<s>[INST] Show that $\frac 1x \ge 3 - 2\sqrt{x}$ for all positive real numbers $x$. Describe when we have equality. Show that $\frac 1x \ge 3 - 2\sqrt{x}$ for all positive real numbers $x$. Describe when we have equality.
I tried manipulating the equation but nothing helps.  Any answer is greatly appreciated.
 [/INST] $$
\frac{1}{x} \geq 3 - 2\sqrt{x} \iff \frac{1+2x\sqrt{x} -3x}{x} \geq 0 \iff \frac{(\sqrt{x}-1)^2(\sqrt{x} + \frac{1}{2})}{x} \geq 0
$$
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 $x \neq 0$. 
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