Let $k$ be a given natural number. Prove that for any positive numbers $x; y; z$ with 
the sum $1$ the following inequality holds: 
\[\frac{x^{k+2}}{x^{k+1}+y^{k}+z^{k}}+\frac{y^{k+2}}{y^{k+1}+z^{k}+x^{k}}+\frac{z^{k+2}}{z^{k+1}+x^{k}+y^{k}}\geq \frac{1}{7}.\]
When does equality occur?
1. **Initial Setup and Inequality Transformation:**

   Given the inequality:
   \[
   \sum_{cyc} \frac{x^{k+2}}{x^{k+1} + y^k + z^k} \geq \frac{1}{7}
   \]
   where \(x, y, z\) are positive numbers such that \(x + y + z = 1\).

2. **Applying Generalized T2's Lemma:**

   Using the Generalized T2's Lemma, we have:
   \[
   \sum_{cyc} \frac{x^{k+2}}{x^{k+1} + y^k + z^k} \geq \frac{\left(\sum_{cyc} x\right)^{k+2}}{3^k \left(\sum_{cyc} x^{k+1} + 2 \sum_{cyc} x^k\right)}
   \]
   Since \(x + y + z = 1\), the inequality becomes:
   \[
   \frac{(x + y + z)^{k+2}}{3^k \left(\sum_{cyc} x^{k+1} + 2 \sum_{cyc} x^k\right)} \geq \frac{1}{7}
   \]
   Simplifying, we get:
   \[
   \frac{1}{3^k \left(\sum_{cyc} x^{k+1} + 2 \sum_{cyc} x^k\right)} \geq \frac{1}{7}
   \]
   which implies:
   \[
   7 \geq 3^k \left(\sum_{cyc} x^{k+1} + 2 \sum_{cyc} x^k\right)
   \]

3. **Defining Functions and Lagrange Multipliers:**

   Define the function:
   \[
   f(x, y, z) = x^{k+1} + y^{k+1} + z^{k+1} + 2x^k + 2y^k + 2z^k
   \]
   and the constraint:
   \[
   g(x, y, z) = x + y + z - 1
   \]
   Using Lagrange multipliers, define:
   \[
   L = f(x, y, z) + \lambda g(x, y, z)
   \]

4. **Partial Derivatives and Equations:**

   Compute the partial derivatives:
   \[
   \frac{\partial L}{\partial x} = (k+1)x^k + 2kx^{k-1} + \lambda = 0
   \]
   \[
   \frac{\partial L}{\partial y} = (k+1)y^k + 2ky^{k-1} + \lambda = 0
   \]
   \[
   \frac{\partial L}{\partial z} = (k+1)z^k + 2kz^{k-1} + \lambda = 0
   \]

5. **Equating Partial Derivatives:**

   From the partial derivatives, we get:
   \[
   (k+1)x^k + 2kx^{k-1} = (k+1)y^k + 2ky^{k-1}
   \]
   \[
   (k+1)y^k + 2ky^{k-1} = (k+1)z^k + 2kz^{k-1}
   \]

6. **Assuming \(x \neq y\) and Deriving Contradictions:**

   Assume \(x \neq y\):
   \[
   k(x^k - y^k) + 2k(x^{k-1} - y^{k-1}) = -(x^k - y^k)
   \]
   Simplifying, we get:
   \[
   k + \frac{2k(x^{k-1} - y^{k-1})}{x^k - y^k} = -1
   \]
   \[
   1 + \frac{2(x^{k-1} - y^{k-1})}{x^k - y^k} = -\frac{1}{k}
   \]
   \[
   \frac{2(x^{k-1} - y^{k-1})}{x^k - y^k} = -\frac{1}{k} - 1
   \]
   \[
   \frac{x^{k-1} - y^{k-1}}{x^k - y^k} = -\frac{1+k}{2k}
   \]
   This implies \(\frac{x^{k-1} - y^{k-1}}{x^k - y^k} < 0\), leading to contradictions in both cases \(x^{k-1} < y^{k-1}\) and \(x^k < y^k\).

7. **Concluding \(x = y = z\):**

   Therefore, \(x = y = z\). Given \(x + y + z = 1\), we have \(x = y = z = \frac{1}{3}\).

8. **Verifying the Equality:**

   Substituting \(x = y = z = \frac{1}{3}\) into the function:
   \[
   f\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right) = 3 \left(\frac{1}{3}\right)^{k+1} + 2 \cdot 3 \left(\frac{1}{3}\right)^k = \frac{1}{3^k} + \frac{2}{3^{k-1}} = \frac{7}{3^k}
   \]

Thus, the inequality holds, and equality occurs when \(x = y = z = \frac{1}{3}\).

The final answer is \(\boxed{\frac{1}{7}}\)