An inequality where we have to show that is works for all positive x, and describe equality and what Show that $\frac 1x \ge 3 - 2\sqrt{x}$ for all positive real numbers $x$. Describe when we have equality.
I was thinking about multiplying both sides by x and then squaring both sides, but I don't think I have the right idea... 
 A: Multiplying by $x$ and squaring both sides gives 
$$
4x^3 + 4x^{\frac{3}{2}} + 1 \geq 9x^2 \Leftrightarrow
$$
$$
4x^3 + 4x^{\frac{3}{2}} + 1 - 9x^2 \geq 0. 
$$
I do not see an easy way to factor this, but wolfram tells us that the factors are
$$
(\sqrt{x} - 1)^2(2\sqrt{x} + 1)(2x^{\frac{3}{2}} + 3x + 1) > 0. 
$$

I'd suggest the following approach instead.

We have that
$$
\frac{1}{x} \geq 3 - 2\sqrt {x} \Leftrightarrow
$$
$$
\frac{1}{x} + 2\sqrt{x} \geq 3.
$$
Let
$$
f(x) = \frac{1}{x} + 2\sqrt{x}, \ x \in (0,\infty)
$$
then
$$
f' (x) = \frac{-1}{x^2} + \frac{1}{\sqrt{x}}.
$$
Set equal to zero, which gives
$$
x^{\frac{3}{2}} = 1
$$
$$
f'(1/2) < 0 , f(2) > 0
$$
Hence we have a min, at $x=1$ for $x \in (0, \infty)$ 
$$
f(1) = 1 + 2\sqrt{1} = 3. 
$$
A: The function
$$f(x):={1\over x}+2\sqrt{x}-3\qquad(x>0)$$
has $f(1)=f'(1)=0$ and is certainly $>0$ when $\sqrt{x}>{3\over2}$. Furthermore
$$f''(x)={2\over x^3}-{1\over2x^{3/2}}={1\over 2x^{3/2}}\left({4\over x^{3/2}}-1\right)$$
is $>0$ when $0<\sqrt{x}<4^{1/3}=1.587$, hence $f$ is convex in this range. Altogether this shows that $f(x)\geq0$ for all $x>0$ with equality only when $x=1$.
