Let $x$, $y$, $z$ be arbitrary positive numbers such that $xy+yz+zx=x+y+z$.
 Prove that $$\frac{1}{x^2+y+1} + \frac{1}{y^2+z+1} + \frac{1}{z^2+x+1} \leq 1$$. 
When does equality occur?
[i]Proposed by Marko Radovanovic[/i]
1. Given the condition \(xy + yz + zx = x + y + z\), we start by noting that this implies \(xy + yz + zx \geq 3\). This follows from the AM-GM inequality, which states that for any non-negative real numbers \(a, b, c\), we have:
   \[
   \frac{a + b + c}{3} \geq \sqrt[3]{abc}
   \]
   Applying this to \(x, y, z\), we get:
   \[
   \frac{x + y + z}{3} \geq \sqrt[3]{xyz}
   \]
   Since \(xy + yz + zx = x + y + z\), we can infer that \(x + y + z \geq 3\).

2. Next, we use the given condition to apply Hölder's Inequality. Hölder's Inequality in this context states that for positive real numbers \(a_i, b_i, c_i\):
   \[
   \left( \sum_{i=1}^n a_i \right) \left( \sum_{i=1}^n b_i \right) \left( \sum_{i=1}^n c_i \right) \geq \left( \sum_{i=1}^n \sqrt[3]{a_i b_i c_i} \right)^3
   \]
   Applying this to our problem, we set:
   \[
   a_i = \frac{1}{x^2 + y + 1}, \quad b_i = 1, \quad c_i = 1
   \]
   Then, we have:
   \[
   \sum_{cyc} \frac{1}{x^2 + y + 1} \leq \sum_{cyc} \frac{1 + y + z^2}{(x^2 + y + 1)(1 + y + z^2)}
   \]
   Simplifying the right-hand side, we get:
   \[
   \sum_{cyc} \frac{1 + y + z^2}{(x^2 + y + 1)(1 + y + z^2)} \leq \sum_{cyc} \frac{1 + y + z^2}{(x + y + z)^2}
   \]

3. To prove the inequality, it suffices to show that:
   \[
   (x + y + z)^2 \geq \sum_{cyc} (1 + y + z^2)
   \]
   Expanding the right-hand side, we get:
   \[
   \sum_{cyc} (1 + y + z^2) = 3 + x + y + z + x^2 + y^2 + z^2
   \]
   Therefore, we need to show:
   \[
   (x + y + z)^2 \geq 3 + x + y + z + x^2 + y^2 + z^2
   \]

4. Expanding the left-hand side, we have:
   \[
   (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)
   \]
   Given \(xy + yz + zx = x + y + z\), we substitute to get:
   \[
   x^2 + y^2 + z^2 + 2(x + y + z) \geq 3 + x + y + z + x^2 + y^2 + z^2
   \]
   Simplifying, we find:
   \[
   2(x + y + z) \geq 3 + x + y + z
   \]
   Which simplifies to:
   \[
   x + y + z \geq 3
   \]
   This is true by our initial assumption.

5. Therefore, the inequality holds, and equality occurs when \(x = y = z = 1\).

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