Find all reals $ k$ such that
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \plus{} 1\geq k(a \plus{} b \plus{} c \plus{} d)
\]
holds for all $ a,b,c,d\geq \minus{} 1$.

[i]Edited by orl.[/i]
To find all reals \( k \) such that 
\[ a^3 + b^3 + c^3 + d^3 + 1 \geq k(a + b + c + d) \]
holds for all \( a, b, c, d \geq -1 \), we need to analyze the inequality under different conditions.

1. **Case 1: \( a = b = c = d = -1 \)**

   Substitute \( a = b = c = d = -1 \) into the inequality:
   \[
   (-1)^3 + (-1)^3 + (-1)^3 + (-1)^3 + 1 \geq k(-1 + -1 + -1 + -1)
   \]
   Simplify the left-hand side:
   \[
   -1 - 1 - 1 - 1 + 1 = -3
   \]
   Simplify the right-hand side:
   \[
   k(-4) = -4k
   \]
   Thus, we have:
   \[
   -3 \geq -4k \implies k \geq \frac{3}{4}
   \]

2. **Case 2: \( a = b = c = d = \frac{1}{2} \)**

   Substitute \( a = b = c = d = \frac{1}{2} \) into the inequality:
   \[
   \left(\frac{1}{2}\right)^3 + \left(\frac{1}{2}\right)^3 + \left(\frac{1}{2}\right)^3 + \left(\frac{1}{2}\right)^3 + 1 \geq k\left(\frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2}\right)
   \]
   Simplify the left-hand side:
   \[
   4 \cdot \frac{1}{8} + 1 = \frac{4}{8} + 1 = \frac{1}{2} + 1 = \frac{3}{2}
   \]
   Simplify the right-hand side:
   \[
   k \cdot 2 = 2k
   \]
   Thus, we have:
   \[
   \frac{3}{2} \geq 2k \implies k \leq \frac{3}{4}
   \]

3. **Combining the results from both cases:**

   From Case 1, we have \( k \geq \frac{3}{4} \).
   From Case 2, we have \( k \leq \frac{3}{4} \).

   Therefore, the only value of \( k \) that satisfies both conditions is:
   \[
   k = \frac{3}{4}
   \]

4. **Verification:**

   We need to verify that \( k = \frac{3}{4} \) satisfies the original inequality for all \( a, b, c, d \geq -1 \).

   Consider the expression:
   \[
   a^3 + b^3 + c^3 + d^3 + 1 \geq \frac{3}{4}(a + b + c + d)
   \]

   We can rewrite it as:
   \[
   a^3 + b^3 + c^3 + d^3 + 1 - \frac{3}{4}(a + b + c + d) \geq 0
   \]

   For \( a = b = c = d = x \), we have:
   \[
   4x^3 + 1 \geq 3x
   \]
   \[
   4x^3 - 3x + 1 \geq 0
   \]

   The polynomial \( 4x^3 - 3x + 1 \) is non-negative for \( x \geq -1 \). This can be verified by checking the critical points and the behavior of the polynomial.

   Therefore, \( k = \frac{3}{4} \) is indeed the correct value.

The final answer is \( \boxed{\frac{3}{4}} \)