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'm kinda stuck on how to do this problem. Any help would be appreciated. Thank you in advance!
By AM-GM, we have
$$
\sqrt{x}+\sqrt{x}+\frac{1}{x}\geq3\sqrt[3]{\sqrt{x}\sqrt{x}\frac{1}{x}}=3.
$$
Equality is iff $\sqrt{x}=\frac{1}{x}$ iff $x=1$.
We want to prove: 

$(1)$ $$\forall x&gt;0, \frac{1}{x} \geq 3-2\sqrt{x}  $$

Let's consider two cases:

Case $1$: $3-2\sqrt{x} \geq 0 \Rightarrow x \leq \frac{9}{4}$

Then both sides of the inequality are positive.

$\frac{1}{x} \geq 3 - 2\sqrt{x} \Leftrightarrow  3x - 2x^{1.5}\leq 1 \Leftrightarrow x&gt; 0$ 

Therefore the inequality $(1)$ holds for $0&lt; x\leq \frac{9}{4}$

Case $2$: $x&gt;\frac{9}{4}$

Then we clearly see that $3-2\sqrt{x} &lt; 0 $ and $\frac{1}{x} &gt; 0 \Rightarrow \frac{1}{x} &gt; 3-2\sqrt{x}$

So generalising to all $x&gt;0$, we get that 

$\frac{1}{x} \geq 3-2\sqrt{x}$ as required.