$2x^3+ax^2+bx+4=0$, $(a,b \in R^+)$ has three real roots. Then :
A. $a\geqslant 4.2^{\frac 1 3}$
B. $a\geqslant 1.2^{\frac 1 3}$
C. $a\geqslant 6.2^{\frac 1 3}$
D. $a\geqslant 2.2^{\frac 1 3}$
The only way I know how to do this is to find roots of its derivative($x_1,x_2$) and check opposite signs of $f(x_1),f(x_2)$ or one of them being $0$(Repeated roots).
But that would be too messy and ugly. Finding roots and putting them back... Also I think it would involve b. Any suggestions?
If $2x^3+ax^2+bx+4$ with $a,b\gt0$ has three real roots, then all three must be negative, since the polynomial is clearly positive when $x\ge0$, so we can write
$$x^3+{a\over2}x^2+{b\over2}x+2=(x+r)(x+s)(x+t)$$
with $r,s,t\gt0$. Expanding out the right hand side and equating coefficients allows us to apply the arithmetic-geometric mean inequality to conclude
$${a\over6}={r+s+t\over3}\ge\sqrt[3]{rst}=\sqrt[3]2$$