Find the greatest positive real number $ k$ such that the inequality below holds for any positive real numbers $ a,b,c$:
\[ \frac ab \plus{} \frac bc \plus{} \frac ca \minus{} 3 \geq k \left( \frac a{b \plus{} c} \plus{} \frac b{c \plus{} a} \plus{} \frac c{a \plus{} b} \minus{} \frac 32 \right).
\]
1. **Assume without loss of generality that \( c = \max(a, b, c) \).**

2. **Rewrite the given inequality:**
   \[
   \frac{a}{b} + \frac{b}{c} + \frac{c}{a} - 3 \geq k \left( \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} - \frac{3}{2} \right)
   \]

3. **Express the left-hand side in a different form:**
   \[
   \frac{a}{b} + \frac{b}{c} + \frac{c}{a} - 3 = \frac{(a - b)^2}{ab} + \frac{(c - a)(c - b)}{ac}
   \]

4. **Express the right-hand side in a different form:**
   \[
   \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} - \frac{3}{2} = \frac{(a - b)^2}{(a + c)(b + c)} + \frac{(a + b + 2c)(c - a)(c - b)}{2(a + b)(b + c)(c + a)}
   \]

5. **Combine the two expressions to form the equivalent inequality:**
   \[
   \left( \frac{1}{ab} - \frac{k}{(a + c)(b + c)} \right)(a - b)^2 + \left( \frac{1}{ac} - \frac{k(a + b + 2c)}{2(a + b)(b + c)(c + a)} \right)(c - a)(c - b) \geq 0
   \]

6. **Show that \( k = 1 \) is a valid solution:**
   - Since \( c = \max(a, b, c) \), we have:
     \[
     \frac{1}{ab} - \frac{1}{(a + c)(b + c)} \geq 0
     \]
     \[
     \frac{1}{ac} - \frac{a + b + 2c}{2(a + b)(b + c)(c + a)} \geq 0
     \]
     \[
     (c - a)(c - b) \geq 0
     \]
   - Therefore, the inequality holds for \( k = 1 \).

7. **Show that \( k > 1 \) is not valid:**
   - Let \( k = m + 1 \) where \( m > 0 \).
   - Consider the function:
     \[
     f(c) = \left( \frac{1}{ab} - \frac{m + 1}{(c - a)(c - b)} \right)(a - b)^2 - \left( \frac{1}{ac} - \frac{(a + b + 2c)}{2(a + b)(b + c)(c + a)} \right)(c - a)(c - b)
     \]
   - Choose \( a \) and \( b \) such that \( b < ma \).
   - As \( c \to \infty \):
     \[
     \lim_{c \to \infty} \left( \frac{1}{ab} - \frac{m + 1}{(c - a)(c - b)} \right)(a - b)^2 = \frac{(a - b)^2}{ab}
     \]
     \[
     \lim_{c \to \infty} \left( \frac{1}{ab} - \frac{m + 1}{(c - a)(c - b)} \right)(a - b)^2 - \left( \frac{2(b - ma)c^2 + (a + b)((1 - m)a + 2b)c + 2ab(a + b)}{2(a + b)(b + c)(c + a)} \right)(c - a)(c - b) = -\infty
     \]
   - Thus, \( \lim_{c \to \infty} f(c) = -\infty \). Therefore, there exists a triple \( (a, b, c) \) such that the inequality does not hold for \( k > 1 \).

8. **Conclusion:**
   The greatest positive real number \( k \) such that the inequality holds for any positive real numbers \( a, b, c \) is \( k = 1 \).

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