Let the elements of $S$ be $a_1, a_2, \ldots, a_{21}$ where $a_1 < a_2 < \ldots < a_{21}$. Assume there does not exist three elements $a, b, c \in S$ for which $bc < 2a^2 < 4bc$. Now consider terms $a_i < a_{i+1} < a_{i+2} \in S$. Edit: The way I used the cases was definitely unclear. Let me rewrite this. Assume $a_i < a_{i+1} < a_{i+2} \le 2a_i$. Then, $a_i a_{i+2} \le 2a_i^2$ but $a_i < a_{i+1}$ implies $2a_i^2 < 2a_{i+1}^2$ so $a_i a_{i+2} < 2a_{i+1}^2$. Since we are assuming there do not exist three elements $a, b, c \in S$ for which $bc < 2a^2 < 4bc$, and since $a_i a_{i+2} < 2a_{i+1}^2,$ that implies $2a_{i+1}^2 \ge 4bc$ which implies $2a_{i+1}^2 \ge 4a_i a_{i+2}$ which implies $\displaystyle \frac{a_{i+1}}{a_i} > \frac{2a_{i+2}}{a_{i+1}}$. But since $a_{i+2} > a_{i+1}, \displaystyle \frac{2a_{i+2}}{a_{i+1}} > 2$ which implies $a_{i+1} > 2a_i$ but $a_{i+1} < a_{i+2} \le 2a_i$ contradicts this. From here we see that for $bc < 2a^2 < 4bc$ to never be true, $2a_i < a_{i+2}$ is always true, and from here we consider the minimal element of $S$, $a_1$. $2a_1 < a_3, 2a_3 < a_5, \ldots 2a_{19} < a_{21}$, so clearly $1024a_1 < a_{21}$ Since $a_{21} \le 2046$, this implies $a_1 = 1$. However, since $2a_i$ is STRICTLY less than $a_{i+2}$ for all $i$, we have $a_1 = 1$ implies $a_3 \ge 3$ implies $a_5 \ge 7$ $\ldots$ implies $a_{21} \ge 2047$. But the maximal possible element of $S$ is $2046$, so this is a contradiction and we are done.

Forgive me, but i have a little problem understanding your solution. Since you showed that a_21 >= 2047, and that contradicts the assumption that 2046 is the largest element. Then that implies a_(i+2) <= 2a_i for some i. How would we make use of that fact to show that there must exist bc < 2a^2 < 4bc? Sorry to bother you,

You're right, the previous solution was incoherent, as I didn't bother to draft up an on paper solution before trying to write an online one, I managed to leave out a chunk where I explain how I applied the cases. Since I had to edit the solution anyway, I figured I might as well write it out the more elegant way. The cases themselves were incorrectly stated (the way I had it typed up, with the additional information I had on paper it made sense).

Thanks for the editing

Partition $S$ into $10$ subsets as following: $S_i =\{2^{i+1} -2 ,2^{i+1} -3,\ldots,2^{i+1} - (2^i +1) = 2^i -1\}$ for all $i = 1,2,\ldots,10$. All $21$ elements are in the $10$ sets, so there exist $3$ elements $a <b<c$ within one set. Now it's very easy to prove that $ac <2 b^2 <4ac$.

namanhams wrote: Divide S into 10 subsets as following: S_i ={2^(i+1) -2 ,.....,2^i -1 } for all i = 1,2,....10 All 21 elements are in 10 sets ,so there exist 3 elements a <b<c which are in one set . Now it's very easy to prove that ac <2 b^2 <4ac I'll be happy if you prove it... Thanks

It's really great, namanhams. Congratulation.

Partition $S$ into $10$ subsets as following: $S_1 = \{1, 2, 3\}$ $S_i = \{2^{i},\ldots,2^{i+1}-1\}, i=2,\ldots,9$ $S_{10} = \{2^{10},\ldots,2^{11}-2\}$ Lemma: if $a, b, c \in S_i$ and $b<a<c$ then $bc<2a^2<4bc$ Proof: $a, b, c \in S_i$ and $b<a<c$ $\Rightarrow c<2a$, and $(b<a, c<2a) \Rightarrow bc<2a^2 $ (*) Similarily, $a, b, c \in S_i$ and $b<a<c$ $\Rightarrow a{\le}2b$, and $(a{\le}2b, a<c) \Rightarrow a^2<2bc \Rightarrow 2a^2<4bc$ (**) (*) and (**) $\Rightarrow bc<2a^2<4bc$ If we select 21 members of the set $\{1,2,3,\ldots,2046\}$ at least 3 of them will be from one of the $S_i$ sets. According to the lemma above the inequality holds for these three elements. Q.E.D.

I agree! Same solution, but worded differently: Let $S$ be the chosen set, and let set $T$ ($|T|=11$)consist of the smallest number in $S$, the third smallest number in $S$, the fifth smallest, etc. We will prove that there must be a pair of numbers $(m,n)$ in $T$ such that $m<n\leq 2m$. Suppose for the sake of contradiction that every pair of numbers in $T$ with $m<n$ has $n>2m$. Then, let $t_i$ be the $i$th smallest element in $T$. We have that $t_1\geq 2^1-1$, and then by an easy induction we obtain that $t_i\geq 2^i-1$. But this implies that $t_{11}\geq 2047$, contradiction. Thus, take one such pair $(m,n)$, and let $A$ be the number in $S$ greater than $m$ and less than $n$. Then, we have \[\frac n2\leq m<A< n\leq 2m,\]so $(a,b,c)=(A,m,n)$ is one such triple, as desired.