$\exists c>0$, $\forall A\subseteq \Bbb R_{\ne 0}$ s.t. $|A|=n$, $\exists B\subseteq A$ s.t. $|B|>cn$ & $b_1+2b_2=2b_3+2b_3$ has no solutions in $B$. 
EXERCISE 2.7.2 fron Alon and Spencer's The Probabilistic Method.
Prove that there is a positive constant $c$ so that every set $A$ of $n$ nonzero reals
contains a subset $B\subseteq A$ of size $|B| > cn$  so that there are no $b_{1},b_{2},b_{3},b_{4}\in B$ satisfying $$b_{1}+2b_{2}=2b_{3}+2b_{4}\,.$$

My idea is to regard the elements of $A$ in modulo $m$ for some $m>0$, then I must find numbers which they are not true in above relation in modulo $m$.  This is just the Idea but my problem is to make rigid and exact.
 A: I know this is an old question, but I'm in the process of writing solutions for Alon and Spencer, and thought I'd post what I wrote.

A set $A$ is called $(*)$-free if $\nexists~a_1,a_2,a_3,a_4 \in A$ s.t. \begin{equation*}
a_1 + 2a_2 = 2a_3 + 2a_4.
\end{equation*} 
Then we note that if $A \mod 1$ is $(*)$-free $\mod 1$, then $A$ must be $(*)$-free as well.  
We now consider the interval $S := (\frac{3}{7},\frac{4}{7})$ $\mod 1$.  We note that for all $a_1, a_2,a_3,a_4 \in S$, $a_1 + 2a_2 \in \left(\frac{9}{7},\frac{12}{7}\right)$ and $2a_3 + 2a_4 \in \left(\frac{12}{7},\frac{16}{7}\right)$.  Since these two intervals are disjoint $\mod 1$, $S$ is $(*)$-free.  
Fix $a \in A$; now, for $x$ uniformly distributed in $[1,R]$ for some $R$, $xa \mod 1$ is uniformly distributed between $a \mod 1$ and $Ra \mod 1$.  For any given $m$, there is a sufficiently large $R$, $xa \mod 1$ wraps around $\geq m$ times as $x$ varies between $1$ and $R$.  Thus, for any given $m$, there exists an $R$ s.t. $xa \mod 1$ wraps around $\geq m$ times for each $a \in A$.  Now, let $\chi_a$ be the indicator for the event that $xa \mod 1 \in S$.  Then \begin{equation*}
\mathbb{E}[\chi_a] \geq \frac{m-1}{7m}
\end{equation*}
Thus, the expected number of $xa \mod 1 \in S$ is \begin{equation*}
\sum \mathbb{E}[\chi_a] \geq n\frac{m-1}{7m}.
\end{equation*}
Taking $m$ arbitrarily large, we have that the expected number of $xa \mod 1 \in S$ is $\geq \frac{n}{7}$.  Thus, there exists some $x$ s.t. there are at least $\frac{n}{7}$ elements of $xA \mod 1$ in $S$.  Let $B$ be the elements  $a\in A$ s.t. $ax \mod 1 \in S$.  Since  $xB$ is $(*)$-free $\mod 1$, so is $xB$.  Since $xB$ are $(*)$-free, so is $B$, yielding the desired result.
