Find all functions $f : \mathbb{N} \rightarrow \mathbb{R}$ such that for all triples $a,b,c$ of positive integers the following holds :

$$f(ac)+f(bc)-f(c)f(ab) \ge 1$$

Proposed by [i]Mojtaba Zare[/i]
1. Let \( P(a, b, c) \) be the inequality:
   \[
   f(ac) + f(bc) - f(c)f(ab) \geq 1 \quad \forall \, a, b, c \in \mathbb{N}.
   \]

2. Consider \( P(1, 1, 1) \):
   \[
   f(1 \cdot 1) + f(1 \cdot 1) - f(1)f(1 \cdot 1) \geq 1 \implies 2f(1) - f(1)^2 \geq 1.
   \]
   Solving the quadratic inequality:
   \[
   f(1)^2 - 2f(1) + 1 \leq 0 \implies (f(1) - 1)^2 \leq 0 \implies f(1) = 1.
   \]

3. Consider \( P(a, b, 1) \):
   \[
   f(a \cdot 1) + f(b \cdot 1) - f(1)f(ab) \geq 1 \implies f(a) + f(b) - f(ab) \geq 1 \quad \forall \, a, b \in \mathbb{N}.
   \]

4. Consider \( P(a, 1, b) \):
   \[
   f(ab) + f(b) - f(b)f(a) \geq 1 \quad \forall \, a, b \in \mathbb{N}.
   \]

5. Adding the two inequalities from steps 3 and 4:
   \[
   f(a) + f(b) - f(ab) + f(ab) + f(b) - f(b)f(a) \geq 2 \implies f(a) + 2f(b) - f(a)f(b) \geq 2.
   \]
   Rearranging:
   \[
   (f(a) - 2)(f(b) - 1) \leq 0 \quad \forall \, a, b \in \mathbb{N}.
   \]
   This implies:
   \[
   1 \leq f(x) \leq 2 \quad \forall \, x \in \mathbb{N}.
   \]

6. Now, we will prove \( f \equiv 1 \). Let \( k \leq 2 \) be an upper bound of \( f \).

7. Setting \( f(c) = f(ab) = k \) in the original equation:
   \[
   f(ac) + f(bc) \geq k^2 + 1.
   \]
   Since \( 1 \leq f(x) \leq 2 \), we have:
   \[
   k^2 + 1 \leq 4 \implies k \leq \sqrt{3} \implies 1 \leq f(x) \leq \sqrt{3}.
   \]

8. Similarly, setting \( f(c) = f(ab) = k \) in the original equation:
   \[
   f(ac) + f(bc) \geq k^2 + 1 \implies k^2 + 1 \leq 2\sqrt{3} \implies k \leq \sqrt{2\sqrt{3} - 1}.
   \]
   Hence:
   \[
   1 \leq f(x) \leq \sqrt{2\sqrt{3} - 1}.
   \]

9. Let \( g(x) = \sqrt{2x - 1} \). Then, by induction, it holds:
   \[
   k \leq \lim_{n \rightarrow \infty} g^n(2).
   \]
   In fact, the value of such limit is \( 1 \).

   Proof: Let \( h(n) = g^{(n)}(2) \). Clearly, \( h \) is decreasing and lowerly bounded \((g \geq 0 \implies h \geq 0)\). Hence, \( h \) converges in \( \mathbb{R} \). Let \( r \) be the value to which it converges. Then:
   \[
   r = \lim_{n \rightarrow \infty} h(n) = \lim_{n \rightarrow \infty} g^{(n)}(2) = \lim_{(n-1) \rightarrow \infty} g(g^{(n-1)}(2)) = g(\lim_{(n-1) \rightarrow \infty} g^{(n-1)}(2)) = g(r).
   \]
   Thus:
   \[
   r = g(r) \implies r = \sqrt{2r - 1} \implies r = 1.
   \]

10. Now, we know:
    \[
    \max f = k \leq 1.
    \]
    But indeed:
    \[
    1 \leq f(x) \leq 2.
    \]

Therefore, \( \boxed{f \equiv 1} \), which indeed fits.