numbers from $1$ to $2046$ We have randomly taken $21$ integers from $1$ to $2046$. 
Show that we can take $a$, $b$ and $c$ from the previous $21$ integers in a way such that the following inequality holds 
\begin{equation}
bc<2a^2<4bc  \end{equation} 
My Attempt:
I have found $3$ triplets of such integers, but I do not think the above inequalities hold for every triplet. 
Any hints ?

Note that for any three positive integers $a,b,c$, if $2^k\leqslant b<a<c<2^{k+1}$ for some $k\in\mathbb{Z}_{>0}$, then
$$bc<ac<a\cdot 2^{k+1}=2a\cdot 2^k\leqslant 2a\cdot a=2a^2,$$
and
$$bc>ba\geqslant 2^ka=\frac{1}{2}\cdot 2^{k+1}a>\frac{1}{2}\cdot a\cdot a=\frac{1}{2}a^2.$$
Thus, $2^k\leqslant b<a<c<2^{k+1}$ for some $k\in\mathbb{Z}_{>0}\Rightarrow bc<2a^2<4bc.$
And $b=1,a=2,c=3$ also satisfies the requirement.
Note that we can divide the integers from $1$ to $2046$ into ten parts: 
\begin{align*}
& A_1=\{1,2,3<2^2\}\\
& A_2=\{2^2=4,5,6,7<2^3\}\\
& \dots \\
& A_{10}=\{2^{10}=1024, 1025,\dots, 2046<2^{11}\}
\end{align*}
If we choose $21=10\times 2+1$ integers in $[1,2046]$,we can always find three of them $x_1<x_2<x_3$ falls in the same part $A_i$. Then we can let $b=x_1,a=x_2,c=x_3$. 
