<s>[INST] Solve inequality with $x$ in the denominator Solve for $x$ when it is in the denominator of an inequality
$$\frac{4}{x+4}\leq2$$ 
I believe the first step is the multiply both side by $(x+4)^2$
$$4(x+4)\leq 2(x+4)^2$$
$$4x+16\leq 2(x^2+8x+16)$$
$$4x+16\leq 2x^2+16x+32$$
$$0 \leq 2x^2+12x+16$$
$$0 \leq (2x+8)(x+2)$$
Stuck here.
 [/INST] Why not just do it this way:
For $x + 4 > 0$ the following is legal:
$$
\begin{align*}
\frac{4}{x + 4} &\le 2 \\
\frac{x + 4}{4} &\ge \frac{1}{2} \\
x + 4 &\ge 2 \\
x &\ge -2
\end{align*}
$$
If $x + 4 < 0$, i.e. $x < -4$, the inequality is satisfied.
The solution is $(-\infty, -4) \cup [-2, \infty)$.
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