Let $x, y$ and $z$ be positive real numbers such that $x \geq y+z$. 
Proof that
$$\frac{x+y}{z} + \frac{y+z}{x} +\frac{z+x}{y} \geq 7$$
When does equality occur?

(Walther Janous)
Given \( x, y, \) and \( z \) are positive real numbers such that \( x \geq y + z \). We need to prove that:
\[ \frac{x+y}{z} + \frac{y+z}{x} + \frac{z+x}{y} \geq 7 \]
and determine when equality occurs.

1. **Initial Setup and Simplification:**
   We start by considering the given inequality:
   \[ \frac{x+y}{z} + \frac{y+z}{x} + \frac{z+x}{y} \]
   Since \( x \geq y + z \), we can use this condition to simplify and manipulate the terms.

2. **Applying the Condition \( x \geq y + z \):**
   We know that \( x \geq y + z \). Let's rewrite the inequality in a more manageable form:
   \[ \frac{x+y}{z} + \frac{y+z}{x} + \frac{z+x}{y} \geq 7 \]

3. **Using the AM-GM Inequality:**
   We can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality to help us. Recall that for any non-negative real numbers \( a, b, \) and \( c \):
   \[ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq 3 \]
   However, we need to adapt this to our specific problem.

4. **Rewriting the Terms:**
   Let's consider the terms individually:
   \[ \frac{x+y}{z} = \frac{x}{z} + \frac{y}{z} \]
   \[ \frac{y+z}{x} = \frac{y}{x} + \frac{z}{x} \]
   \[ \frac{z+x}{y} = \frac{z}{y} + \frac{x}{y} \]

5. **Combining the Terms:**
   Combining these, we get:
   \[ \frac{x}{z} + \frac{y}{z} + \frac{y}{x} + \frac{z}{x} + \frac{z}{y} + \frac{x}{y} \]

6. **Using the Given Condition:**
   Since \( x \geq y + z \), we can infer that:
   \[ \frac{x}{z} \geq \frac{y+z}{z} = 1 + \frac{y}{z} \]
   Therefore:
   \[ \frac{x}{z} + \frac{y}{z} \geq 1 + \frac{y}{z} + \frac{y}{z} = 1 + 2\frac{y}{z} \]

7. **Simplifying Further:**
   Similarly, we can simplify the other terms:
   \[ \frac{y}{x} + \frac{z}{x} \leq \frac{y}{y+z} + \frac{z}{y+z} = 1 \]
   \[ \frac{z}{y} + \frac{x}{y} \geq \frac{z}{y} + \frac{y+z}{y} = 1 + \frac{z}{y} + \frac{z}{y} = 1 + 2\frac{z}{y} \]

8. **Combining All Terms:**
   Combining all these, we get:
   \[ \frac{x}{z} + \frac{y}{z} + \frac{y}{x} + \frac{z}{x} + \frac{z}{y} + \frac{x}{y} \geq 1 + 2\frac{y}{z} + 1 + 1 + 2\frac{z}{y} \]

9. **Final Simplification:**
   Since \( x \geq y + z \), we can conclude that:
   \[ \frac{x+y}{z} + \frac{y+z}{x} + \frac{z+x}{y} \geq 7 \]

10. **Equality Condition:**
    Equality occurs when \( y = z \) and \( x = y + z = 2y \).

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