Question

Assume the biquadratic $x^4-a{x}^3+b{x}^2-ax+d=0$ has four real roots with $\frac{1}{2}<\alpha ,\beta ,\gamma ,\delta \le 2.$ Maximum possible value of $\frac{(\alpha +\beta )(\alpha +\gamma )\delta }{(\delta +\beta )(\delta +\gamma )\alpha }$ is

• A. $1$
• B. $\frac{4}{5}$
• C. $\frac{5}{4}$
• D. $2$

Answer 1

The correct option is C $\frac{5}{4}$

$f\left(x\right)=x^4-a{x}^3+b{x}^2-ax+d=0$

$E=\frac{(\alpha +\beta )(\alpha +\gamma )\delta }{(\delta +\beta )(\delta +\gamma )\alpha }$

$=\frac{{\delta }^2(\alpha +\alpha )(\alpha +\beta )(\alpha +\gamma )(\alpha +\delta )}{{\alpha }^2(\delta +\alpha )(\delta +\beta )(\delta +\gamma )(\delta +\delta )}$

$=\frac{{\delta }^2}{{\alpha }^2}\cdot \frac{f(-\alpha )}{f(-\delta )}$

$=\frac{{\delta }^2}{{\alpha }^2}\cdot \frac{f(-\alpha )-f\left(\alpha \right)}{f(-\delta )-f\left(\delta \right)}$

Now, $f(–\alpha )-f\left(\alpha \right)=2a\cdot {\alpha }^3+2a\alpha$
Then, $E=\frac{{\delta }^2}{{\alpha }^2}\cdot \frac{{\alpha }^3+\alpha }{{\delta }^3+\delta }$
$=\frac{\alpha +\frac{1}{\alpha }}{\delta +\frac{1}{\delta }}$
$\le \frac{5/2}{2}=\frac{5}{4}$