take $f(x)=\frac {x}{1-x}$

the graph shows that the function is concave within $(1,\infty)$.

take $\lambda_i =\frac 1n$

Therefore, by jensen's inequality:

$\Rightarrow \displaystyle\sum_{i=1}^n \lambda_i f(x_i) \leq f\bigg(\displaystyle\sum_{i=1}^n \lambda_i x_i\bigg)$

$\Rightarrow \frac 1n \cdot \displaystyle\sum_{i=1}^n \frac{x_i}{1-x_i} \leq f\bigg(\frac{\displaystyle\sum_{i=1}^n  x_i}{n}\bigg)$

$\Rightarrow \frac 1n \cdot \displaystyle\sum_{i=1}^n \frac{x_i}{1-x_i} \leq \frac{\frac{\displaystyle\sum_{i=1}^n  x_i}{n}}{1-\frac{\displaystyle\sum_{i=1}^n  x_i}{n}}$

$\Rightarrow \displaystyle\sum_{i=1}^n \frac{x_i}{1-x_i} \leq n \cdot \frac{\displaystyle\sum_{i=1}^n  x_i}{n-\displaystyle\sum_{i=1}^n  x_i} $