \\
& \iff x_ix_j < (n-1)^2.
\end{align}
Unfortunately, it is possible to have $x_ix_j \geq (n-1)^2$. Fortunately, this issue can be circumvented. W.L.O.G. let $x_1 \leq x_2 ... \leq x_n , x_1 < 1 < x_n$. There are $2$ cases:
Case $1$:
$x_1x_n \geq (n-1)^2 \Rightarrow \dfrac{x_1+x_n+2(n-1)}{(n-1)^2 + (n-1)(x_1+x_n) + x_1x_n } \leq \dfrac{x_1+x_n+2(n-1)}{2(n-1)^2 + (n-1)(x_1+x_n)} = \dfrac{1}{n-1}.$ Since $\dfrac{1}{n-1+x_n}$ is positive, we conclude that $\dfrac{1}{n-1+x_1} < \dfrac{1}{n-1}$. But the largest term in the sum $ \displaystyle \sum_{i=1}^{n} \dfrac{1}{n-1+x_i}$ is $\dfrac{1}{n-1+x_1}$. Thus,
$$\displaystyle \sum_{i=1}^{n} \dfrac{1}{n-1+x_i} \leq \dfrac{1}{n-1} + \displaystyle \sum_{i=2}^{n-1} \dfrac{1}{n-1+x_i} < \dfrac{1}{n-1} + \dfrac{n-2}{n-1} =1. $$
Case $2$: $x_1x_n < (n-1)^2 \Rightarrow \dfrac{1}{n-1+x_1} + \dfrac{1}{n-1+x_n} < \dfrac{1}{n} + \dfrac{1}{n-1+x_1x_n}$. Letting $x_1'=1, x_n'=x_1x_n$ and $x_j'=x_j \ \forall \ j \notin \{1,n\},$ we have $\displaystyle \sum_{i=1}^{n} \dfrac{1}{n-1+x_i} < \displaystyle  \sum_{i=1}^{n} \dfrac{1}{n-1+x_i'}.$ And we are done by a repeated application of this "smoothing" procedure! (At each stage,  W.L.O.G. re-order the  variables $x_i$s from smallest to largest, since the previous round of "smoothing" disrupted the original order.) If, at any stage, the product of the smallest term with the largest term is greater than or equal to $(n-1)^2$, we are done by Case $1$. Otherwise, we continually increase the L.H.S. and eventually end up with all the $x_is$ equal to $1$, giving the equality case L.H.S. $=1$.
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