let $n\ge 3,n\in N$, and $x_{1},x_{2},x_{3},\cdots,x_{n}$ are positive numbers,and such that $$\sum_{i=1}^{n}\dfrac{1}{x_{i}+1}=1,$$
show that: for any real numbers $\alpha\ge 1$,we have $$\sum_{i=1}^{n}\dfrac{1}{x^{\alpha}_{i}+1}\ge\dfrac{n}{(n-1)^{\alpha}+1}$$
With change of variable $a_i = \frac{1}{x_i +1}$ we have to prove $\sum \frac{a_i^\alpha}{(1-a_i)^\alpha+a_i^\alpha}\geq\frac{n}{(n-1)^\alpha+1}$ under the constraint $\sum a_i =1$. Note that $$\sum_i \frac{a_i^\alpha}{(1-a_i)^\alpha+a_i^\alpha}=\sum_i \frac{a_i^\alpha}{(\sum _{j\neq i} a_j)^\alpha+a_i^\alpha}\\ \geq \sum_i \frac{a_i^\alpha}{(n-1)^{\alpha-1}(\sum _{j\neq i} a_j^\alpha)+a_i^\alpha}=:S,$$ where the power means inequality is applied to the sum in each denominator to obtain the last inequality. Using the fact that $\left\lbrace \frac{a_i^\alpha}{(n-1)^{\alpha-1}(\sum _{j\neq i} a_j^\alpha)+a_i^\alpha}\right\rbrace$ and $\left\lbrace(n-1)^{\alpha-1}(\sum _{j\neq i} a_j^\alpha)+a_i^\alpha\right\rbrace$ have opposite sorting order, we can apply the Chebyshev's sum inequality and obtain $$\frac{S}{n}\left(\frac{1}{n}\sum_i (n-1)^{\alpha-1}(\sum _{j\neq i} a_j^\alpha)+a_i^\alpha\right) \geq \frac{1}{n}\sum_i a_i^\alpha.$$ Therefore, $$S\geq\frac{n\sum_i a_i^\alpha}{\left((n-1)^\alpha+1\right)\sum_i a_i^\alpha}\\=\frac{n}{(n-1)^\alpha +1},$$ which completes the proof.
