Prove that $\frac{x_1}{1+x_2+x_3+\ldots+x_n}+\frac{x_2}{1+x_1+x_3+\ldots+x_n}+\ldots+\frac{x_n}{1+x_1+x_2+\ldots+x_{n-1}}\ge\frac{n}{2n-1}$.

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.

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.