# 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!!

By AM-GM, we have

$$\frac{\frac{1}{x}+\sqrt{x}+\sqrt{x}}{3}\ge\sqrt[3]{\frac{1}{x}\sqrt{x}\sqrt{x}}=1$$

Hence $$\frac{1}{x}+2\sqrt{x}\ge3$$

as desired.

• Oh!! How could I miss this.Thanks!! Aug 15, 2014 at 14:30

$$2\sqrt{x}-\left(3-\frac1x\right)=\left(2\sqrt{x}+1\right)\cdot\left(1-\frac1{\sqrt{x}}\right)^2\geqslant0$$

Just for sake of completeness:

Let $$f(x)=2\sqrt{x}-3+\frac1x$$

Now,$$f'(x)=\frac1{\sqrt{x}}-\frac1{x^2}\begin{cases}\ge0\;x\ge1\\\le0,\;x\in(0,1]\end{cases}$$

Also, $$\lim_{x\to 0^+}f(x)>0,f(1)=0,\lim_{x\to \infty}f(x)>0$$

Thus, $$f(x)\ge0\;\forall\;x\in\mathbb R^+$$

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, $$2 \sqrt{x} \geq 3 - \frac{1}{x}.$$

• oops, repeated. Aug 15, 2014 at 14:38