Find the values of $a,b \in \Bbb R$ (if exists) such that $-5 \le \frac{x^2+ax+b}{x^2+2x+3} \le 4$ for all $x \in \Bbb R$ Find the values of $a,b \in \Bbb R$ (if exists) such that $$-5 \le \frac{x^2+ax+b}{x^2+2x+3} \le 4$$ for all $x \in \Bbb R$
My try:
I noticed that  $x^2+2x+3 > 0$, so i can divide the inequality in 2 parts:
$-5(x^2+2x+3)\le x^2+ax+b $ $\space$ $\land$ $\space$ $4(x^2+2x+3) \ge x^2+ax+b$
After that i tried to manipulate the discriminant of both inequalities (both are quadratic equations) but i found nothing.
Any hints? 

Hint: $$x^2+2x+3=(x+1)^2+2\ge2$$
