Find the Range of $y=\frac{x+a}{x^2+bx+c^2}$ 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?
 A: 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...
A: 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 $(-\infty, \infty)$.
