Đặt $\sqrt{x}=a(a\ge 0)$
Bất phương trình trở thành $4a+\dfrac{2}{a}<2a^2+\dfrac{1}{2a^2}+2$
$\Leftrightarrow 2a^2+\dfrac{1}{2a^2}+2-4a-\dfrac{2}{a}>0$
$\Leftrightarrow 4a^2-8a^3+4a^2-4a+1>0$
$\Leftrightarrow (2a^2-4a+1)(2a^2+1)>0$
Vì $2a^2+1>0$ nên ta có $2a^2-4a+1>0\Rightarrow \left[ \begin{array}{l}a>\dfrac{2+\sqrt{2}}{2}\\a<\dfrac{2-\sqrt{2}}{2}\end{array} \right.$
$\Rightarrow \left[ \begin{array}{l}a\ge \dfrac{3+\sqrt{2}}{2}\\0\le a<\dfrac{3-\sqrt{2}}{2}\end{array} \right.$
