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!!
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4 Answers
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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.
answered Aug 15, 2014 at 14:28
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$$2\sqrt{x}-\left(3-\frac1x\right)=\left(2\sqrt{x}+1\right)\cdot\left(1-\frac1{\sqrt{x}}\right)^2\geqslant0$$
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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, $$f(x)\ge0\;\forall\;x\in\mathbb R^+$$
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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, $$2 \sqrt{x} \geq 3 - \frac{1}{x}. $$
