Find all second degree polynomials $P(x)$ such that for all $a \in\mathbb{R} , a \geq 1$, then 
$P(a^2+a) \geq a.P(a+1)$ 
 
1. Let \( P(x) = ux^2 + vx + w \). The given inequality is:
   \[
   P(a^2 + a) \geq a \cdot P(a + 1) \quad \forall a \geq 1
   \]
   Substituting \( P(x) \) into the inequality, we get:
   \[
   u(a^2 + a)^2 + v(a^2 + a) + w \geq a \cdot (u(a + 1)^2 + v(a + 1) + w)
   \]

2. Expanding both sides:
   \[
   u(a^2 + a)^2 + v(a^2 + a) + w = u(a^4 + 2a^3 + a^2) + v(a^2 + a) + w
   \]
   \[
   = ua^4 + 2ua^3 + ua^2 + va^2 + va + w
   \]
   \[
   a \cdot (u(a + 1)^2 + v(a + 1) + w) = a \cdot (u(a^2 + 2a + 1) + v(a + 1) + w)
   \]
   \[
   = a \cdot (ua^2 + 2ua + u + va + v + w)
   \]
   \[
   = aua^2 + 2ua^2 + ua + ava + av + aw
   \]
   \[
   = ua^3 + 2ua^2 + ua + va^2 + va + aw
   \]

3. Simplifying the inequality:
   \[
   ua^4 + 2ua^3 + ua^2 + va^2 + va + w \geq ua^3 + 2ua^2 + ua + va^2 + va + aw
   \]
   \[
   ua^4 + 2ua^3 + ua^2 + va^2 + va + w - ua^3 - 2ua^2 - ua - va^2 - va - aw \geq 0
   \]
   \[
   ua^4 + (2u - u)a^3 + (u - 2u)a^2 - ua - aw + w \geq 0
   \]
   \[
   ua^4 + ua^3 - ua^2 - (u + w)a + w \geq 0
   \]

4. To ensure the inequality holds for all \( a \geq 1 \), we need \( u > 0 \). Let \( t = \frac{w}{u} \):
   \[
   a^4 + a^3 - a^2 - (t + 1)a + t \geq 0 \quad \forall a \geq 1
   \]

5. Factorizing the polynomial:
   \[
   (a - 1)(a^3 + 2a^2 + a - t) \geq 0 \quad \forall a \geq 1
   \]

6. Since \( a^3 + 2a^2 + a - t \) is increasing for \( a \geq 1 \), we need:
   \[
   a^3 + 2a^2 + a - t \geq 0 \quad \forall a \geq 1
   \]

7. Evaluating at \( a = 1 \):
   \[
   1^3 + 2 \cdot 1^2 + 1 - t \geq 0
   \]
   \[
   1 + 2 + 1 - t \geq 0
   \]
   \[
   4 - t \geq 0
   \]
   \[
   t \leq 4
   \]

8. Therefore, \( t = \frac{w}{u} \leq 4 \), which implies \( w \leq 4u \).

The final answer is \( \boxed{ u > 0 \text{ and } w \leq 4u } \)