An algebra problem by Anish Roy

We write the inequality in the form $2x^2 + 4y^2 + 1 - 4xy - 2x ≤ 0.$ Thus $(x^2 - 4xy + 4y^2) + (x^2 - 2x + 1) ≤ 0$ . Hence $(x - 2y)^2 + (x - 1)^2 ≤ 0$ . Since $x, y$ are real, we know that $(x - 2y)^2 ≥ 0$ and $(x - 1)^2 ≥ 0$ . Hence it follows that $(x - 2y)^2 = 0$ and $(x - 1)^2 = 0$ . Therefore $x = 1$ and $y = \frac{1}{2}$ . Thus we get $2x+4y=2(1)+4(\frac{1}{2})=\boxed{4}$