Find all functions $ f:R \implies R $ , such for all $x,y,z$ 
$f(xy)+f(xz)\geq f(x)f(yz) + 1$
1. **Define the assertion:**
   Let \( P(x, y, z) \) be the assertion of the given functional inequality:
   \[
   f(xy) + f(xz) \geq f(x) f(yz) + 1
   \]

2. **Evaluate \( P(0, 0, 0) \):**
   \[
   P(0, 0, 0) \implies f(0 \cdot 0) + f(0 \cdot 0) \geq f(0) f(0 \cdot 0) + 1
   \]
   Simplifying, we get:
   \[
   2f(0) \geq f(0)^2 + 1
   \]
   Let \( f(0) = a \). Then:
   \[
   2a \geq a^2 + 1
   \]
   Rearrange to form a quadratic inequality:
   \[
   a^2 - 2a + 1 \leq 0
   \]
   This simplifies to:
   \[
   (a - 1)^2 \leq 0
   \]
   Therefore:
   \[
   a = 1
   \]
   Hence, \( f(0) = 1 \).

3. **Evaluate \( P(1, 1, 1) \):**
   \[
   P(1, 1, 1) \implies f(1 \cdot 1) + f(1 \cdot 1) \geq f(1) f(1 \cdot 1) + 1
   \]
   Simplifying, we get:
   \[
   2f(1) \geq f(1)^2 + 1
   \]
   Let \( f(1) = b \). Then:
   \[
   2b \geq b^2 + 1
   \]
   Rearrange to form a quadratic inequality:
   \[
   b^2 - 2b + 1 \leq 0
   \]
   This simplifies to:
   \[
   (b - 1)^2 \leq 0
   \]
   Therefore:
   \[
   b = 1
   \]
   Hence, \( f(1) = 1 \).

4. **Evaluate \( P(x, 0, \frac{1}{x}) \) for \( x \neq 0 \):**
   \[
   P(x, 0, \frac{1}{x}) \implies f(x \cdot 0) + f(x \cdot \frac{1}{x}) \geq f(x) f(0 \cdot \frac{1}{x}) + 1
   \]
   Simplifying, we get:
   \[
   f(0) + f(1) \geq f(x) f(0) + 1
   \]
   Using \( f(0) = 1 \) and \( f(1) = 1 \):
   \[
   1 + 1 \geq f(x) \cdot 1 + 1
   \]
   Therefore:
   \[
   2 \geq f(x) + 1
   \]
   Hence:
   \[
   f(x) \leq 1
   \]

5. **Evaluate \( P(1, 0, x) \):**
   \[
   P(1, 0, x) \implies f(1 \cdot 0) + f(1 \cdot x) \geq f(1) f(0 \cdot x) + 1
   \]
   Simplifying, we get:
   \[
   f(0) + f(x) \geq f(1) f(0) + 1
   \]
   Using \( f(0) = 1 \) and \( f(1) = 1 \):
   \[
   1 + f(x) \geq 1 \cdot 1 + 1
   \]
   Therefore:
   \[
   1 + f(x) \geq 2
   \]
   Hence:
   \[
   f(x) \geq 1
   \]

6. **Combine the results:**
   From steps 4 and 5, we have:
   \[
   f(x) \leq 1 \quad \text{and} \quad f(x) \geq 1
   \]
   Therefore:
   \[
   f(x) = 1 \quad \forall x \in \mathbb{R}
   \]

Thus, the only function that satisfies the given inequality is \( f(x) = 1 \) for all \( x \).

The final answer is \( \boxed{ f(x) = 1 } \) for all \( x \in \mathbb{R} \).