Hi! Math Help! Please Hurry!

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

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

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