Sum of two uniforms at least twice their gap

Let $X$ and $Y$ be independent uniform random variables on $(0,1)$. Find the probability that $X+Y\ge 2\,|X-Y|$.

$\tfrac23$. By symmetry take $Y\ge X$ and double. The condition becomes $Y\le 3X$, so we keep the half-square above the diagonal except the triangle with corners $(0,0),(0,1),(\tfrac13,1)$, of area $\tfrac16$. That leaves $\tfrac12-\tfrac16=\tfrac13$, and doubling gives $\tfrac23$.