Romanian Mathematical Olympiad

From $x^2 - x + 1 = (x - \frac{1}{2})^2 + \frac{3}{4} \ge \frac{3}{4}$, $3y^2 - 2y + 3 = 3(y - \frac{1}{3})^2 + \frac{8}{3} \ge \frac{8}{3}$
we derive that $(x^2 - x + 1)(3y^2 - 2y + 3) \ge \frac{3}{4} \cdot \frac{8}{3} = 2$, for any real numbers $x$ and $y$. Equality holds for $(3y - 1)^2 = 0$ and $(x - \frac{1}{2})^2 = 0$, that is $x = \frac{1}{2}$ and $y = \frac{1}{3}$.