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 $x \in \Bbb R$

Question

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?

Answer 1

Let $a,b\in\Bbb{R}$ be such that the inequalities hold. Then clearing denominators as you did yields the inequalities $$6x^2+(a+10)x+(b+15)\geq0,$$ $$3x^2+(8-a)x+(12-b)\geq0,$$ for all $x\in\Bbb{R}$. This means both polynomials in $x$ have nonpositive discriminants, i.e. that $$(a+10)^2-24(b+15)\leq0,$$ $$(8-a)^2-12(12-b)\leq0.$$ Isolating $b$ from both inequalities yields $$\frac{(a+10)^2}{24}-15\leq b\leq 12-\frac{(8-a)^2}{12}.$$ Moreover, a bit of algebra shows that for $a$ we then have $$3a^2-12a-532\leq0.$$ By the quadratic formula this is equivalent to $$|a-2|\leq4\sqrt{\tfrac{34}{3}}.$$ Can you finish from here?

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For completeness, a nice and grainy plot of the solution set in the $(a,b)$-plane:

Answer 2

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