additive discrepancy under a multiplicative constraint Consider four sequences of numbers, $0 \le a_i, b_i, c_i, d_i \le 1$, suppose they satisfy the following constraints:
(1). $\sum_{i=1}^K a_i \ge 1/2 + \epsilon$;
(2). $\sum_{i=1}^K d_i \le 1/2 - \epsilon$;
(3). $a_i d_i = b_i c_i$ for all $i=1, \ldots, K$.
Is it true that
\begin{equation}
\sum_{i=1}^K (|a_i - b_i| + |a_i - c_i|) = \Omega(\epsilon) ?
\end{equation}
Added remark after seeing the answer: Initially my Condition (1) was,
(1). $\sum_{i=1}^K a_i, \sum_{i=1}^K b_i, \sum_{i=1}^K c_i \ge 1/2 + \epsilon$.
But it turns out (see the answer below) that the inequality is true even without the requirement on b_i and c_i.
 A: $\newcommand{\ep}{\epsilon}$
It is not hard to show (see the proof at the end of this answer) that for any real $a,b,c,d\ge0$ such that $ad=bc$ we have 
\begin{equation}\tag{1}
 |a-b|+|a-c|\ge a-d. 
\end{equation}
Replacing here $a,b,c,d$ by $a_i,b_i,c_i,d_i$ and summing in $i$, we have 
\begin{equation}
 \sum_i (|a_i - b_i| + |a_i - c_i|)\ge\sum_i a_i-\sum_i d_i\ge2\ep, 
\end{equation}
as desired. (The conditions that $\sum_i b_i, \sum_i c_i \ge 1/2 + \ep$ were not needed or used here.)

Proof of (1). If $a=0$, then $a-d\le0$, so that (1) holds. So, without loss of generality (wlog), $a>0$, whence $d=bc/a$ and $a-d=(a^2-bc)/a$. So, wlog $a^2>bc$ and hence $a\ge b\wedge c$. Also, wlog $c\le b$. 
So, one of the following two cases takes place. 
Case 1: $0\le c\le b\le a$. Here (1) can be rewritten as 
$$f(a,b,c):=a-b-c+bc/a\ge0,$$
which follows because $f(a,b,c)$ is nonincreasing in $b$ (given that $c/a\le1$) and hence $f(a,b,c)\ge f(a,a,c)=0$. So, (1) holds in Case 1. 
Case 2: $0\le c\le a\le b$. Here (1) can be rewritten as 
$$g(a,b,c):=b-c+bc/a-a\ge0,$$
which follows because $g(a,b,c)$ is nondecreasing in $b$ and hence $g(a,b,c)\ge g(a,a,c)=0$. So, (1) holds in Case 2 as well. 
Inequality (1) is completely proved. 
