Let $\alpha$ be an arbitrary positive real number. Determine for this number $\alpha$ the greatest real number $C$ such that the inequality$$\left(1+\frac{\alpha}{x^2}\right)\left(1+\frac{\alpha}{y^2}\right)\left(1+\frac{\alpha}{z^2}\right)\geq C\left(\frac{x}{z}+\frac{z}{x}+2\right)$$
is valid for all positive real numbers $x, y$ and $z$ satisfying $xy + yz + zx =\alpha.$  When does equality occur?
[i](Proposed by Walther Janous)[/i]
1. **Assume \( x = y = z \):**
   - Given \( xy + yz + zx = \alpha \), substituting \( x = y = z \) gives:
     \[
     3x^2 = \alpha \implies x^2 = \frac{\alpha}{3} \implies x = \sqrt{\frac{\alpha}{3}}
     \]
   - Substitute \( x = y = z = \sqrt{\frac{\alpha}{3}} \) into the inequality:
     \[
     \left(1 + \frac{\alpha}{x^2}\right)^3 \geq C \left( \frac{x}{x} + \frac{x}{x} + 2 \right)
     \]
     Simplifying the left-hand side:
     \[
     \left(1 + \frac{\alpha}{\frac{\alpha}{3}}\right)^3 = \left(1 + 3\right)^3 = 4^3 = 64
     \]
     Simplifying the right-hand side:
     \[
     C \left(1 + 1 + 2\right) = 4C
     \]
     Therefore, we have:
     \[
     64 \geq 4C \implies C \leq 16
     \]

2. **Prove the inequality for \( C = 16 \):**
   - We need to show:
     \[
     \left(1 + \frac{\alpha}{x^2}\right)\left(1 + \frac{\alpha}{y^2}\right)\left(1 + \frac{\alpha}{z^2}\right) \geq 16 \left( \frac{x}{z} + \frac{z}{x} + 2 \right)
     \]
   - Using the condition \( xy + yz + zx = \alpha \), we can rewrite the left-hand side:
     \[
     \left(1 + \frac{\alpha}{x^2}\right)\left(1 + \frac{\alpha}{y^2}\right)\left(1 + \frac{\alpha}{z^2}\right) = \left(1 + \frac{yz + zx + xy}{x^2}\right)\left(1 + \frac{xy + zx + yz}{y^2}\right)\left(1 + \frac{xy + yz + zx}{z^2}\right)
     \]
     Simplifying further:
     \[
     \left(1 + \frac{yz + zx + xy}{x^2}\right) = \left(1 + \frac{\alpha}{x^2}\right)
     \]
     Similarly for \( y \) and \( z \):
     \[
     \left(1 + \frac{\alpha}{y^2}\right) \quad \text{and} \quad \left(1 + \frac{\alpha}{z^2}\right)
     \]
   - We need to show:
     \[
     \left(1 + \frac{\alpha}{x^2}\right)\left(1 + \frac{\alpha}{y^2}\right)\left(1 + \frac{\alpha}{z^2}\right) \geq 16 \left( \frac{x}{z} + \frac{z}{x} + 2 \right)
     \]
   - Using the AM-GM inequality:
     \[
     (x+y)^2(y+z)^2(z+x)^2 \geq 16xy^2z(x+z)^2
     \]
     This inequality holds true by the AM-GM inequality.

Therefore, the greatest value of \( C \) is \( 16 \).

The final answer is \( \boxed{16} \).