if $x_{i}>0$,show that $$\sum_{i=1}^{n}\dfrac{x_{i}}{(n-1)x_{i+1}+1}+\dfrac{1}{\sum_{i=1}^{n}x_{i}}\ge\dfrac{n+1}{n}$$
I have proven $n=2$and $n=3$
$n=3$ case let $x,y,z>0$.prove $$\dfrac{x}{2y+1}+\dfrac{y}{2z+1}+\dfrac{z}{2x+1}+\dfrac{1}{x+y+z}\ge\dfrac{4}{3}$$
By Cauchy-Schwarz inequality we have $$\sum_{cyc}\dfrac{x}{2y+1}\ge \dfrac{(x+y+z)^2}{2xy+2yz+2zx+x+y+z}\ge\dfrac{t^2}{\dfrac{2}{3}t^2+t}=\dfrac{3t}{2t+3}$$ where $t=x+y+z>0$ so we must prove $$\dfrac{3t}{2t+3}+\dfrac{1}{t}\ge\dfrac{4}{3}$$ the last inequality it is clear true!
A proof for $n=4$.
Let $x+y+z+t=4u$.
Thus, by C-S and AM-GM we obtain: $$\sum_{cyc}\frac{x}{3y+1}+\frac{1}{x+y+z+t}=\sum_{cyc}\frac{x^2}{3xy+x}+\frac{1}{x+y+z+t}\geq$$ $$\geq\frac{(x+y+z+t)^2}{3\sum\limits_{cyc}xy+x+y+z+t}+\frac{1}{x+y+z+t}=\frac{16u^2}{3(x+z)(y+t)+4u}+\frac{1}{4u}\geq$$ $$\geq \frac{16u^2}{3\left(\frac{x+z+y+t}{2}\right)^2+4u}+\frac{1}{4u}=\frac{4u}{3u+1}+\frac{1}{4u}.$$ Id est, it's enough to prove that: $$\frac{4u}{3u+1}+\frac{1}{4u}\geq\frac{5}{4}$$ or $$(u-1)^2\geq0$$ and we are done.
For $n\geq5$ this way does not help.