Prove the inequality $2\sqrt{x}\ge3-\frac1x $ Given that $x\gt0$   prove the following inequality:
$2\sqrt{x}\ge3-\frac1x $
I have done it using calculus but how can I do it using elementary methods?Thanks!!

Note that $$2 \sqrt{x} + \frac{1}{x} = \sqrt{x} + \sqrt{x} + \frac{1}{x} \\
\geq 3 \sqrt[3]{\sqrt{x} \times \sqrt{x} \times \frac{1}{x}} =3.$$
That is,
$$2 \sqrt{x} \geq 3 - \frac{1}{x}. $$
