$\dfrac{x + \sqrt x + 1}{\sqrt x}$
$= \sqrt x + 1 + \dfrac{1}{\sqrt x}$
Ta có:
$\sqrt x + \dfrac{1}{\sqrt x} \geq 2\sqrt{(\sqrt x)\cdot \left(\dfrac{1}{\sqrt x}\right)} = 2$
$\Rightarrow \sqrt x + 1 + \dfrac{1}{\sqrt x} \geq 2 + 1 = 3$
Dấu = xảy ra $\Leftrightarrow \sqrt x = \dfrac{1}{\sqrt x} \Leftrightarrow x = 1$
Vậy $min\left(\dfrac{x + \sqrt x + 1}{\sqrt x}\right) = 3$ tại $ x= 1$
