Problem 1528

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 $\boxed{\frac{1}{7}}$