<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] It might be useful to simplify the inequality before trying to solve:
$$\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4} \Leftrightarrow x^2-2 \leq \frac{x^3+2x}{x+4}$$
Now, note that (using polynomial division)


*

*$\frac{x^3+2x}{x+4} = x^2 - 4x+18 -\frac{72}{x+4}$


So, the inequality turns into 
$$x^2-2 \leq x^2 - 4x+18 -\frac{72}{x+4} \Leftrightarrow \frac{72}{x+4} \leq -4x+20 = 4(5-x) \Leftrightarrow \boxed{\frac{18}{x+4} \leq 5-x}$$
Taking into consideration that the inequality sign flips when multiplying by a negative number you get two cases:
$$\frac{18}{x+4} \leq 5-x \Leftrightarrow \begin{cases} 18 \leq (5-x)(x+4) & \mbox{for } x >-4 \\ 18 \geq (5-x)(x+4) & \mbox{for }x<-4\end{cases} \Leftrightarrow \boxed{\begin{cases} (x+1)(x-2) \leq 0 & \mbox{for } x >-4 \\ (x+1)(x-2) \geq 0 & \mbox{for }x<-4\end{cases}}$$
Note that $(x+1)(x-2)$ corresponds to an upwards open parabola with zeros at $x=-1$ and $x=2$. So, you get
$$\begin{cases} (x+1)(x-2) \leq 0 & \mbox{for } x >-4 \\ (x+1)(x-2) \geq 0 & \mbox{for }x<-4\end{cases} \Leftrightarrow \begin{cases} x \in [-1,2] \cap (-4, \infty)\\ x \in ((-\infty,-1] \cup [2,\infty)) \cap ( -\infty,-4) \end{cases} \Leftrightarrow \boxed{x\in ([-1,2] \cup ( -\infty,-4))}$$
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