Yet another inequality

$x_1, x_2, \dots, x_n$ are $n$ positive real numbers with sum $S$, $n \gt 1$.

Show that

$$ \sum_{i=1}^{n} \frac{x_i}{S - x_i} \ge \frac{n}{n-1}$$


Solution [Click here to expand/collapse]

Using AM >= GM we have that

$\sum y_i \sum \frac{1}{y_i} \ge n^2$

Applying this to (wlog, assuming $S = 1$) $$\left(\sum_{i=1}^{n} \frac{1}{1-x_i}\right) \left(\sum_{i=1}^{n} (1 - x_i)\right) \ge n^2$$ i.e $$ \sum_{i=1}^{n} \frac{1}{1-x_i} \ge \frac{n^2}{n-1}$$ And so $$ \sum_{i=1}^{n} \left(\frac{1}{1-x_i} - 1\right) \ge \frac{n^2}{n-1} -n$$ Which gives us what we want to prove.