jmc

If the quadratic $x^2+ax+b$ has no real roots, then $x^2+ax+b>0$ for all $x,$ which means the given inequality is equivalent to $x+c\le 0,$ and the solution is $(-\infty ,-c].$ The solution given in the problem is not of this form, so the quadratic $x^2+ax+b$ must have real roots, say $r$ and $s,$ where $r<s.$

Then $x^2+ax+b=(x-r)(x-s),$ and the inequality becomes $$\frac{x+c}{(x-r)(x-s)}\le 0.$$This inequality is satisfied for sufficiently low values of $x,$ but is not satisfied for $x=-1,$ which tells us that $r=-1.$ The inequality is now $$\frac{x+c}{(x+1)(x-s)}\le 0.$$The inequality is then satisfied for $x=1,$ which tells us $c=-1.$ Then the inequality is not satisfied for $x=2,$ which tells us $s=2.$ Thus, the inequality is $$\frac{x-1}{(x+1)(x-2)}=\frac{x-1}{x^2-x-2}\le 0,$$so $a+b+c=(-1)+(-2)+(-1)=\boxed{-4}.$