For every integer $ n \geq 2$ determine the minimum value that the sum $ \sum^n_{i\equal{}0} a_i$ can take for nonnegative numbers $ a_0, a_1, \ldots, a_n$ satisfying the condition $ a_0 \equal{} 1,$ $ a_i \leq a_{i\plus{}1} \plus{} a_{i\plus{}2}$ for $ i \equal{} 0, \ldots, n \minus{} 2.$
1. We start with the given conditions:
   - \( a_0 = 1 \)
   - \( a_i \leq a_{i+1} + a_{i+2} \) for \( i = 0, \ldots, n-2 \)

2. To find the minimum value of \( \sum_{i=0}^n a_i \), we need to find the values of \( a_i \) that satisfy the given conditions and minimize the sum.

3. Consider the system of equations derived from the condition \( a_i \leq a_{i+1} + a_{i+2} \):
   \[
   \begin{cases}
   a_1 + a_2 = 1 \\
   a_2 + a_3 = a_1 \\
   a_3 + a_4 = a_2 \\
   \vdots \\
   a_{n-1} + a_n = a_{n-2} \\
   a_n = 0
   \end{cases}
   \]

4. We observe that the sequence \( a_i \) resembles the Fibonacci sequence. Let's denote the Fibonacci sequence by \( f_i \), where \( f_0 = 0 \), \( f_1 = 1 \), and \( f_{i+2} = f_{i+1} + f_i \).

5. We propose that \( a_i = \frac{f_{n-i}}{f_n} \) for \( i = 0, \ldots, n \). This satisfies the initial condition \( a_0 = 1 \) because:
   \[
   a_0 = \frac{f_n}{f_n} = 1
   \]

6. We need to verify that this choice satisfies the inequality \( a_i \leq a_{i+1} + a_{i+2} \):
   \[
   a_i = \frac{f_{n-i}}{f_n} \leq \frac{f_{n-(i+1)}}{f_n} + \frac{f_{n-(i+2)}}{f_n} = \frac{f_{n-i-1} + f_{n-i-2}}{f_n} = \frac{f_{n-i}}{f_n}
   \]
   This holds true because \( f_{n-i} = f_{n-i-1} + f_{n-i-2} \).

7. Now, we calculate the sum \( \sum_{i=0}^n a_i \):
   \[
   \sum_{i=0}^n a_i = \sum_{i=0}^n \frac{f_{n-i}}{f_n} = \frac{1}{f_n} \sum_{i=0}^n f_{n-i} = \frac{1}{f_n} \sum_{j=0}^n f_j
   \]
   where \( j = n-i \).

8. The sum of the first \( n+1 \) Fibonacci numbers is given by:
   \[
   \sum_{j=0}^n f_j = f_{n+2} - 1
   \]

9. Therefore, the minimum value of the sum is:
   \[
   \sum_{i=0}^n a_i = \frac{f_{n+2} - 1}{f_n}
   \]

10. For large \( n \), the ratio \( \frac{f_{n+2}}{f_n} \) approaches the golden ratio \( \phi \approx 1.618 \). However, for exact values, we use the Fibonacci sequence directly.

The final answer is \( \boxed{ \frac{f_{n+2} - 1}{f_n} } \)