$ĐK:x > 0\\ x-\sqrt{x}+1\\ =x-\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4} > 0 \, \forall \, x>0\\ \Rightarrow A> 0 \, \forall \, x>0\\ (\sqrt{x}-1)^2 \ge 0 \, \forall \, x>0\\ \Leftrightarrow x-2\sqrt{x}+1 \ge 0 \, \forall \, x>0\\ \Leftrightarrow x-\sqrt{x}+1 \ge \sqrt{x} \, \forall \, x>0\\ \Rightarrow A=\dfrac{x-\sqrt{x}+1}{\sqrt{x}} \ge 1 \, \forall \, x>0\\ \Rightarrow A\ge\sqrt{A}\, \forall \, x>0$