Determine all polynomials $P$ with real coeffients such that $1 + P(x) = \frac12 (P(x -1) + P(x + 1))$ for all real $x$.
1. **Given Equation:**
   We start with the given functional equation for the polynomial \( P(x) \):
   \[
   1 + P(x) = \frac{1}{2} (P(x-1) + P(x+1))
   \]

2. **Using Derivatives:**
   To explore the properties of \( P(x) \), we take the derivative of both sides with respect to \( x \):
   \[
   P'(x) = \frac{1}{2} (P'(x-1) + P'(x+1))
   \]

3. **Simplifying the Derivative Equation:**
   Rearrange the derivative equation:
   \[
   P'(x+1) - P'(x) = P'(x) - P'(x-1)
   \]
   This implies that the difference between consecutive derivatives is constant. Let this constant be \( k \):
   \[
   P'(x+1) - P'(x) = k
   \]
   This is a first-order linear difference equation.

4. **Solving the Difference Equation:**
   The general solution to this difference equation is:
   \[
   P'(x) = kx + d
   \]
   where \( k \) and \( d \) are constants.

5. **Integrating to Find \( P(x) \):**
   Integrate \( P'(x) \) to find \( P(x) \):
   \[
   P(x) = \int (kx + d) \, dx = \frac{k}{2} x^2 + dx + c
   \]
   where \( c \) is the constant of integration.

6. **Substituting Back into the Original Equation:**
   Substitute \( P(x) = \frac{k}{2} x^2 + dx + c \) back into the original equation to determine the constants:
   \[
   1 + \frac{k}{2} x^2 + dx + c = \frac{1}{2} \left( \frac{k}{2} (x-1)^2 + d(x-1) + c + \frac{k}{2} (x+1)^2 + d(x+1) + c \right)
   \]

7. **Expanding and Simplifying:**
   Expand the right-hand side:
   \[
   \frac{1}{2} \left( \frac{k}{2} (x^2 - 2x + 1) + d(x-1) + c + \frac{k}{2} (x^2 + 2x + 1) + d(x+1) + c \right)
   \]
   \[
   = \frac{1}{2} \left( \frac{k}{2} x^2 - kx + \frac{k}{2} + dx - d + c + \frac{k}{2} x^2 + kx + \frac{k}{2} + dx + d + c \right)
   \]
   \[
   = \frac{1}{2} \left( kx^2 + 2dx + 2c + k \right)
   \]
   \[
   = \frac{k}{2} x^2 + dx + c + \frac{k}{2}
   \]

8. **Equating Both Sides:**
   Equate the coefficients of \( x^2 \), \( x \), and the constant terms on both sides:
   \[
   1 + \frac{k}{2} x^2 + dx + c = \frac{k}{2} x^2 + dx + c + \frac{k}{2}
   \]
   This implies:
   \[
   1 = \frac{k}{2}
   \]
   Therefore, \( k = 2 \).

9. **Final Form of \( P(x) \):**
   Substitute \( k = 2 \) back into the polynomial:
   \[
   P(x) = \frac{2}{2} x^2 + dx + c = x^2 + dx + c
   \]

Conclusion:
The polynomial \( P(x) \) that satisfies the given functional equation is:
\[
P(x) = x^2 + bx + c
\]
where \( b \) and \( c \) are real numbers.

The final answer is \( \boxed{ P(x) = x^2 + bx + c } \).