Find all real polynomials $ f$ with $ x,y \in \mathbb{R}$ such that
\[ 2 y f(x \plus{} y) \plus{} (x \minus{} y)(f(x) \plus{} f(y)) \geq 0.
\]
1. We start with the given inequality for all \( x, y \in \mathbb{R} \):
   \[
   2 y f(x + y) + (x - y)(f(x) + f(y)) \geq 0.
   \]
2. Let \( x = a \) and \( y = b \). Then the inequality becomes:
   \[
   2 b f(a + b) + (a - b)(f(a) + f(b)) \geq 0.
   \]
3. Consider the case when \( a = x - y \) and \( b = y \). Substituting these into the inequality, we get:
   \[
   2 y f(x) + (x - 2y)(f(x - y) + f(y)) \geq 0.
   \]
4. Now, consider the case when \( a = y \) and \( b = x - y \). Substituting these into the inequality, we get:
   \[
   2 (x - y) f(x) + (y - (x - y))(f(y) + f(x - y)) \geq 0,
   \]
   which simplifies to:
   \[
   2 (x - y) f(x) + (2y - x)(f(y) + f(x - y)) \geq 0.
   \]
5. We now have two inequalities:
   \[
   2 y f(x) + (x - 2y)(f(x - y) + f(y)) \geq 0,
   \]
   and
   \[
   2 (x - y) f(x) + (2y - x)(f(y) + f(x - y)) \geq 0.
   \]
6. Divide both inequalities by \( x - 2y \) and \( 2y - x \) respectively, and let \( y \rightarrow \infty \). This implies that \( f(x - y) + f(y) \rightarrow f(x) \).
7. Since \( f \) is a polynomial, it is not hard to see that \( f(x - y) + f(y) = f(x) \).
8. Since \( x \) is arbitrary, this is a form of Cauchy's functional equation, which implies that \( f \) must be linear.
9. Therefore, \( f(x) = cx \) for some constant \( c \).
10. To satisfy the original inequality, we need \( c \geq 0 \).

The final answer is \( \boxed{ f(x) = cx } \) with \( c \geq 0 \).