5. (5-7,8) There are 10 segments, the length of each of which is expressed as an integer not exceeding some $N$. a) Let $N=100$. Provide an example of a set of 10 segments such that no three of them can form a triangle. b) Find the maximum $N$ for which it can be guaranteed that there will be three segments that can form a triangle.
Answer: a) $1,1,2,3,5,8,13,21,34,55 ;$ b) $N=54$.

Solution: a) If each new segment is chosen to be equal to the sum of the two largest of the remaining ones, then it is impossible to form a triangle with its participation.

b) From the previous point, it is clear that for $N=55$ such a sequence can be constructed.

We will show that for $N=54$ it is already impossible to construct such a sequence. Let's order the lengths of the segments in ascending order: $a_{1} \leqslant a_{2} \leqslant \ldots \leqslant a_{10} \leqslant 54$. If a triangle cannot be formed from segments of lengths $a_{i}, a_{i+1}, a_{i+2}$, this means that the triangle inequality is violated, i.e., $a_{i+2} \geqslant a_{i}+a_{i+1}$. Thus, since it is obvious that $a_{1}, a_{2} \geqslant 1$, by sequentially applying this inequality for $i=1,2, \ldots, 8$, we get: $a_{3} \geqslant a_{1}+a_{2} \geqslant 2, a_{4} \geqslant a_{2}+a_{3} \geqslant 3, \ldots a_{10} \geqslant a_{8}+a_{9} \geqslant 55$, which leads to a contradiction.