Range of $\frac{c x^2(1-c)^2}{(x^2 +c^3)(x^2+c)}$? I have the following problem:
Let $f(x) = \frac{c x^2(1-c)^2}{(x^2 +c^3)(x^2+c)}$ with $c>0$ and $x \in \mathbb{R}$. How to prove that $0 \leq f(x) \leq 1$?
I'm not sure how to check the range, I tried plotting the function but that's not as formal as the problem asks. Though, at least I'm sure that the range is correct. Thanks for the help.

First to check if $0 \leq f(x)$ everywhere:  $c>0,$ anything squared is $\geq 0$, so yes.  The minimum is reached at $x=0$ only, because the numerator of a fraction must be $0$ for the fraction to be $0$.
If $c=1$ then the function is $0$ everywhere.  If $c$ is large we have roughly $c^3/c^4 = 1/c$ so that will be $\leq 1$.  We could try small values of $c$, but I don't see any values of $x$ that will bring the function over $1$.
The calculus to check for extrema expanded out is straightforward but messy: $21$ terms.  So we simplify by pulling out constants and expand the denominator to get $$\frac{x^2}{x^4 + x^2(c^3+c) + c^4}$$ and then we set $y=x^2$ and relabel to get
$$\frac{y}{y^2+ay+b}$$ which is far more tractable, and has extrema at $y^2 = b$.  Substituting $x^2 = y$ and $c^4 = b$, we get $x = \pm c$ as locations of the extrema.
For the values there, we get $$\frac{(1-c)^2}{(1+c)^2} = \frac{(c-1)^2}{(1+c)^2}$$ which is less than $1$ for all $c>0$.
