Question

If $2x^3+a{x}^2+bx+4=0$ (a and b are positive real numbers) has $3$ real roots, then prove that $a+b\ge 6(2^{1/3}+4^{1/3})$.

Answer 1

$2x^3+9x^2+bx+4=0$

Let the three real roots are $\alpha ,\beta ,\gamma$

Sum of roots, $\alpha ,\beta ,\gamma =\frac{-a}{2}$

$\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =\frac{b}{2}$

$\Rightarrow \alpha \beta \gamma =-2$

If $\alpha ,\beta ,\gamma$ are three numbers then,

$AM\ge GM$

$\frac{\alpha +\beta +\gamma }{3}\ge {\left(\alpha \beta \gamma \right)}^{1/3}$

$\frac{-a}{2}\times \frac{1}{3}\ge {(-2)}^{1/3}$

$\frac{-a}{b}\ge -{\left(2\right)}^{1/3}\to \left(I\right)$

$\&$ If $\alpha \beta ,\beta \gamma ,\gamma \alpha$ are three numbers then,

$AM\ge GM$

$\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{3}\ge {\left({\alpha }^2{\beta }^2{\gamma }^2\right)}^{1/3}$

$\frac{b}{2}\times \frac{1}{3}\ge {(-2^2)}^{1/3}$

$\frac{b}{6}\ge -{\left(4\right)}^{1/3}\to \left(II\right)$

From equation $II$ $-$ equation $I$

$\frac{b}{6}+\frac{a}{b}\ge 4^{1/3}+2^{1/3}$

$a+b\ge 6(2^{1/3}+4^{1/3})$

Hence, the answer is $a+b\ge 6(2^{1/3}+4^{1/3}).$