11.3. The sequence $a_{n}$ is constructed as follows: $a_{1}, a_{2}$ are arbitrary real numbers, and for $n \geq 3$, the number $a_{n}$ is the smallest of the numbers $\left|a_{i}-a_{j}\right|, 1 \leq$ $i<j \leq n-1$. For example, if $a_{1}=6, a_{2}=\frac{19}{2}$, then we get the sequence $6, \frac{19}{2}, \frac{7}{2}, \frac{5}{2}, 1,1,0,0,0 \ldots$ For some choice of $a_{1}, a_{2}$, the sequence resulted in $a_{10}=1$. Find the smallest possible value of $a_{3}$ in such a sequence.
Solution. Let $S_{n}$ be the set of all numbers $\left|a_{i}-a_{j}\right|, 1 \leq i<j \leq n-1, \quad n \geq 3$. Then $a_{n}$ is the smallest number in the set $S_{n}$. Since $S_{n} \subset S_{n+1}$, then $a_{n} \geq a_{n+1}$, i.e., starting from $a_{3}$, the sequence is non-increasing. Therefore, $a_{9} \geq a_{10}=1$. Further, from the definition of the sequence, it follows that $a_{10} \leq\left|a_{8}-a_{9}\right|=a_{8}-a_{9} \Rightarrow a_{8} \geq a_{9}+a_{10} \geq 2$. Similarly, $a_{7} \geq a_{8}+a_{9} \geq 2+1=3, a_{6} \geq a_{7}+a_{8} \geq 3+2=5, \ldots, a_{3} \geq a_{4}+a_{5} \geq 13+8=21$. This value is achieved if we take $a_{1}=55, a_{2}=34$.

Criteria.

(-.) It is strictly proven that $a_{n} \geq a_{n+1}$ for $n \geq 3$.

$(-/+)$ Answer + incomplete example: it is stated that $\min \left(a_{3}\right)=21$, and for this case, all terms of the sequence except $a_{1}, a_{2}$ are written out.

$(+/ 2)$ Answer + example (including $\left.a_{1}, a_{2}\right)$, or it is proven that $a_{3} \geq 21$, but without proof, it is used that $a_{n} \geq a_{n+1}$, an example of the sequence for $a_{3}=21$ is missing, or answer + incomplete example (without $a_{1}, a_{2}$) + significant progress in justifying the minimality.

$(+/-)$ It is strictly and without gaps proven that $a_{3} \geq 21$, the example is missing, or answer + example (including $\left.a_{1}, a_{2}\right)+$ justification of minimality with minor gaps.

(+.) Logically correct and complete solution with no significant gaps except the following:

- incomplete example (not explicitly written $a_{1}, a_{2}$),
- the fact that $a_{n} \geq a_{n+1}$ for $n \geq 3$ is used without proof,
- an arithmetic error is made in calculating $a_{3}$.

(+) It is strictly and without gaps proven that $a_{3} \geq 21$, and $a_{1}, a_{2}$ are explicitly written out, for which $a_{3}=21$.