A segment of unit length is cut into eleven smaller segments, each with length of no more than $a$. For what values of $a$, can one guarantee that any three segments form a triangle?

[i](4 points)[/i]
To solve this problem, we need to ensure that any three segments can form a triangle. This requires satisfying the triangle inequality for any three segments \(a_i, a_j, a_k\) such that \(a_i + a_j > a_k\), \(a_i + a_k > a_j\), and \(a_j + a_k > a_i\).

1. **Initial Setup:**
   - We have a unit length segment divided into 11 smaller segments.
   - Each segment has a length of at most \(a\).
   - We need to find the range of \(a\) such that any three segments can form a triangle.

2. **Sum of Segments:**
   - The total length of all segments is 1.
   - If each segment has a maximum length \(a\), then the sum of the lengths of the 11 segments is \(11a\).
   - Therefore, \(11a \leq 1\), which gives \(a \leq \frac{1}{11}\).

3. **Triangle Inequality:**
   - For any three segments to form a triangle, the sum of the lengths of any two segments must be greater than the length of the third segment.
   - Consider the worst-case scenario where 9 segments have length \(a\) and the remaining 2 segments have lengths summing to \(1 - 9a\).

4. **Worst-Case Analysis:**
   - Let the lengths of the two remaining segments be \(b\) and \(c\) such that \(b + c = 1 - 9a\).
   - The longest segment among the 11 segments is \(a\).
   - For the triangle inequality to hold, the sum of the two smallest segments must be greater than the largest segment.

5. **Applying the Triangle Inequality:**
   - The sum of the two smallest segments (which are \(b\) and \(c\)) must be greater than the largest segment \(a\):
     \[
     b + c > a
     \]
   - Since \(b + c = 1 - 9a\), we have:
     \[
     1 - 9a > a
     \]
   - Solving for \(a\):
     \[
     1 > 10a \implies a < \frac{1}{10}
     \]

6. **Combining Conditions:**
   - From the sum of segments, we have \(a \leq \frac{1}{11}\).
   - From the triangle inequality, we have \(a < \frac{1}{10}\).

Therefore, combining these conditions, the value of \(a\) must satisfy:
\[
\frac{1}{11} \leq a < \frac{1}{10}
\]

The final answer is \(\boxed{\frac{1}{11} \leq a < \frac{1}{10}}\).