ĐK: $x,y\geq 0$
$C=x+2\sqrt{x}-4\sqrt{xy}+y$
=$x+2\sqrt{x}(1-2\sqrt{y})+(1-2\sqrt{y})^2-(3y-4\sqrt{y}+1)$
=>$3C=3(\sqrt{x}+1-2\sqrt{y})^2-(9y-12\sqrt{y}+4)+1$
=>$3C=3(\sqrt{x}+1-2\sqrt{y})^2-(3\sqrt{y}-2)^2+1$ $\geq 1 \forall x,y\in \mathbb{R}$
=>$C\geq \frac{1}{3}$
Dấu "=" xảy ra khi:
$\left\{\begin{matrix} \sqrt{x}+1-2\sqrt{y}=0 & & \\ 3\sqrt{y}-2=0 & & \end{matrix}\right.$
Khi đó sẽ tìm được $x=\frac{1}{9}$; $y=\frac{4}{9}$.