Find the Range of $y=\frac{x+a}{x^2+bx+c^2}$

Question

Given that $x^2-4cx+b^2 \gt 0$ $\:$ $\forall$ $x \in \mathbb{R}$ and $a^2+c^2-ab \lt 0$

Then find the Range of $$y=\frac{x+a}{x^2+bx+c^2}$$

My try:

Since $$x^2-4cx+b^2 \gt 0$$ we have Discriminant

$$D \lt 0$$ $\implies$

$$b^2-4c^2 \gt 0$$

Also

$$x^2+bx+c^2=(x+a)^2+(b-2a)(x+a)+a^2+c^2-ab$$

Hence

$$y=\frac{1}{(x+a)+b-2a+\frac{a^2+c^2-ab}{x+a}}$$

But Since $a^2+c^2-ab \lt 0$ We can't Use AM GM Inequality

Any way to proceed?

Answer 1

This is not a complete solution. Just my opinion on the possible solution.

Cross multiplying and subtracting $x+a$ gives an equation:

$x^2y+x(by-1)+(yc^2-a)=0$

The discriminant turns out to be: $y^2(b^2-4c^2)+y(4a-2b)+1$

Further the discriminant of the above equation is:

$16(a^2+c^2-ab)$

Which is certainly less than 0 according to info provided. I believe in a similar manner the range shall pop up. But certainly, I might be wrong...

Answer 2

If a number $p$ is in the range of the $y$ you have given, it means $\frac{x+a}{x^2+bx+c^2} = p$ for some $x$.

Or, in other words, that equation has a real solution.

=> $px^2+bpx+pc^2 = x+a$ For a real value of $x$

=> the discriminant of $px^2+(bp-1)x+pc^2-a$ is non-negative.

After some rearrangement we have any $p$ where the following inequality stands true is in the range of y.

=> $p^2(b^2-4c^2) + p(4a-2b) + 1 \geq 0$

The discriminant upon expansion turns out to be $16(a^2+c^2-ab)$ which is given to be strictly negative. As you have noted, the leading coefficient is positive. This means the given expression is always positive so every value of $p$ satisfies the condition. So, $p$ takes all real values. This means the range of the given function is $(-\infty, \infty)$.