I claim the answer is $4 / 9$. Let $s=x+y, p=x y$, so $x$ and $y$ are $\frac{s \pm \sqrt{s^{2}-4 p}}{2}$. Since $x$ and $y$ are real, $s^{2}-4 p \geq 0$. If one of the three quantities is less than or equal to $1 / 9$, then at least one of the others is at least $4 / 9$ by the pigeonhole principle since they add up to 1. Assume that $s-2 p<4 / 9$, then $s^{2}-4 p<(4 / 9+2 p)^{2}-4 p$, and since the left side is non-negative we get $0 \leq p^{2}-\frac{5}{9} p+\frac{4}{81}=\left(p-\frac{1}{9}\right)\left(p-\frac{4}{9}\right)$. This implies that either $p \leq \frac{1}{9}$ or $p \geq \frac{4}{9}$, and either way we're done. This minimum is achieved if $x$ and $y$ are both $1 / 3$, so the answer is $\frac{4}{9}$, as claimed.