Prove that $\sum_{1\le iLet $n \ge 2$ be a an integer and $x_1,...,x_n$ are positive reals such that $$\sum_{i=1}^nx_i=\frac{1}{2}$$
Prove that
$$\sum_{1\le i<j\le n}\frac{x_ix_j}{(1-x_i)(1-x_j)}\le \frac{n(n-1)}{2(2n-1)^2}$$
Here is the source of the problem (in french) here
Edit:
I'll present my best bound yet on $$\sum_{1\le i<j\le 1}\frac{x_ix_j}{(1-x_i)(1-x_j)}=\frac{1}{2}\left(\sum_{k=1}^n\frac{x_k}{1-x_k}\right)^2-\frac{1}{2}\sum_{k=1}^n\frac{x_k^2}{(1-x_k)^2}$$
This formula was derived in @GCab's Answer.
First let $a_k=x_k/(1-x_k)$
so we want to prove
$$\left(\sum_{k=1}^na_k\right)^2-\sum_{k=1}^na_k^2\le \frac{n(n-1)}{(2n-1)^2}$$
But since $$\frac{x_k}{1-x_k}<2x_k\implies \sum_{k=1}^na_k<1 \quad (1)$$
Hence $$\left(\sum_{k=1}^na_k\right)^2\le \sum_{k=1}^na_k$$
Meaning $$\left(\sum_{k=1}^na_k\right)^2-\sum_{k=1}^na_k^2\le\sum_{k=1}^na_k(1-a_k)$$
Now consider the following function $$f(x)=\frac{x}{1-x}\left(1-\frac{x}{1-x}\right)$$
$f$ is concave on $(0,1)$ and by the tangent line trick we have $$f(x)\le f'(a)(x-a)+f(a)$$
set $a=1/2n$ to get $$a_k(1-a_k)\le\frac{4n^2\left(2n-3\right)}{\left(2n-1\right)^3}\left(x_k-\frac{1}{2n}\right)+ \frac{2(n-1)}{(2n-1)^2}$$
Now we sum to finish $$\sum_{k=1}^na_k(1-a_k)\le \frac{2n(n-1)}{(2n-1)^2}$$
Maybe by tweaking $(1)$ a little bit we can get rid of this factor of $2$

Hint:
Indicating as $S_n , \, T_n$
$$
\begin{array}{l}
 T_n  = \sum\limits_{1 \le i < j \le n} {a_i a_j }  \\ 
 S_n  = \sum\limits_{1 \le i,j \le n} {a_i a_j }
  = \sum\limits_{1 \le i \le n} {\sum\limits_{1 \le j \le n} {a_i a_j } }
  = \sum\limits_{1 \le i \le n} {a_i } \sum\limits_{1 \le j \le n} {a_j }
  = \left( {\sum\limits_{1 \le i \le n} {a_i } } \right)^2  \\ 
 \end{array}
$$
then you also have that
$$
\begin{array}{l}
 S_n  = \sum\limits_{1 \le i,j \le n} {a_i a_j }
  = \sum\limits_{1 \le i < j \le n} {a_i a_j }  + \sum\limits_{1 \le i = j \le n} {a_i a_j }
  + \sum\limits_{1 \le j < i \le n} {a_i a_j }  =  \\ 
  = 2T_n  + \sum\limits_{1 \le i \le n} {a_i ^2 }  \\ 
 \end{array}
$$
