jmc

By AM-GM, $$2\sqrt{x}+\frac{1}{x}=\sqrt{x}+\sqrt{x}+\frac{1}{x}\ge 3\sqrt[3]{\sqrt{x}\cdot \sqrt{x}\cdot \frac{1}{x}}=3.$$Equality occurs when $x=1,$ so the minimum value is $\boxed{3}.$