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# Hi! Math Help! Please Hurry!

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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!

Oct 24, 2018

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

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.

Oct 24, 2018
#2
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Alright, i was hoping for something a little more detailed, but thanks for trying to help!

FencingKat  Oct 25, 2018