<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] An idea:
$$\frac{4}{x+4}\le 2\stackrel{\text{div. by 2}}\iff \frac{2}{x+4}\le 1\iff \frac{2}{x+4}-1\le 0\stackrel{\text{common denom.}}\iff $$
$$\iff \frac{2-x-4}{x+4}\le 0\stackrel{\text{mult. by (-1)}}\iff \frac{x+2}{x+4}\ge0\stackrel{\text{mult. by}\; (x+4)^2}\iff(x+2)(x+4)\ge 0\iff$$
$$\iff x<-4\,\,\vee\,\,x\ge-2\;\ldots$$ 
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