Given positive numbers $x_1,...,x_n$ such that $x_1\cdots x_n=1$ prove that $\frac{1}{n-1+x_1}+\cdots+\frac{1}{n-1+x_n}\le 1$

According to page 8, Problem 25 in [1], the equivalent problem was published by Vasile Cirtoaje on Gazeta Matematica, Seria B, No. 10, 1991.

Five solutions are given in [1]. Here is one of them:

Let $r = \frac{n-1}{n}$. By AM-GM, we have, for $i=1, 2, \cdots, n$, $$\sum_{j\ne i} x_j^r \ge (n-1) \left(\prod_{j\ne i} x_j\right)^{r/(n-1)} = (n-1) x_i^{r - 1}$$ which results in $x_1^r + x_2^r + \cdots + x_n^r \ge (n-1)x_i^{r - 1} + x_i^r$ and $$\frac{x_i}{x_i + n-1} \ge \frac{x_i^r}{x_1^r + x_2^r + \cdots + x_n^r}. \tag{1}$$ Summing up (1) for $i = 1, 2, \cdots, n$, we have $$\sum_{i=1}^n \frac{x_i}{x_i + n-1} \ge 1$$ which is written as $$\sum_{i=1}^n \frac{1}{x_i + n-1} \le 1.$$ We are done.

Reference

[1] Vasile Cirtoaje, “Algebraic Inequalities: Old and New Methods”, 2006.

One can use Lagrange multipliers for: $$L = \sum_{k=1}^{n} 1/(n-1 + x_{k}) - \lambda(\prod_{k=1}^{n}x_{k} - 1)$$ One has: $$\frac{dL}{dx_{k}} = -\frac{x_{k}}{(n-1+ x_{k})^{2}} -\lambda\prod_{j\neq k}x_{j}$$ And when this is zero, one can multiply both sides by $x_{k}$ to get: $$\frac{x_{k}^{2}}{(n-1+ x_{k})^{2}} =\lambda\prod_{k}x_{k}$$ Because of condition on product of $x_{k}'s$ all $x_{k}$ must be the same and equal to 1 at the maximum.

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}