<s>[INST] Prove that: $\sum_{1\le j\le n}{\frac{x_j}{\sum_{1\le i\le n} x_i-x_j}}\ge\frac{n}{n-1}$ 
$\color{red}{\textbf{Problem:}}$
Let, $x_i>0,1\le i\le n$, then
Prove that:
$$\sum_{1\le j\le n}{\frac{x_j}{\sum_{1\le i\le n} x_i-x_j}}\ge\frac{n}{n-1}$$

$\color{red}{\textbf{Proof:}}$
Using AM-GM inequality we have,
\begin{align}&\sum_{1\le i<j\le n} (x_i^2+x_j^2)=(n-1)\sum_{1\le i\le n} x_i^2\ge 2\left(\sum_{1\le i < j \le n}x_ix_j\right)\\
\implies &\sum_{1\le i\le n} x_i^2\ge \frac 2{n-1}\left(\sum_{1\le i < j \le n}x_ix_j\right)\\ 
\implies &\left(\sum_{1\le i\le n} x_i\right)^2\ge\left( 2+\frac 2{n-1}\right)\left(\sum_{1\le i < j \le n}x_ix_j\right)\\
&\qquad\qquad\quad =\frac{2n}{n-1}\left(\sum_{1\le i < j \le n}x_ix_j\right)\end{align}
Finally, applying Cauchy-Schwars, we obtain
\begin{align}\left(\sum_{1\le j\le n}{\frac{x_j}{\sum_{1\le i\le n} x_i-x_j}}\right)\left(\sum_{1\le j\le n} \left({x_j}{\sum_{1\le i\le n} x_i-x_j}\right)\right)&\ge \frac{ \left(\sum_{1\le j\le n} x_i\right)^2}{2\left(\sum_{1\le i < j \le n}x_ix_j\right)}\\
&\ge \frac {\frac{2n}{n-1}\sum_{1\le i < j \le n}x_ix_j}{2\sum_{1\le i < j \le n}x_ix_j}\\
&=\boxed {\frac{n}{n-1}.}\end{align}
I need to know if there is something wrong with my solution.
 [/INST] OP's work is correct however $x_j>0$ needs to be mentioned. Here is another approach:
Let $s=\sum_{j=1}^{n} x_j,$ then
$$f(x_j)=\frac{x_j}{s-x_j} \implies f''(x)=\frac{2s}{(s-x_j)^3}>0,\quad  0<x_j<s.$$
So by Jensen's inequality
$$\frac{1}{n}\sum_{j=1}^{n} f(x_j) \ge f(\frac{s}{n}) \implies \sum_{j=1}^{n} \frac{x_j}{s-x_j}\ge n \frac{s/n}{s-s/n}=\frac{n}{n-1}. $$
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