TITLE: prove $\sum_{cyc}\frac{n-1+x_{2}x_{3}\cdots x_{n}}{1+(n-1)x_{1}}\ge n$
QUESTION [9 upvotes]: let $x_{i}>0(i=1,2,\cdots,n)$,prove or disprove
$$\dfrac{n-1+x_{2}x_{3}\cdots x_{n}}{1+(n-1)x_{1}}+\dfrac{n-1+x_{1}x_{3}\cdots x_{n}}{1+(n-1)x_{2}}+\cdots+\dfrac{n-1+x_{1}x_{2}\cdots x_{n-1}}{1+(n-1)x_{n}}\ge n$$
I try to use Cauchy-Schwarz inequality
$$LHS\cdot\left(n+(n-1)\sum_{i=1}^{n}x_i\right)\ge \left(\sum_{cyc}\sqrt{n-1+x_{2}x_{3}\cdots x_{n}}\right)^2$$
we need show
$$\left(\sum_{cyc}\sqrt{n-1+x_{2}x_{3}\cdots x_{n}}\right)^2\ge n^2+n(n-1)\sum_{i=1}^{n}x_{i}$$
it seem this last inequality not right,because from Maclaurin's inequality it's  Reverse inequality

REPLY [3 votes]: Partial answer that perhaps someone can complete it :
First define :
$$f\left(x\right)=\frac{\left(\left(n-1\right)x+a\right)}{x+\left(n-1\right)x^{2}}$$
For $a\geq 0.25$ and $n\geq 3$ we have for $x\in(-\infty,\infty)$ :
$$h\left(x\right)=\frac{d}{dx}\left(\frac{d}{dx}\left(f\left(e^{x}\right)\right)\right)>0$$
For details on the second derivatives see Wolfram alpha .
Now conclude using Jensen's inequality as the function $h(x)$ is convex . The rest is smooth because it's a simple one variable inequality .
In fact we can completes it partially using RCF corollary see https://link.springer.com/content/pdf/10.1186/1029-242X-2011-101.pdf .
For that we can use a little $\varepsilon$ such that $x_1x_2\cdots x_n\geq \varepsilon>0$
To conclude definitively we can use Jensen's inequality on $f(x)$:
$$f''(x)>0$$
For $0<a=x_1x_2\cdots x_n\leq \varepsilon $ and $n\geq 3 $  .
I can add details if necessary .