Maximum value of expression Let the maximum value of the expression $y=\frac{x^4-x^2}{x^6+2x^3-1}$ for $x>1$ is $\frac p q$,where $p$ and $q$ are relatively prime positive integers.Find $p+q$.
My attempt: $y=\frac{x^2(x^2-1)}{(x^3+1)^2-2}$, then I could not think of any way to find maximum value of $f(x)$. Can someone help me in solving for $p$ and $q$?

Let $t:=x-\frac{1}{x}$, which can take any value in $\mathbb{R}_{>0}$  Then, $y=\frac{t}{t^3+3t+2}$.  To maximize $y$, we have to minimize $\frac{1}{y}=t^2+3+\frac{2}{t}$.  By the AM-GM Inequality,
$$\frac{1}{y}=\left(t^2+\frac{1}{t}+\frac{1}{t}\right)+3\geq 3\sqrt[3]{t^2\cdot\frac{1}{t}+\frac{1}{t}}+3=6\,.$$
That is, $y \leq \frac{1}{6}$.  The equality holds if and only if $t^2=\frac{1}{t}$, or $t=1$.  Hence, the maximum of $y$ is $\frac{1}{6}$, which is attained at $x=\frac{1+\sqrt{5}}{2}$.
