For the sake of convenience, let $S$ denotes the above expression. First, before demonstrating that $S\leqslant 1$. We need to define a new sequence of real numbers $(y_k)_{k=0}^{n+1}$ with $y_0=0,y_1=1$ and $x_{k+1}=y_0+y_1+\dots+y_k$ for every $k\in\{1,2,\dots,n+1\}$. Since $x_1<x_2<\cdots<x_{n+2}$, implies $y_1,y_2,\dots,y_{n+1}$ are positive reals. Obviously, $$S=\left(\frac{(y_1+\cdots+y_n)+(y_2+\cdots+y_{n+1})}{(y_1+\cdots+y_{n})(y_2+\cdots+y_{n+1})}\right)\left(\sum_{i=1}^{n}\frac{y_{i+1}y_i}{y_{i+1}+y_i}\right)$$Now, consider the expression \begin{align*} \left(\sum_{i=1}^{n}\frac{y_{i+1}y_i}{y_{i+1}+y_i}\right)&=\frac{y_2y_1}{y_2+y_1}+\frac{y_3y_2}{y_3+y_2}+\cdots+\frac{y_{n+1}y_n}{y_{n+1}+y_n}\\ &=\left(y_1-\frac{y_1^2}{y_1+y_2}\right)+\cdots+\left(y_n-\frac{y_n^2}{y_{n}+y_{n+1}}\right)\\ &=\sum_{i=1}^{n}y_i-\sum_{i=1}^{n}\frac{y_i^2}{y_i+y_{i+1}} \end{align*}According to Cauchy-Schwarz inequality, $$\left(\sum_{i=1}^{n}\frac{y_i^2}{y_i+y_{i+1}}\right)\left(\sum_{i=1}^{n}(y_i+y_{i+1})\right)\geqslant (y_1+y_2+\cdots+y_{n})^{2}$$Plugging back into the above equation, \begin{align*} \left(\sum_{i=1}^{n}\frac{y_{i+1}y_i}{y_{i+1}+y_i}\right)&\leqslant \sum_{i=1}^{n}y_i-\frac{(y_1+\cdots+y_n)^2}{(y_1+\cdots+y_n)+(y_2+\cdots+y_{n+1})}\\ &=\left(\sum_{i=1}^{n}y_i\right)\left(1-\frac{y_1+\cdots+y_n}{(y_1+\cdots+y_n)+(y_2+\cdots+y_{n+1})}\right)\\ &=\left(\sum_{i=1}^{n}y_i\right)\left(\frac{y_2+\cdots+y_{n+1}}{(y_1+\cdots+y_n)+(y_2+\cdots+y_{n+1})}\right)\\ &=\frac{(y_1+\cdots+y_n)(y_2+\cdots+y_{n+1})}{(y_1+\cdots+y_n)+(y_2+\cdots+y_{n+1})} \end{align*}Therefore, $$S\leqslant \left(\frac{(y_1+\cdots+y_n)+(y_2+\cdots+y_{n+1})}{(y_1+\cdots+y_{n})(y_2+\cdots+y_{n+1})}\right)\left(\frac{(y_1+\cdots+y_n)(y_2+\cdots+y_{n+1})}{(y_1+\cdots+y_n)+(y_2+\cdots+y_{n+1})}\right)=1.$$With equality if and only if $$\frac{y_1^2}{(y_1+y_2)^2}=\frac{y_2^2}{(y_2+y_3)^2}=\cdots=\frac{y_n^2}{(y_n+y_{n+1})^2}=a^2$$for some constant $0<a<1$. Which is equivalent to $$\frac{y_{i+1}}{y_i}=\frac{1}{a}-1\Rightarrow y_i=r^{i-1}\text{ where }r=\frac{1}{a}-1\text{ for every }i=1,2,\dots,n+1.$$Consequently, equality holds when $x_i=1+r+r^2+\cdots+r^{i-2}=\frac{r^{i-1}-1}{r-1}$ for all $i=2,3,\dots,n+2$. In conclusion, the maximum value of $S$ is $\boxed{1}$.