<s>[INST] Solve the inequality $\frac{2}{x} > 3 x$ I've been slowly losing my mind on how to do this question, any help would be greatly appreciated.

Solve the inequality $$\frac{2}{x} > 3 x.$$

 [/INST] To solve the equation
$$
\frac{2}{x} > 3x
$$
we could multiply be $x$ to obtain
$$
2 > 3x^2.
$$
But you have to be careful!
If $x < 0$, the inequality is reversed.
In other words:
$$
\frac{2}{x} > 3x 
\iff \begin{cases}
2 < 3x^2 & x < 0 \\
2 > 3x^2 & x > 0.
\end{cases}
$$
Notice the equation isn't defined for $x = 0$, because you can't divide by $0$.
Case 1: $x > 0$.
Then we have
$$
2> 3x^2
\iff x^2 < \frac{2}{3}
\iff x \in \left(- \sqrt{ \frac{2}{3}}, \sqrt{ \frac{2}{3}}\right).
$$
But, since $x > 0$, our first solution interval is $I_1 := \left(0, \sqrt{ \frac{2}{3}}\right)$.
Case 2: $x < 0$.
Then we have
$$
2 < 3x^2
\iff x^2 > \frac{3}{2}
\iff x > \sqrt{\frac{3}{2}} \quad \text{and} \qquad x < -\sqrt{\frac{3}{2}}
$$
But again, since $x < 0$, our second (and final) solution interval is $I_2 := \left(- \infty, -\sqrt{\frac{3}{2}}\right)$.
So our solution is $I_1 \cup I_2 =\left(- \infty, -\sqrt{\frac{3}{2}}\right) \cup \left(0, \sqrt{ \frac{2}{3}}\right)$.
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