Question

If $2x^3+a{x}^2+bx+4=0,a,b>0$ has $3$ real roots and the minimum value of $a+b$ is $m(x^{1/3}+4^{1/3})$, then the value of $m+x$ is

Answer 1

Let $\alpha ,\beta ,\gamma$ be the roots of
$2x^3+a{x}^2+bx+4=0$
As all the coefficients are positive, so all the roots will be negative
Assuming
${\alpha }_1=-\alpha ,{\alpha }_2=-\beta ,{\alpha }_3=-\gamma$
Now,
$\alpha +\beta +\gamma =-\frac{a}{2}\Rightarrow {\alpha }_1+{\alpha }_2+{\alpha }_3=\frac{a}{2}\cdots \left(1\right)$
$\alpha \beta +\beta \gamma +\alpha \gamma =\frac{b}{2}\Rightarrow {\alpha }_1{\alpha }_2+{\alpha }_2{\alpha }_3+{\alpha }_3{\alpha }_1=\frac{b}{2}\cdots \left(2\right)$
$\alpha \beta \gamma =-\frac{4}{2}\Rightarrow {\alpha }_1{\alpha }_2{\alpha }_3=2\cdots \left(3\right)$

Using
$AM\ge GM\Rightarrow \frac{{\alpha }_1+{\alpha }_2+{\alpha }_3}{3}\ge ({\alpha }_1{\alpha }_2{\alpha }_3{)}^{1/3}$
From equation $\left(1\right)$ and $\left(3\right)$,
$\Rightarrow a\ge 6\times 2^{1/3}\cdots \left(4\right)$
Also,
$\frac{{\alpha }_1{\alpha }_2+{\alpha }_2{\alpha }_3+{\alpha }_1{\alpha }_3}{3}\ge ({\alpha }_1{\alpha }_2{\alpha }_3{)}^{2/3}$
From equation $\left(2\right)$ and $\left(3\right)$,
$\Rightarrow b\ge 6\times 2^{2/3}\Rightarrow b\ge 6\times 4^{1/3}\cdots \left(5\right)$

Therefore, using equation $\left(4\right)$ and $\left(5\right)$,
$a+b\ge 6(2^{1/3}+4^{1/3})$
The minimum value
$a+b=6(2^{1/3}+4^{1/3})\Rightarrow m(x^{1/3}+4^{1/3})=6(2^{1/3}+4^{1/3})\Rightarrow m=6,x=2\therefore m+x=8$