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).
 A: For any $x,y \in (0,2)$, we have
$$\frac{x}{2-x} - \frac{y}{2-y} = \frac{2(x-y)}{(2-x)(2-y)}
= 2\left(\frac{x-y}{2-y}\right)\left[\left(\frac{1}{2-x} - \frac{1}{2-y}\right) + \frac{1}{2-y}\right]
= 2\left(\frac{x-y}{2-y}\right)\left[\frac{x-y}{(2-x)(2-y)} + \frac{1}{2-y}\right]
= \frac{2(x-y)^2}{(2-x)(2-y)^2} + \frac{2(x-y)}{(2-y)^2}
\ge \frac{2(x-y)}{(2-y)^2}
$$
Substitute $x$ by $x_i$ and $y$ by $\frac{1}{n}$ and then sum over $i$, we get
$$
\sum_{i=1}^n \frac{x_i}{2-x_i} - \frac{n}{2n-1}
=   \sum_{i=1}^n \left(\frac{x_i}{2-x_i} -\frac{\frac{1}{n}}{2-\frac{1}{n}}\right)
\ge \frac{2}{(2-\frac{1}{n})^2}\sum_{i=1}^n \left( x_i - \frac{1}{n}\right)
=   \frac{2}{(2-\frac{1}{n})^2} \left(\sum_{i=1}^n x_i - 1 \right)
= 0$$
The steps look almost magical. How do I come up with that? The answer is we are using Jensen's inequality from behind. 
Jensen's inequality is actually more generic and applicable beyond differentiable functions. If you have any function $f(x)$ whose graph curved up on both side at every point, then Jensen's inequality works. Such functions are called 
convex functions. For any convex
function $f(x)$, Jensen's inequality tells us:
$$\frac{1}{n} \sum_{i=1}^n f(x_i)\;\;\ge\;\; f(\frac{1}{n} \sum_{i=1}^n x_i)$$
This means if you want to verify $\sum_{i=1}^n f(x_i) \ge$ some number $M$. You just
need to check what happens when all $x_i$ are equal to each other.
In this case, we are told we cannot use Jensen's inequality. So we expand our
target function $f(x_i)$ around $\frac{1}{n}$, the mean of the $x_i$'s, and 
we are sure after we cancel the linear parts, the rest of the expansion will be non-negative.
A: 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.
A: Hint: Use Jensen's inequality for f(x) = x/(2 -x)
