# disjoint sum-free subsets

I'd like to prove that for a set of nonzero integers, call it $$S$$, there exist two disjoint subsets $$S_1$$ and $$S_2$$ such that $$S_1$$ and $$S_2$$ are sum-free (i.e., if $$a,b \in S_i$$, then $$a+b \notin S_i$$), and $$|S_1 \cup S_2| > 2 \frac{|S|}{3}.$$ I've seen a proof of the fact that any set of nonzero integers $$S$$ has a sum-free subset of cardinality at least $$\frac{|S|}{3}$$ using a probabilistic argument (e.g. pages 1-2 here), but I'm not sure how to adapt/generalize the proof to the problem above.

Any help or hint appreciated!

Pick a prime $$p > \max_{s \in S}\{|s|\}$$ such that $$p \equiv 1 \mod 6$$, write $$p = 6k + 1$$ and let $$S_1 = \{2k+1, 2k+2, \ldots, 4k+1\}$$, $$S_2 = \{k+1,\ldots,2k\}$$ and $$S_3 = \{4k+2,\ldots,5k+1\}$$. Note that $$S_1$$ is sum free as before. By a simple calculation, so is $$S_2 \cup S_3$$ (consider all cases of taking two elements from either set). Finally, write $$T = S_1 \cup S_2 \cup S_3$$ (note: $$|T| = 4k+1$$).
For a fixed $$s \in S$$ and $$\alpha \in \{1,\ldots, p -1\} = \mathbb{Z}^*_p$$ chosen uniformly at random, the random variable $$\alpha \cdot s$$ distributes uniformly over $$\mathbb{Z}^*_p$$ (as $$p > |s|$$) and so, letting $$\mathbb{I}_\alpha(s)$$ denote an indicator random variable which is $$1$$ iff $$\alpha s \in T$$ we get $$\mathbb{E}_\alpha[\mathbb{I}_\alpha(s)] = |T|/(p-1) > 2/3$$.
By linearlity of expectation: $$\mathbb{E}_\alpha[\sum_{s} \mathbb{I}_\alpha(s)] > |S|2/3$$ and for an $$\alpha$$ which attains the expected value (or above), the elements mapped to $$S_1$$ and the elements mapped to $$S_2 \cup S_3$$ prove the claim.