<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] EDIT: As pointed out, this answer contains a flawed application of Cauchy-Schwarz. In particular, the L.H.S. of the inequality can be zero, while the R.H.S. is positive, an absurdity. I tried to remedy this flaw but ultimately failed. Hence, I have come up with a new approach that proceeds by a "smoothing principle" argument. Any feedback on the new proof will be much appreciated.
Note: In your post, you claimed the following:
$x_k=a_k^n=a_ka_k^{n-1}=\frac{a_k^{n-1}}{a_1...a_n}.$
I do not see how the last equality follows. If it were true, then we would have $a_k=\dfrac{1}{a_1a_2...a_n}=1$, which is not necessarily the case. But anyways, it is possible to solve the problem using AM-GM and Cauchy-Schwarz.
Firstly, the case where $x_1=x_2=...=x_n=1$ is trivial. We only consider the case where $\exists \ i \in \{1,2,..,n\}$ such that $x_i \neq 1$. Note that:
$$\dfrac{1}{n-1+x_1} + \dfrac{1}{n-1+x_2} + ... + \dfrac{1}{n-1+x_n} \leq 1$$
\begin{align}
 & \iff \left(\dfrac{1}{n}- \dfrac{1}{n-1+x_1}\right) + \left(\dfrac{1}{n}- \dfrac{1}{n-1+x_2}\right) + ... + \left(\dfrac{1}{n}- \dfrac{1}{n-1+x_n}\right) \geq 0 \\
& \iff \dfrac{x_1-1}{n(n-1+x_1)} + \dfrac{x_2-1}{n(n-1+x_2)} + ... + \dfrac{x_n-1}{n(n-1+x_n)} \geq 0 \\
& \iff \dfrac{x_1-1}{n-1+x_1} + \dfrac{x_2-1}{n-1+x_2} + ... + \dfrac{x_n-1}{n-1+x_n} \geq 0 \ (\because \dfrac{1}{n} > 0 ) \\
\end{align}
By Cauchy-Schwarz,
$$\left(\dfrac{x_1-1}{n-1+x_1} + \dfrac{x_2-1}{n-1+x_2} + ... + \dfrac{x_n-1}{n-1+x_n} \right) \left[(x_1-1)(n-1+x_1) + ...  + (x_n-1)(n-1+x_n) \right]$$
$$\geq \left[(x_1-1)+(x_2-1) + ... + (x_n-1) \right]^2$$
The R.H.S. of the preceding inequality is clearly non-negative. It thus suffices to prove that:
\begin{align}
& (x_1-1)(n-1+x_1) + (x_2-1)(n-1+x_2) + ... + (x_n-1)(n-1+x_n) > 0 \\
& \iff n(x_1-1)+(x_1-1)^2 + n(x_2-1)+(x_2-1)^2 + ... +  \ n(x_n-1)+(x_n-1)^2 > 0 \\
& \iff (x_1-1)^2 + ... + (x_n-1)^2 + n\left[(x_1-1) + ... + (x_n-1) \right] > 0. \\
\end{align}
Since $\exists \ i \in \{1,2,..,n\}$ such that $x_i \neq 1$, it is obvious that $(x_1-1)^2 + ... + (x_n-1)^2 > 0$. Now, we claim that $n\left[(x_1-1) + ... + (x_n-1) \right] \geq 0 $, which concludes our proof. To see why this is true, note that:
\begin{align}
n\left[(x_1-1) + ... + (x_n-1) \right] & = n(x_1 + ... + x_n -n) \\
& \geq n(n \sqrt[^n]{x_1...x_n} - n) \ (\text{By applying AM-GM inequality})\\
& = n(n-n) \ (\because x_1...x_n=1) \\
&=0.
\end{align}
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