Prove that $\frac{x_1}{1+x_2+x_3+\ldots+x_n}+\frac{x_2}{1+x_1+x_3+\ldots+x_n}+\ldots+\frac{x_n}{1+x_1+x_2+\ldots+x_{n-1}}\ge\frac{n}{2n-1}$. If $x_1,x_2,\ldots,x_n>0$ and $x_1+x_2+\ldots+x_n=1$, prove that $$\frac{x_1}{1+x_2+x_3+\ldots+x_n} + \frac{x_2}{1+x_1+x_3+\ldots+x_n} +\ldots + \frac{x_n}{1+x_1+x_2+\ldots +x_{n-1}} \ge \frac {n}{2n-1}$$ 
This can easily be simplified: $$\frac{x_1}{2-x_1} + \frac{x_2}{2-x_2} +\ldots + \frac{x_n}{2-x_n} \ge \frac {n}{2n-1}$$
I could try using the Cauchy-Schwarz inequality: $$\frac{x_1}{2-x_1} + \frac{x_2}{2-x_2} +\ldots + \frac{x_n}{2-x_n} = \frac{x_1^2}{2x_1-x_1^2} + \frac{x_2^2}{2x_2-x_2^2} +\ldots + \frac{x_n^2}{2x_n-x_n^2} \ge \frac{(x_1+x_2+\ldots+x_n)^2}{2(x_1+x_2+\ldots+x_n)-(x_1^2+x_2^2+\ldots+x_n^2)}=\frac{1}{2-(x_1^2+x_2^2+\ldots+x_n^2)}=\frac{n}{2n-(x_1^2+x_2^2+\ldots+x_n^2)n}$$
It's left to prove that $$(x_1^2+x_2^2+\ldots+x_n^2)n \ge 1$$
or $$(x_1^2+x_2^2+\ldots+x_n^2)n=1$$I can continue using the Cauchy-Schwarz inequality again: $$(x_1^2+x_2^2+\ldots+x_n^2)n=(x_1^2+x_2^2+\ldots+x_n^2)(1+1+\ldots+1)\ge(x_1+x_2+\ldots+x_n)^2=1$$
So I've proved it myself. (I've shown this proof after editing, I didn't post the question with the solution in the details. I found the proof a while after putting the question here).

Writing ${\displaystyle {x_i \over 2 - x_i} = -1 + {2 \over 2 - x_i}}$, you have
 $$\frac{x_1}{2-x_1} + \frac{x_2}{2-x_2} +\ldots + \frac{x_n}{2-x_n} = -n + \frac{2}{2-x_1} + \frac{2}{2-x_2} +\ldots + \frac{2}{2-x_n}$$
So it suffices to show
$$ \frac{2}{2-x_1} + \frac{2}{2-x_2} +\ldots + \frac{2}{2-x_n} \geq n + \frac{n}{2n-1}$$
This is the same as 
$$\frac{1}{2-x_1} + \frac{1}{2-x_2} +\ldots + \frac{1}{2-x_n} \geq \frac{n^2}{2n - 1}$$
Letting $y_i = 2 - x_i$, this is the same as showing
$$\frac{1}{y_1} + \frac{1}{y_2} +\ldots + \frac{1}{y_n} \geq \frac{n^2}{2n - 1}$$
The condition $\sum_i x_i = 1$ translates into $\sum_i y_i = 2n - 1$.
By the arithmetic-harmonic mean inequality, 
$$n\bigg(\frac{1}{y_1} + \frac{1}{y_2} +\ldots + \frac{1}{y_n}\bigg)^{-1} \leq {1 \over n}\sum_i y_i $$
$$= {2n - 1 \over n}$$
This is equivalent to what you want to show.
