Progetto Olimpiadi della Matematica

The answer is (B). Indeed, the quantity $|x| + 2|y|$ is non-negative and its square equals $x^{2} + 4 y^{2} + 4|x||y| \geq x^{2} + 4 y^{2} = 1$, so also $|x| + 2|y| \geq 1$. This value can actually be achieved (exactly when $|x||y| = 0$, that is if $x = \pm 1, y = 0$ or $x = 0, y = \pm \frac{1}{2}$), and it is therefore the minimum possible.