HMMT February

We can rewrite the expression as $$\frac{x_{4}^{2}}{x_{1}^{2}} \cdot \frac{\left(x_{1}+x_{1}\right)\left(x_{1}+x_{2}\right)\left(x_{1}+x_{3}\right)\left(x_{1}+x_{4}\right)}{\left(x_{4}+x_{1}\right)\left(x_{4}+x_{2}\right)\left(x_{4}+x_{3}\right)\left(x_{4}+x_{4}\right)} \frac{x_{4}^{2}}{x_{1}^{2}} \cdot \frac{f\left(-x_{1}\right)}{f\left(-x_{4}\right)}$$ where $f(x)$ is the quartic. We attempt to find a simple expression for $f\left(-x_{1}\right)$. We know that $$f\left(-x_{1}\right)-f\left(x_{1}\right)=2 a \cdot x_{1}^{3}+2 a \cdot x_{1}$$ Since $x_{1}$ is a root, we have $$f\left(-x_{1}\right)=2 a \cdot x_{1}^{3}+2 a \cdot x_{1}$$ Plugging this into our previous expression: $$\frac{x_{4}^{2}}{x_{1}^{2}} \cdot \frac{x_{1}^{3}+x_{1}}{x_{4}^{3}+x_{4}} \frac{x_{1}+\frac{1}{x_{1}}}{x_{4}+\frac{1}{x_{4}}}$$ The expression $x+\frac{1}{x}$ is maximized at $x=2, \frac{1}{2}$ and minimized at $x=1$. We can therefore maximize the numerator with $x_{1}=2$ and minimize the denominator with $x_{4}=1$ to achieve the answer of $\frac{5}{4}$. It can be confirmed that such an answer can be achieved such as with $x_{2}=x_{3}=\frac{\sqrt{10}-1}{3}$.