First, consider the case where $9$ segments has length $a$ and the other $2$ segments sum up to length $1-9a$. By the triangle inequality, the sum of the $2$ smallest segments has to be strictly greater than the size of the largest segment. Decreasing the size of any of the the $9$ segments will just increase the sum of the other $2$ segments and make it easier to form a triangle, since the longest size won't increase as it's at most $a$ anyway and the other $2$ segments just increase in their sum of the lengths. Also, the longest segment has length $a$, so the sum of $11$ of the segments is at most $11a$, which is given by the problem to be equal to 1, so that means that $a$ is at least $\frac{1}{11}$, and also by the triangle inequality, $2-18a>a$, which means that $2>20a$ or $a<\frac{1}{10}$, and for $a$ larger than $\frac{1}{9}$ just make each of the segment length $a$ until out of length and make the remaining segments random, which will sum up to less than $a$, so the answer is $\boxed{\frac{1}{11}\leq{a}<\frac{1}{10}}$