Show that \(\frac 1x \ge 3 - 2\sqrt{x}\) for all positive real numbers x. Describe when we have equality.

Hi! Please provide a full solution.

Thanks!

FencingKat Oct 24, 2018

#1**+11 **

\(\frac{1}{x} \geq 3 - 2\sqrt{x} \iff \frac{1+2x\sqrt{x} -3x}{x} \geq 0 \iff \frac{(\sqrt{x}-1)^2(\sqrt{x} + \frac{1}{2})}{x} \geq 0\)

We can cancel off the \((\sqrt{x}-1)^2\), and the \(x\) on the bottom, since \(x\) must be positive, and squares are always positive.

This leaves us with \(\sqrt{x} + \frac{1}{2} \geq 0\) which is true for all positive \(x\).

Hence this inequality also applies for all positive \(x\).

Equality will apply when \(x = 1\), since \(\sqrt{x}+\frac{1}{2}\) is strictly positive, and \(\frac{1}{x}\) is strictly positive for \(x \neq 0\).

Source:

1) https://math.stackexchange.com/questions/1929829/show-that-frac-1x-ge-3-2-sqrtx-for-all-positive-real-numbers-x-descri

2) https://math.stackexchange.com/questions/1929829/show-that-frac-1x-ge-3-2-sqrtx-for-all-positive-real-numbers-x-descri/1929838

If it is urgent, this is all I have. If you need something more detailed, I'm sure someone will be able to better assist you soon.

KnockOut Oct 24, 2018

#2**0 **

Alright, i was hoping for something a little more detailed, but thanks for trying to help!

FencingKat
Oct 25, 2018