Every set of n integers contains a sum-free subset of more than n/3 elements

Idea. Multiply by a random unit modulo a prime and keep the elements landing in a sum-free "middle third".

A suitable prime. There are infinitely many primes $p\equiv2\pmod3$. Choose such a prime $p=3k+2$ with $p>|a|$ for all $a\in A$; then no $a$ is divisible by $p$.

The middle third. Let $S=\{k+1,k+2,\dots,2k+1\}\subset\mathbb Z_p$, so $|S|=k+1$. If $u,v\in S$, then $2k+2\le u+v\le4k+2$. Modulo $p=3k+2$ these sums are the residues $2k+2,\dots,3k+1$ and $0,\dots,k$. None lies in $S$. So $S$ is sum-free modulo $p$.

Averaging. For $x$ uniform in $\{1,\dots,p-1\}$ and fixed $a\in A$, $ax\bmod p$ is uniform on the non-zero residues. So
$$\Pr[ax\in S]=\frac{k+1}{3k+1}>\frac13.$$
By linearity, the expected number of $a\in A$ with $ax\bmod p\in S$ is more than $\frac n3$. Hence some $x$ gives $B=\{a\in A:ax\bmod p\in S\}$ with more than $\frac n3$ elements.

$B$ is sum-free. If $a+b=c$ with $a,b,c\in B$, then multiplying by $x$ gives $ax+bx\equiv cx\pmod p$ with $ax\bmod p,\ bx\bmod p,\ cx\bmod p\in S$, contradicting the sum-freeness of $S$. ∎

Lesson. Randomise by a dilation in a finite field to bring a hard structure into a nice fixed set, and use linearity of expectation.