Let $n$ be an integer number greater than $2$, let $x_{1},x_{2},\ldots ,x_{n}$ be $n$ positive real numbers such that
\[\sum_{i=1}^{n}\frac{1}{x_{i}+1}=1\]
and let $k$ be a real number greater than $1$. Show that:
\[\sum_{i=1}^{n}\frac{1}{x_{i}^{k}+1}\ge\frac{n}{(n-1)^{k}+1}\]
and  determine the cases of equality.
1. **Lemma 1:** We start by proving the inequality:
   \[
   n^{k-1}\sum_{i=1}^{n} a_i^k \ge (\sum_{i=1}^{n} a_i)^k
   \]
   **Proof:** By Hölder's Inequality, we have:
   \[
   \left( \sum_{i=1}^{n} 1^{\frac{k}{k-1}} \right)^{\frac{k-1}{k}} \left( \sum_{i=1}^{n} a_i^k \right)^{\frac{1}{k}} \ge \sum_{i=1}^{n} a_i
   \]
   Simplifying the left-hand side, we get:
   \[
   n^{\frac{k-1}{k}} \left( \sum_{i=1}^{n} a_i^k \right)^{\frac{1}{k}} \ge \sum_{i=1}^{n} a_i
   \]
   Raising both sides to the power of \(k\), we obtain:
   \[
   n^{k-1} \sum_{i=1}^{n} a_i^k \ge \left( \sum_{i=1}^{n} a_i \right)^k
   \]
   This completes the proof of Lemma 1.

2. **Substitution:** Let \( a_i = \frac{1}{x_i + 1} \). This implies \( x_i = \frac{1 - a_i}{a_i} \). Given the condition:
   \[
   \sum_{i=1}^{n} \frac{1}{x_i + 1} = 1
   \]
   we have:
   \[
   \sum_{i=1}^{n} a_i = 1
   \]

3. **Transforming the inequality:** We need to show:
   \[
   \sum_{i=1}^{n} \frac{1}{x_i^k + 1} \ge \frac{n}{(n-1)^k + 1}
   \]
   Substituting \( x_i = \frac{1 - a_i}{a_i} \) into the inequality, we get:
   \[
   \sum_{i=1}^{n} \frac{1}{\left( \frac{1 - a_i}{a_i} \right)^k + 1} = \sum_{i=1}^{n} \frac{a_i^k}{a_i^k + (1 - a_i)^k}
   \]

4. **Applying Lemma 1 and Cauchy-Schwarz Inequality:**
   \[
   \sum_{i=1}^{n} \frac{a_i^k}{a_i^k + (1 - a_i)^k} \ge \sum_{i=1}^{n} \frac{a_i^k}{a_i^k + \left( \sum_{j \neq i} a_j \right)^k}
   \]
   Using Lemma 1, we know:
   \[
   \left( \sum_{j \neq i} a_j \right)^k \le (n-1)^{k-1} \sum_{j \neq i} a_j^k
   \]
   Therefore:
   \[
   \sum_{i=1}^{n} \frac{a_i^k}{a_i^k + (n-1)^{k-1} \sum_{j \neq i} a_j^k}
   \]

5. **Simplifying the expression:**
   \[
   \sum_{i=1}^{n} \frac{a_i^k}{a_i^k + (n-1)^{k-1} \sum_{j \neq i} a_j^k} \ge \frac{(\sum_{i=1}^{n} a_i^k)^2}{\sum_{i=1}^{n} a_i^{2k} + (n-1)^{k-1} \sum_{i \neq j} a_i^k a_j^k}
   \]

6. **Using the fact that:**
   \[
   (n-1) \sum_{i=1}^{n} a_i^{2k} \ge \sum_{i \neq j} a_i^k a_j^k
   \]
   we get:
   \[
   \frac{(\sum_{i=1}^{n} a_i^k)^2}{\sum_{i=1}^{n} a_i^{2k} + (n-1)^{k-1} \sum_{i \neq j} a_i^k a_j^k} \ge \frac{n}{(n-1)^k + 1}
   \]

7. **Conclusion:**
   \[
   \sum_{i=1}^{n} \frac{1}{x_i^k + 1} \ge \frac{n}{(n-1)^k + 1}
   \]

The final answer is \(\boxed{\frac{n}{(n-1)^k + 1}}\)