<s>[INST] Easy inequality going wrong 
Question to solve: $$\frac{3}{x+1} + \frac{7}{x+2} \leq \frac{6}{x-1}$$

My method:
$$\implies \frac{10x + 13}{(x+1)(x+2)} - \frac{6}{x-1} \leq 0$$
$$\implies \frac{4x^2 -15x-25}{(x-1)(x+1)(x+2)} \leq 0$$
$$\implies (x-5)(4x+5)(x-1)(x+1)(x+2) \leq 0$$
Using method of intervals, I get:
For $x\leq-2, -5/4\leq x \leq -1$ and $-1\leq x\leq1$, x is less than or equal to zero.
But, this range i got is incorrect. Where did I go wrong then?
 [/INST] After you reached the stage
$$ (x-5)(4x+5)(x-1)(x+1)(x+2)\le 0\stackrel{\div 4}\iff$$
$$ (x+2)\left(x+\frac54\right)(x+1) (x-1)(x-5)\le0$$
and since all the factors are with odd exponent, you can apply the snake method, and get the solution
$$x<-2\;\;,\;\;\;or\;\;\;-\frac54\le x<-1\;\;,\;\;\;or\;\;\;1<x\le 5$$
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