<s>[INST] Solving $\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4}$. Solve the inequality
$\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4} \Leftrightarrow$ $\frac{x^2-2}{x^2+2}  - \frac{x}{x+4} \leq 0 \Rightarrow$ $\frac{x^3+4x^2-2x-8-x^3-2x}{(x^2+2)(x+4)} \geq 0 \Leftrightarrow$ $4x^2-4x-8 \geq 0 \Rightarrow$ $x \geq \frac{1}{2} \pm \sqrt{2.25} \Rightarrow$ $x_1 \geq -1, \; x_2 \geq 2$ 
We notice that $x_2 \geq 2$ is a false root and testing implies that the solutions of the inequality lies within the interval $-1 \leq x \leq 2$.
Problem: But $x<-4$ also solves the inequality, so I must have omitted or done something wrong? And also, am I using implication and equivalence symbols correctly when doing the calculations?
Thank you for your help!
 [/INST] You reduced the inequality to $$\frac{4(x+1)(x-2)}{(x^2+2)(x+4)} \ge 0$$
which is equivalent to $$\frac{(x+1)(x-2)}{x+4} \ge 0$$
This is true if and only if an even number of terms $(x+1), (x-2), (x+4)$ are $\le 0$. There are four options:


*

*$x+1 \ge 0$, $x-2 \ge 0$, $x+4 > 0$ which gives $x \in [2, +\infty\rangle$

*$x+1 \le 0$, $x-2 \le 0$, $x+4 > 0$ which gives $x \in \langle -4, -1]$

*$x+1 \le 0$, $x-2 \ge 0$, $x+4 < 0$ which gives no solutions

*$x+1 \ge 0$, $x-2 \le 0$, $x+4 < 0$ which gives no solutions


Therefore the solutions are $x \in \langle -4, -1] \cup [2, +\infty\rangle$.
</s>