
2. Find all real numbers $x$ and $y$ such that

$$
x^{2}+2 y^{2}+\frac{1}{2} \leq x(2 y+1)
$$


Solution: We write the inequality in the form

$$
2 x^{2}+4 y^{2}+1-4 x y-2 x \leq 0
$$

Thus $\left(x^{2}-4 x y+4 y^{2}\right)+\left(x^{2}-2 x+1\right) \leq 0$. Hence

$$
(x-2 y)^{2}+(x-1)^{2} \leq 0
$$

Since $x, y$ are real, we know that $(x-2 y)^{2} \geq 0$ and $(x-1)^{2} \geq 0$. Hence it follows that $(x-2 y)^{2}=0$ and $(x-1)^{2}=0$. Therefore $x=1$ and $y=1 / 2$.
