- #1

issacnewton

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## Homework Statement

For ##x> 1##, prove that $$2\sqrt{x} > 3 - \frac 1 x$$

## Homework Equations

Definition of increasing function

## The Attempt at a Solution

Let ##f(x) = 2\sqrt{x} + \frac 1 x## defined on domain ##[1, \infty)##. Function is increasing if ## f'(x) > 0## on some interval ##I##. This leads to ##\frac{1}{\sqrt{x}} - \frac{1}{x^2} > 0##. So we have ## \frac{1}{\sqrt{x}} > \frac{1}{x^2} ##. Since ## x \in [1, \infty)##, we can flip this ## \sqrt{x} < x^2 ##. Therefore, ##x < x^4##. Hence ##x (x^3-1) > 0##. ##\therefore x(x-1)(x^2+x+1) > 0##. Since ## x > 0 ## and ##(x^2+x+1) > 0## if ## x \in [1, \infty)##, we have ## x > 1##. So ##f(x)## is increasing on ##(1,\infty)##. Its also known that ##f(1) = 3##. How do I proceed from here on ?

Thanks